Top
SCIENCE. STORIES. IMPACT.

iBiology

Bringing the World's Best Biology to You

Mechanisms of DNA Repair

Part 1: Mechanisms of DNA Repair by Recombination

We were unable to retrieve the video

00:00:03.02 Hi. My name is Jim Haber. I'm a professor of Biology at Brandeis University and I'm director of the Rosenstiel
00:00:10.15 Basic Medical Sciences Research Center there.
00:00:12.25 My lab has been interested for a long time in how cells are able to repair DNA damage most notably broken chromosomes.
00:00:21.06 We study this using the simple model organism budding yeast saccharomyces cerevisiae but
00:00:27.29 because of the very strong evolutionary conservation of these mechanisms much of what we have learned is applicable
00:00:34.01 in understanding what goes on in human cells as well. One of the primary motivations for the work that we've
00:00:40.23 carried out is to understand the remarkable genome instability of human cancer cells. This is illustrated here in comparison
00:00:49.03 to the normal spectral karyotype - the false color images of a human chromosome
00:00:55.04 so there are 23 pairs of chromosomes in a normal human and in comparison to that in this cancer cell
00:01:01.15 what one can see is an astonishing number of chromosomal alterations. There are translocations, and inversions, and deletions
00:01:09.16 there are truncations and in fact there are lots of other alterations which at this level of analysis
00:01:17.00 you can't see such as individual mutations. And one of the things that we would like to understand is why cells that have this kind of
00:01:27.20 instability have emerged from the background of normal human cells. Some of these translocations turn out to be
00:01:35.18 very important in terms of the origins of cancer themselves. Many of these alterations
00:01:41.05 probably just come along for the ride and are not particularly important. Here's an example of one chromosome which is
00:01:48.06 very important. A rearrangement called the Philadelphia chromosome found because identified in Philadelphia
00:01:55.18 A translocation that joins a small part of chromosome 22 to the end of chromosome 9. This translocation
00:02:06.11 joins a normal c-ABL gene which is a protein kinase to the region that is controlled by the BCR gene
00:02:14.08 and this fusion places the c-ABL gene under the control of the wrong promoter and enhancer sequences
00:02:23.11 so that it is expressed at the wrong time and place relative to the normal c-ABL gene and it is this mis-expression of this
00:02:30.25 protein which is associated with the origin of these cancers. Another thing that you can't see when you look
00:02:38.15 at these kinds of karyotypes is what is known as loss of heterozygosity. The example I'd like to give is of the retinoblastoma
00:02:48.04 tumor suppressor gene. When the mutations in the retinoblastoma gene were first found it was thought
00:02:55.05 that the retinoblastoma gene was a dominant mutation because people who inherited a single copy of this
00:03:01.03 mutation were likely to end up with tumors of the eye which is what retinoblastoma causes
00:03:07.07 but in fact it is a recessive mutation and it is only manifested in those cells which have lost the heterozygosity that was implicit in the original
00:03:19.07 parent chromosomes and as I've illustrated here what happens is there is a recombination event
00:03:27.08 which I'll specify a little bit more in a second which has caused this chromosome to lose
00:03:34.02 the dominant marker which is shown here the capitol RB marker and become homozygous for the rb tumor suppressor mutation
00:03:43.05 This happens by what is called a mitotic crossing over event - a recombination event in which there's been
00:03:52.29 an exchange of arms of these chromosomes as you see here and then after this exchange has taken place there is a hybrid
00:04:00.23 chromosome so part of it is black and part of it is blue and depending on how these chromatids
00:04:07.13 separate after DNA replication and mitosis you will end up with cases which are illustrated here where the cells have become homozygous
00:04:18.04 for this marker and indeed homozygous for markers all along the rest of the
00:04:22.17 segment of the chromosome and this is an example of the loss of heterozygosity. Loss of heterozygosity
00:04:29.26 can also show up in other ways. For example just by deleting a region of the wildtype chromosome so that the normal RB gene here
00:04:38.27 has been lost and now the only copy of RB that is being expressed is the mutant copy and I'll talk about mechanisms
00:04:47.04 that also cause deletions of this sort to expose this recessive mutation.
00:04:54.06 Now what's clear about all of these defects that I'm gonna talk about is that normal cells have very accurate
00:05:00.20 mechanisms to repair DNA damage and it is really in the absence of the repair of those damage that we see the kind of karyotypes
00:05:10.07 that I illustrated with this tumor suppressor gene. So as a general rule what I'm gonna say is that cells
00:05:17.01 that show this kind of instability either have defects in the way they repair chromosomal breaks or they have defects in what is known
00:05:25.04 as the DNA damage checkpoint about which I'm not gonna spend much time today
00:05:31.14 The DNA damage checkpoint is a mechanism by which cells sense that they have broken chromosomes, they arrest
00:05:38.03 their ability to progress through the cell cycle blocking progression prior to mitosis effectively giving the cell much
00:05:45.12 more time in order to be able to repair the DNA damage and so either because these cells have a defect in the repair
00:05:53.23 or in this checkpoint they lead to this instability that we're interested in understanding more about. Okay so
00:06:02.15 an example of genes in humans that are involved - which are defective in their repair of double strand breaks
00:06:11.07 are two that I'll mention a little bit more. One is called BRCA2 it's one of the two familially inherited breast cancer defects
00:06:20.04 and the other is BLM which stands for the Bloom syndrome protein about which I'll talk about and in both
00:06:26.22 of these cases we know that people who have a high predisposition for cancer are defective in one of these two
00:06:34.09 mutations which implicates that there are problems in DNA repair. There are also people who have defects in these checkpoint genes
00:06:44.12 two examples being something called Ataxia Telangiectasia Mutated which is ATM
00:06:51.11 and another protein probably the most prevalent mutation in all of human cancers p53
00:06:58.20 and these are defects in this DNA damage monitoring system and again
00:07:03.01 the failure to arrest the cells properly during cell cycle progression turns out to be
00:07:08.20 a major source of the instability that we examine. The other important underlying concept
00:07:15.10 of much of what we work on is the fact that the source of the double strand breaks the broken chromosomes
00:07:21.13 that we study does not come from exogenous sources such as ionizing radiation or chemical exposures it comes from the fragility
00:07:29.13 of the DNA replication process itself. What's illustrated here is a vertebrate cell a chicken DT40 cell
00:07:37.25 which has been depleted for a protein called Rad51 that I'm gonna talk about a lot. Rad51 is a DNA repair protein
00:07:46.17 and in the absence of this DNA repair protein what you see are these large number of chromosome interruptions
00:07:53.18 which are called chromatid breaks. One of the two sister chromatid strands is intact and the other sister chromatid
00:08:00.07 strand has a break and it is the role of the Rad51 recombinase protein to use the information on the sister
00:08:08.00 chromatid which is intact to patch up the break in the broken chromosome and much of what I'm gonna talk
00:08:15.06 about is how that actually happens. So one of the things that we take away from this is the fact that the source of
00:08:22.17 most of this damage is the process of DNA replication itself and that the recombination machinery is in fact
00:08:28.24 essential for life in essentially any organism with a large genome because there are these kinds of breaks that occur
00:08:37.13 all the time. The Rad51 protein in budding yeast that we study is not essential but that's
00:08:44.11 only because the saccharomyces genome is about literally this big and the human genome is two meters long and that the level of damage therefore
00:08:53.24 per individual yeast cell is not large enough that all the cells in the population at any one time are
00:08:59.22 suffering these kinds of breaks. One of the other things that I wanna emphasize is the fact that this problem
00:09:07.01 of chromosome breakage if anything becomes more severe as cells progress toward becoming cancer.
00:09:12.22 This idea was illustrated in an article by Halazonetis Gorgoulis and Bartek in 2008 in Science and the relevant part
00:09:22.22 of the abstract that they published is presented here and in particular one sentence which notes that activated
00:09:32.00 oncogenes which occur in precancerous cells and in cancers actually induce the stalling and collapse of DNA replication forks
00:09:40.13 and lead to the kind of breaks the chromosome breaks that I've been talking about
00:09:44.16 so as cells progress into cancer this problem of trying to repair accurately the chromosome damage becomes an increasing problem
00:09:53.21 . Okay so what I want to start with then is how do cells end up with these chromosome breaks?
00:10:00.25 The simplest way in which chromosome breakage could
00:10:03.23 occur is if the DNA replication fork which starts here and moves down the DNA encounters one of these nicks - a break in the DNA. Here
00:10:13.10 simply an interruption on one of the two strands. This is now one DNA molecule Watson and Crick and as the replication comes to this point
00:10:21.01 it cannot progress beyond this nick and as the replication fork then comes from the other side to this point what one ends up with is a
00:10:30.06 double strand break. This kind of interruption is not simply something that happens where a nick is introduced in the DNA by accident but in fact
00:10:42.27 there are certain drugs anticancer drugs such as camptothecin which cause this kind of interruption by blocking the action of the protein
00:10:53.11 Topoisomerase I and leaving a covalent attachment of the protein to the DNA and as the replication fork reaches this point there is the
00:11:02.01 possibility of creating these kinds of double strand breaks. Stalled replication forks have been produced in a large number of ways. The
00:11:11.24 two that I've illustrated here are places where the DNA replication fork encounters lesions which it cannot go past. The first of these are UV
00:11:22.20 induced photodimers cyclobutane dimers which have formed primarily between two adjacent thymidine residues and these photodimers really
00:11:32.15 block the progression of the DNA replication fork and can lead to the breakage of these molecules by some kind of either torsional stress
00:11:43.18 or in fact by enzymes which cleave the DNA. Another way in which these replication forks are stalled are by unusual DNA structures which form
00:11:56.08 ahead of the DNA replication fork and which make palindromes and other structures which again are difficult for the DNA replication machinery to
00:12:04.29 bypass. These happen in so called triplet repeat sequences such as CTG CTG CTG repeated many many times and which is characteristic of a defect
00:12:18.26 that's seen in Huntington's disease or in other DNA sequences which are characteristic of diseases such as Friedrich's Ataxia and in both
00:12:29.14 of these cases the stalling of the DNA replication fork creates and opportunity for chromosomes to break. There are lots of fragile sites in
00:12:41.04 human chromosomes which are sites where this kind of replication process is exaggerated and so these are sites where in fact you can see
00:12:51.07 frequent chromosome breakage by looking again at the karyotype as I showed you before and in fact one can reveal that these sites are
00:13:01.22 particularly fragile by slowing down the normal DNA replication process using a chemical inhibitor
00:13:09.01 in this case called aphidicolin and if one treats the cells with
00:13:13.18 aphidicolin what one sees here is that there are now many of these chromatid breaks which have occurred very much like I showed you
00:13:21.20 in the absence of the Rad51 protein there are such breaks at a very low level in normal cells but they are exaggerated tremendously by adding
00:13:34.26 this inhibitor to DNA replication. So we know there are double strand breaks that arise themselves and they can be repaired in a variety of ways.
00:13:43.08 The two large categories of mechanisms by which these breaks can be repaired are called either homologous recombination mechanisms
00:13:52.14 in which the ends of the double strand break have to find a template a nearly identical DNA molecule that they can use as a matrix
00:14:01.20 in order to patch up the broken chromosome and there are a number of different ways in which I'm gonna illustrate how this repair can take place
00:14:09.00 and then there are also other mechanisms which are called non-homologous end joining mechanism in which the ends of the DNA are pretty
00:14:17.17 much just joined back together with little or no regard for DNA sequence identity. These joining events are inherently mutagenic because
00:14:27.00 they often involve a loss of a few base pairs or more of DNA but they are an efficient way of also repairing double strand breaks and then some
00:14:37.21 times as I showed you at the beginning there are truncations and many of those truncations involve simply adding new telomere
00:14:44.04 sequences to the ends of double strand breaks. That's a mechanism that I'm not going to talk about
00:14:48.27 in this lecture. I'm gonna focus mostly on homologous
00:14:54.14 recombination. As I said this is a situation where the double strand breaks have occurred and the end of the double strand
00:15:03.25 break has to be able to find a partner in order to be able to carry out this repair.
00:15:09.04 In essence the way that this is going to work is that the end of the DNA
00:15:13.22 molecule is going to be used to search other DNA molecules to find locations in which essentially the same sequence has been located
00:15:22.28 and that can be done by the fact that the end of the DNA molecule is essentially able to make base pairs with the templates and has the same sequence
00:15:32.25 it can make equivalent basepairs to what the original DNA molecule had and this is the mechanism by which homology can be
00:15:40.13 recognized and I'm gonna talk also about how that actually can be seen. The mechanism that I'm gonna start with is called break induced
00:15:51.25 replication and it is really the mechanism that we have been sort of focusing on. We saw the stalled replication forks can lead to DNA breakage the
00:16:03.29 reason that this may be advantageous to cells is it turns out to be a way to restart DNA replication in a recombination dependent fashion
00:16:13.04 and this recombination dependent mechanism we call break induced replication. Now I showed you that DNA molecules stall in replication
00:16:26.18 in the presence of photodimers and said that these can be broken but it turns out that the situation is actually a little bit more complicated
00:16:34.13 than that. The stalled replication fork can actually undergo a little ballet in which the strands of the DNA molecule are rearranged in what is now known
00:16:45.01 as a Holliday junction. This 4-way structure was in fact postulated in 1964 by Robin Holliday who was studying DNA repair and he
00:16:59.00 suggested that this junction would have unusual properties that would be important
00:17:03.12 in recombination and of course he was right. Here what's
00:17:07.22 happened is that the two light blue strands which are the newly replicated strands are being pulled apart and joined to each other to form a duplex
00:17:18.13 molecule and the purpose of doing this in part is to push the DNA replication fork back away from the UV photodimer so that this region
00:17:28.24 of the DNA is exposed and there are DNA repair enzymes that can get at this photodimer and repair it. But in the process of this one forms this
00:17:40.29 remarkable Holliday junction. We know this happens because at least with budding yeast we can arrest cells in replication using a drug
00:17:50.29 called hydroxyurea and if one takes electron micrographs of the DNA from those cells one can actually see these regressed replication forks
00:18:01.20 these Holliday junctions. Sometimes these structures as it's noted here are called chicken feet
00:18:08.01 because of the structures that they resemble and this
00:18:13.00 has all happened because the replication machinery has arrived at these places and then undergone this reversal of the replication fork
00:18:21.03 involving also the dissociation of the DNA replication polymerases so that one can end up with this structure. Here's another picture of this
00:18:32.20 structure which just illustrates the fact that this is a very symmetric structure. The four arms that are generated are in fact symmetrical
00:18:42.14 with respect to each other and is illustrated here all of the base pairs of this structure can be made so that we have four arms of this structure
00:18:53.10 all joined at this common point. This is just another picture which again illustrates that all the base pairs can be formed. Now this Holliday junction
00:19:05.14 has other quite remarkable properties. The most astonishing of these is the fact that the branch can simply migrate up and down the
00:19:16.10 DNA and that this migration which takes place involves at every step breaking a pair of base pairs and forming right next to it a new pair
00:19:27.29 of base pairs and so actually it's energetically neutral because you're breaking as many basepairs as you're forming and this branch
00:19:36.24 can migrate back and forth on DNA with ease. That turns out to be very useful if there are certain places in DNA that are easier to manipulate
00:19:46.06 than others and this branch can migrate to those points. In fact cells facilitate this kind of process by having enzymes that will in fact directly
00:19:58.13 make this branch go down the DNA in one direction or the other in e. coli we know the most about these molecules there's a pair of molecules
00:20:10.24 called ruvA and ruvB the ruvA molecule actually binds to this Holliday junction structure and the ruvB proteins turn out to be hexameric
00:20:20.05 protein motors which can pull the DNA through the structure and actually cause this branch to migrate and the actual motion of this which has
00:20:31.29 been worked out from studying the structure of these molecules crystallographically and enzymatically is shown here. Now there's
00:20:40.24 another property of Holliday junctions which turns out to be equally important beyond branch migration and that is the fact that these
00:20:47.21 molecules have to be at the end of the process of repair or replication these branch molecules have got to be taken apart. If you think about
00:20:59.05 this as a pair of chromosomes and they were ready to undergo mitosis they're gonna be unable to go through mitosis if the two DNA
00:21:06.20 molecules are linked together in this way. They won't be able to separate to the two poles of the mitotic spindle so there has to be a mechanism
00:21:16.06 that can take these Holliday junctions apart and there are in fact these enzymes - resolvases - which can take these branch molecules
00:21:27.19 apart and it's sometimes hard to think about this just because when we draw these structures on the board they don't look symmetrical anymore
00:21:35.11 but if you think about the fact that this was in three dimensions a completely symmetrical structure then it turns out that you can
00:21:44.11 either cleave the outside pair of strands here or you can cleave the inside pair of strands here and these are actually very similar or identical operations
00:21:55.25 by an enzyme because of the symmetry of this structure. If the enzyme cleaves the inner pair of these structures where the cross
00:22:04.05 is then the result of this is you have two chromatids which are in fact completely identical to the way they started. In geneticists parlance
00:22:15.10 big A is still linked to big B and little a is still linked to little b and when these come apart there's no change to the genetic information which
00:22:26.16 has been segregated. In contrast, if the breaks are made on the outer pair of strands here then it turns out that the only way that these
00:22:36.28 DNA molecules are still held together is by the light blue strand - the newly synthesized strand the way we were talking about it before - and now
00:22:47.11 this arm of the chromosome including the big A marker is only linked to the little b marker and conversely little a is only linked to big B.
00:22:58.19 So there has been a crossover which is characteristic of what you probably learned about
00:23:04.21 in studying meiosis there are crossovers of the two DNA
00:23:09.24 molecules which have occurred because of the resolution of one of these Holliday junctions and it looks as if these two segments are
00:23:17.26 very far apart from each other but remember they're really immediately adjacent bases and since the DNA molecule held by only a single strand
00:23:27.14 of DNA is very flexible these things can really be put back together and ligated with no difficulty. So Holliday junction resolvases can take these
00:23:36.27 branch structures and create in some cases crossovers which would be associated with loss of heterozygosity as I illustrated earlier or with
00:23:48.04 the creation of genetic diversity that occurs in meiosis. So if we go back to the stalled replication fork that we were discussing and realize
00:24:01.07 that this regression of the replication fork will lead to the formation of a Holliday junction one of the ways that this regressed replication fork
00:24:11.25 is going to break is in fact because of the action of these Holliday junction resolvases and it doesn't matter whether it cuts in this plane or in the
00:24:21.14 other plane because in both cases there's going to be one intact DNA molecule and there's going to be one broken molecule and I've simply
00:24:30.26 flipped this little piece of DNA over so that you can see now that it's oriented in the same direction and so now essentially what's happened
00:24:40.24 is the DNA replication fork which was coming in this direction has reached this point, arrested, backed up, formed this regressed
00:24:49.15 replication fork, and then broken at the place where this lesion had occurred in order to give the cell a chance to repair this break or this interruption
00:25:01.23 in a different way. And the different way that it is going to use is this homologous recombination mechanism known as break induced
00:25:09.26 replication and I'm gonna go through all the steps of this process here. So we got to the point where we have a broken replication fork. That replication
00:25:22.01 fork is acted upon by enzymes which chew away one of the two strands of the DNA almost in all the physiological cases we know
00:25:31.09 about it is the 5' ended strand which is chewed away by a series of enzymes known as exonucleases and the purpose of this is to generate
00:25:40.26 a large region of single stranded DNA which ends at the 3' end of this molecule. This turns out to be the business end for doing almost all of homologous
00:25:51.12 recombination. This single stranded DNA molecule attracts the recombinase protein here called recA but recA is the bacterial
00:26:03.09 name of the Rad51 protein. So recA or Rad51 binds to this single stranded piece of DNA and forms a filament. This filament formed on single
00:26:15.13 stranded DNA is just an extraordinary machine which is capable of searching the entire space of the genome to find sequences which are homologous
00:26:25.02 to the single stranded piece of DNA inside the filament and when it does find those sequences and here that's pretty easy because these
00:26:35.03 sequences are immediately adjacent as these two sister chromatids are being replicated
00:26:40.02 simultaneously this filament then engages not only the
00:26:45.23 single strand of DNA but also the double stranded DNA template and carries out this exchange of base pairs which I illustrated before in
00:26:54.10 which now the single strand that has invaded is now base paired to the template and there's a displacement or D loop which is created
00:27:04.17 which is the beginning of a recombination event and here in break induced replication this little strand invasion allows time for the whole DNA
00:27:16.09 replication apparatus to be assembled so that now you can start to have both leading and lagging strand DNA polymerization and one has
00:27:26.06 taken what was a broken DNA molecule which was the product of stalled replication forks and re-initiated replication so that now replication can
00:27:37.19 continue in simple organisms it can go all the way to the end of a relatively short
00:27:43.23 - meaning hundreds of thousands of base pairs - chromosome.
00:27:46.12 In big cells we imagine that this occurs and then that replication fork simply meets another replication fork as would normally
00:27:54.27 happen so that the whole molecule can be replicated. Let me just say a little bit more about
00:28:03.11 what's really going on in this process of strand exchange.
00:28:06.12 The Rad51 or recA protein forms a filament on single stranded DNA and when it does so each monomer of recA can bind three nucleotides
00:28:18.23 of the single stranded DNA and it will bind and form this filament and the filament as it binds stretches the single stranded DNA to about one
00:28:30.19 and a half times the normal length. It also stretches double stranded DNA so you probably remember that B form DNA has 10.4 base pairs per
00:28:43.02 helical turn. When double stranded DNA is bound inside the recA filament it has been underwound and extended so that there are now 18.6 basepairs
00:28:53.28 per turn. So there's almost a 50% increase in the length of the molecule and this unwinding exposes the DNA bases in a way which
00:29:05.05 makes it easier for this strand exchange process that I have mentioned to take place. Now what's going on at every point in that strand
00:29:14.05 exchange is pretty simple. Here's double stranded DNA which is held inside the recA or Rad51 protein and here's the single stranded DNA
00:29:23.13 which is in the same protein and all that's happening basically is that we're forming an A-T basepair
00:29:29.21 between one of the two strands of the double stranded DNA
00:29:33.27 and the original single stranded DNA and one releases the other strand of DNA and this is going to happen of course at every
00:29:43.08 basepair along this whole region as long as all the bases are essentially homologous. The way in which the strand exchange happens biochemically
00:29:52.17 can be illustrated here this is a pretty standard assay for the way that people know the recA
00:29:59.01 or Rad51 filament can carry out these strand exchange reactions.
00:30:02.20 Here the recA protein is bound to a single stranded circular piece of DNA and then confronted
00:30:12.23 with a linear double stranded molecule and since the recA will now search for the homologous sequences in the double stranded DNA it begins to cause
00:30:24.17 the exchange of basepairs that I illustrated before. There's some intermediate point where part of this exchange has taken place
00:30:31.27 where one of the two strands has been base paired and the other strand as illustrated here is being kicked out and this process will continue
00:30:41.15 all the way around until the circle is complete and one ends up with a nicked circle which is illustrated NC and with a displaced single
00:30:51.03 strand piece of DNA and those things can be seen here on a agarose gel which is probed for the presence of these particular molecules.
00:31:02.29 So here we start out with single stranded DNA at the bottom. There's the double stranded DNA
00:31:09.23 and then you can see the appearance of this nicked circle
00:31:13.24 which is the product of this reaction and above it kind of smeared out are these joint molecules or the intermediates of the strand exchange.
00:31:22.14 Probably an easier way to sort of see this is by using a different approach which is to use radioactive phosphate
00:31:32.21 to label the ends of the linear double stranded DNA molecule. Here again there's a circle which is the single strand of DNA with Rad51 protein coating it.
00:31:44.00 Again it's going to carry out this reaction to form a nicked circle and a displaced linear strand but you can see all of those things now again separated
00:31:55.01 on a gel at the beginning all of the label is in the linear double stranded DNA and then as time
00:32:01.16 passes what one sees is the appearance of the nicked circle product,
00:32:05.21 these joint molecules that actually precede the appearance of the nicked circle product so over time
00:32:11.28 the nicked circle product is building up and at the same time as the nicked circle is appearing one sees the appearance of the single stranded DNA
00:32:21.27 and there wasn't any single stranded DNA at the beginning but only after the repair reaction is over do we see the appearance of this product.
00:32:30.04 So we know from these experiments biochemically that the recA and Rad51 proteins have this ability to exchange strands using base pairing
00:32:42.17 as the mechanism to sort of catalyze strand invasion which is the key step in all of these homologous recombination processes.
00:32:50.09 Here's just another picture of this reaction from left to right. At the beginning there's
00:33:00.26 single stranded DNA which is coated by the recombination protein recA
00:33:05.25 and there's double stranded DNA here which is going to be all engaged inside this recA filament where you have both single stranded
00:33:15.28 and double stranded DNA. It's still pretty mysterious what goes on exactly inside this filament but coming out the other end of this filament
00:33:24.04 now is a new double stranded DNA which involves the red strand which was originally single stranded and now you see the displacement
00:33:32.19 of one of the two strands of the original duplex so this is the strand exchange process that we talked about and in the case of the cartoons
00:33:42.09 that I've shown you before that would mean that you would have a region where there was strand invasion and the displaced or D loop DNA
00:33:52.07 will be exposed in this same complex. Very recently there has been a real increase
00:34:06.20 in our understanding of the way in which the recA and Rad51 proteins work
00:34:12.07 from the work of Nikola Pavletich's lab where they have been able to study this process in more detail at the xray crystallographic level.
00:34:22.28 To do this Pavletich's lab very cleverly took six recA molecules but they made a synthetic
00:34:31.23 gene in which all six molecules were part of a single open reading frame
00:34:36.21 and they were linked together by putting short protein linkers between each of the six molecules.
00:34:43.17 So now what one has is a very small piece of the recA filament but it's a uniform piece
00:34:49.17 which is therefore able to be analyzed more efficiently by xray crystallography
00:34:55.15 and when they did this kind of analysis it became clear that the single strand of DNA is basically at the axis of this filament
00:35:06.02 - it's being held by all of these molecules - each recA monomer binds to three basepairs
00:35:13.02 and as a whole this recA protein machine now has unwound or stretched
00:35:21.13 the single stranded DNA in the way that I talked about before. So we begin to see how this is really being held inside the recA filament
00:35:31.06 from this picture and then Pavletich has also taken pictures of what's happening during this process of strand exchange.
00:35:39.06 So the darkest strand here - the one that's here - is the single strand of DNA which I showed you before
00:35:48.05 and over here is double stranded DNA - the double stranded DNA helix
00:35:53.07 - and just beginning is that this strand is beginning to be incorporated into this double stranded structure.
00:36:01.15 So we're seeing if you will an intermediate of this strand exchange and then as time goes on one completes the process of this strand exchange
00:36:12.15 leaving behind one of the strands of the original double stranded DNA molecule as the displaced strand. So we're beginning to see this process
00:36:23.22 crystallographically it's still very difficult to understand in terms of the real dynamics of this process how this machine really works.
00:36:32.09 How does it line up double stranded DNA with a single stranded DNA if they just come together
00:36:39.08 the probability they're going to be lined up is exceptionally
00:36:42.18 small so there has to be sliding and lots and lots of comparisons at every step essentially there has to be a question of is this the right pair
00:36:52.09 of base pairs that can be made or should we move over one or move over two and all of these dynamic questions still remain to be understood.
00:36:59.27 To go back to the process that we were talking about so now what we can see is that the recA filament can carry out this strand
00:37:12.14 exchange process that leads to the displacement of the D-loop the strand that used to be part of the original double stranded DNA molecule
00:37:22.21 and there can be also now the establishment of new DNA replication and this theoretically can go all the way to the end of some replicating structure.
00:37:32.25 There's still one problem left in this process and that's that there's still a crossover here which after a little bit of branch migration
00:37:43.01 and ligation turns out to again be Holliday junction and that Holliday junction has to be resolved or there isn't going to be the ability
00:37:51.11 for these two replicating molecules to come apart. So again we need to look at the consequences of Holliday junction resolution.
00:38:00.17 So again as I said Holliday junctions can be cleaved either on the top strands or the cross strands and the two outcomes are different if they cross on the
00:38:10.16 A orientation these two molecules come apart without any exchanges but if they cut in the B orientation then there's going to be a crossover
00:38:20.29 and here we call that crossover a sister chromatid exchange because we're looking particularly at the case where we have restarted DNA replication
00:38:30.10 to allow the formation of two sister chromatids. We can visualize sister chromatid repair
00:38:40.13 and sister chromatid exchange in mammalian cells in a fairly easy fashion.
00:38:45.29 It turns out that you can grow cells in the presence of an analog of deoxythymidine called bromodeoxyuridine
00:38:54.07 and this will essentially incorporate bromodeoxyuridine in many of the sites where deoxythymidine should be placed in DNA replication
00:39:03.03 and one ends up with a DNA molecule after growing the cells in bromodeoxyuridine so that all the strands have bromodeoxyuridine label.
00:39:15.25 If you then take away the bromodeoxyuridine label and allow the cells to go through another round of DNA replication then as Messelson and Stahl
00:39:25.03 first showed for e. coli one ends up with semiconservative replication so one strand
00:39:30.04 has bromodeoxyuridine on it and the other strand has deoxythymidine
00:39:36.27 wherever there are T's in the structure and then if you allow this to replicate again in a second division now only one of the four strands of the products
00:39:49.14 of this replication is going to have bromodeoxyuridine associated with it. If there's no crossing over associated with this process
00:39:58.26 then one's going to end up with a molecule which has a single line of bromodeoxyuridine along the replicated chromosome and that's shown over here.
00:40:09.24 So here is a DNA molecule which has essentially a continuous line of bromodeoxyuridine reflecting the situation on the left.
00:40:21.03 But the situation on the right which is illustrated at this arrow has undergone one of those crossing over events - a sister chromatid exchange
00:40:30.09 - and that can be understood because now instead of having just a continuous line of
00:40:36.13 bromodeoxyuridine on one of the two sister chromatids
00:40:39.08 there's been a crossover and that crossover places bromodeoxyuridine on part of one sister chromatid and the other part
00:40:47.00 of the other sister chromatid which is illustrated here. So one can see evidence for these sister
00:40:53.11 chromatid exchanges by using this bromodeoxyuridine trick.
00:40:57.07 Now normal cells turn out to have quite a number of these events even in the absence of any other
00:41:06.04 perturbations so the normal process of DNA replication
00:41:09.15 as I showed you before at the beginning you take away the Rad51 protein you see all these chromatid breaks which are places where Rad51
00:41:17.28 recombinase is required in order to patch up these regions. Here what we're seeing is another consequence of that process
00:41:26.23 because that means that during the process of sister chromatid repair there are frequently sister chromatid exchanges
00:41:34.02 which are the outcome of that break induced replication process. We can make these chromatids
00:41:42.12 or chromosomes remarkably exchange by adding DNA damaging agents
00:41:48.27 so that if we increase the likelihood that these chromatids need to have DNA repair mechanisms in order to finish the process of replication
00:41:56.01 we end up with these so called harlequin chromosomes. Harlequin was a common character in renaissance comedia del'arte
00:42:05.25 plays in Italy and he wore a costume that had black and white checks on it and so these are called Harlequin chromosomes
00:42:16.11 and you can see every one of these represents another sister chromatid exchange on these chromosomes. Here's a picture of Harlequin
00:42:24.17 that I found that serves to just illustrate this. So what we also learn from this is that if we impose
00:42:34.03 DNA damage on these cells we're going to have an increased frequency
00:42:38.02 of repair and an increase frequency of these sister chromatid exchange events. Okay so what I've so far been describing
00:42:47.17 is one of the mechanisms of homologous replication this is break induced replication. We saw that
00:42:53.11 the ends of the double strand break have to be processed
00:42:56.03 by exonucleases that create a single stranded region of DNA this single strand DNA binds the Rad51 protein or the RecA protein
00:43:06.22 and the Rad51 protein filament then can search to find on the homologous partner in this case a sister chromatid identical sequences
00:43:16.02 and carry out this process of strand exchange and then this allows the initiation of new replication by the establishment of both leading and lagging strand
00:43:25.07 replication and eventually this process can essentially restart this broken and stalled replication fork and can allow this replication
00:43:37.21 to continue and the cells not to have any problems with the integrity of their chromosomes. So any defect in the machinery
00:43:46.10 that causes this process to take place is essentially going to create a situation where there are broken chromosomes that persist and which
00:43:57.19 are candidates for all of those wild number of rearrangements that we saw in the beginning
00:44:04.14 in metastatic cancer cells. There's another place where break induced replication
00:44:12.00 turns out to be very important and that is in a subset of transformed cell lines and cancers. You may know that when we are born
00:44:21.29 most of the cells in our bodies have chromosome ends called telomeres which are repeated sequences of a certain length
00:44:30.12 and as these cells divide telomeres get shorter and shorter and shorter with every replication cycle simply because the enzyme
00:44:40.01 that had added these repeated sequences called telomerase turns out to be turned off in almost
00:44:47.01 all the cells of the body except a small subset of stem cells.
00:44:50.27 In most cancers, when those cells become cancerous they turn back on the telomerase enzyme and cells become immortalized
00:45:03.15 in part because they can maintain chromosome ends with a long telomere and they never reach a point which is known as senescence.
00:45:11.03 There is however a small number of cancers - osteosarcomas being one particular example
00:45:19.09 - and many transformed cell lines never turn telomerase back on
00:45:24.12 and yet they become immortalized either as tumors or as cell lines and they do that by a recombination dependent process
00:45:32.05 which leads to what is called alternative length of telomeres and it turns out that this alternative
00:45:38.20 lengthening of telomeres process seems to be another example
00:45:43.08 of how break induced replication is operating in these cells. So we know for example from work in my own lab that in budding yeast
00:45:53.22 we can take cells and knock out the telomerase enzyme and as we expected in most cases
00:46:00.08 the telomeres get shorter and shorter and shorter and eventually
00:46:03.09 almost all the cells die because they have unprotected chromosome ends, but a small population of survivors emerges
00:46:11.06 which have figured out how to elongate their telomeres in the absence of telomerase
00:46:15.23 and it turns out that they all require proteins that we can identify
00:46:22.02 as being necessary for break induced replication, most notably a nonessential DNA polymerase subunit of pol delta called Pol32.
00:46:31.01 So not only is break induced replication important in restarting DNA replication forks but it's also very
00:46:39.23 important in maintaining chromosome ends in the absence of telomerase.
00:46:43.26 We talked about break induced replication, now I'm gonna start talking about the other mechanisms of homologous recombination
00:46:53.23 and I'm gonna actually start with one that's by far the simplest called single strand annealing.
00:46:58.19 In single strand annealing there's been a double strand break
00:47:02.15 and that double strand break is acted upon by those exonucleases here represented by Pac-Man chewing away the DNA
00:47:12.16 until there are long regions of single stranded DNA and when enough single stranded DNA has been exposed there's a probability
00:47:22.07 that there will be the exposure of sequences which are repeated on both sides of the double strand break.
00:47:28.08 Here this sequence is Watson and this sequence over here is Crick
00:47:34.12 and those two can then anneal which is where the process gets its name - single strand annealing - to form a duplex
00:47:42.29 that has these extra tails which are these additional pieces of DNA. Those tails can be clipped off and we know the enzymes
00:47:51.06 that do that and once they have clipped these regions off DNA polymerases can come and fill in these little gaps and what one ends up with is a deletion
00:48:01.23 which has removed all the DNA sequences from the point where the two repeated sequences are and made a deletion.
00:48:11.23 So this is a potent source of deletions, especially in cells that have much repeated DNA
00:48:18.00 in their genomes, and in yeast we often have to put artificially repeated
00:48:25.03 sequences in places that we want to study this process because there's almost no repeated sequences in the yeast genome,
00:48:32.15 but in humans a huge part of the human genome is made up of repeated sequences. There are, for example, something on the order of 500,000 copies
00:48:43.08 of a 300 basepair sequence called Alu and these Alu sequences can serve as the termini for deletion events between them which are caused
00:48:54.20 by this single strand annealing process. So you might wonder how long this process can occur: we've been able in budding yeast where there
00:49:06.05 aren't a lot of other repeated sequences to show that we can take those blue sequences and put them as far apart as a hundred kilobases
00:49:12.25 and we can actually make a double strand break as I'll show you later and this resection process will go for a hundred kilobases and eventually
00:49:25.00 allow those sequences to be joined. It's been recognized for a long time that Alu sequences are the endpoints for a lot of chromosomal
00:49:34.26 rearrangements that are associated with human disease. This is just one example from a paper
00:49:41.15 from 1990 that suggested that these events arose by crossover
00:49:47.01 events - unequal recombination events - between different Alu repeats in two different copies of the two sister chromatid
00:49:54.23 copies of the same gene that would produce this deletion but I think it is in fact more likely that these have occurred by a break on this chromatid
00:50:05.07 and then the chewing away of the DNA in both cases so that one ends up with single strand annealing which has produced the same deletion.
00:50:15.10 Okay, this brings us now to actually what turns out to be the most common form of double strand break repair both in yeast cells
00:50:26.08 and in mammalian cells and that is homologous recombination known as gene conversion. In gene conversions only a very very small piece
00:50:36.24 of DNA needs to be synthesized. The broken DNA molecule here finds a template. That template
00:50:43.17 could be a sister chromatid but it also could be a homologous
00:50:46.11 chromosome or it could be a little piece of this same sequence located in a different location - a so called ectopic location
00:50:54.20 - and the ends then find this homologous sequence and use a small amount of DNA synthesis to patch up the chromosome break.
00:51:04.26 Sometimes this can be associated with crossing over and that can lead to loss of heterozygosity. A mechanism by which gene conversion
00:51:17.18 takes place is in some ways similar to but different from the mechanism of break induced replication. Here's a mechanism first suggested
00:51:27.23 by Michael Reznick in 1976 and then refined by Szostak, Orr-Weaver, Rothstein, and Stahl in 1983 which has gone through a lot of modifications
00:51:37.10 to the form that I'm showing you now. The idea of this mechanism is again there is a double strand
00:51:43.14 break, the ends of the double strand break are chewed
00:51:46.11 away by those same exonucleases that I talked about before, and then these ends of the double strand break attract Rad51
00:51:53.26 or RecA recombinase and engage in strand invasion, but unlike what happened over here in break induced replication, both ends of the double strand
00:52:04.03 break can engage in this process and one ends up with the structures that are shown here in which essentially both ends of the double strand break
00:52:14.22 have engaged in strand invasion and then there's the initiation of new DNA synthesis but it's again different from what happens
00:52:21.21 in break induced replication. In break induced replication we saw that there was both leading
00:52:27.08 and lagging strand DNA synthesis, but here the two 3' ends of the broken
00:52:34.08 molecule serve as primers and each one of these primers sets up leading strand synthesis so there's no lagging strand synthesis
00:52:42.12 and in fact there is no requirement for lagging strand polymerases during this process and then these two blue lines represent new DNA
00:52:52.04 synthesis which are emanating directly from the 3' ends of the invading strands. In the mechanism
00:53:00.05 that I'm showing you here this leads to the formation
00:53:04.12 of an intermediate structure which is something that we haven't seen before but it is reminiscent of things we have previously seen,
00:53:11.06 namely there is not one Holliday junction, but two Holliday junctions and this double Holliday junction mechanism turns out to have some
00:53:21.01 very remarkable features of its own. As these Holliday junctions are resolved this structure can either be resolved so that there is no crossing
00:53:31.25 over and the red sequences here and the red sequences here remain associated with each other, or there can be crossovers
00:53:39.29 depending on the resolution of these Holliday junctions so that here these red sequences are now joined to blue sequences and vice versa.
00:53:48.09 So this is a different mechanism from break induced replication and it involves very limited amounts of DNA synthesis, and the DNA synthesis
00:53:57.11 is primed by the 3' ends of the invading DNA molecule and it does not involve lagging strand synthesis. The double Holliday junction
00:54:07.11 which is shown here has its own remarkable properties and one of them which is illustrated is on the next slide. Here what happens is that
00:54:19.26 this double Holliday junction can actually be dissolved and this is a property that cannot happen
00:54:26.16 with the single Holliday junction because it has no way to exchange
00:54:30.23 strands in the way that I'm talking about. Here the BLMs helicase - remember that I mentioned BLMs protein
00:54:38.25 early on as being a defect in DNA repair - well this protein turns out to be a helicase which can push
00:54:45.03 on these two cross structures and push them towards each other.
00:54:49.16 When it does that this creates a steric problem because as you do this the DNA in-between will become increasingly overwound
00:54:58.12 and this will resist the action of the BLMs helicase, but the BLMs helicase turns out to be associated
00:55:04.26 with a topoisomerase which you probably know is an enzyme
00:55:09.03 that can nick one strand of DNA and rotate the DNA around the remaining strand, taking up the extra supercoiling
00:55:15.13 that is happening as the BLMs protein pushes these junctions toward each other. So the combined action of the BLMs helicase and the topoisomerase
00:55:25.17 allows these things to come closer and closer together until there are essentially no more double Holliday junctions and when this process
00:55:34.07 occurs in this way one ends up resolving this molecule in a way that can never have a crossing over.
00:55:41.23 So the presence of the BLMs helicase and topoisomerase III
00:55:45.15 takes this double Holliday junction structure and ensures that it will be generating non crossover but repaired DNA molecules
00:55:54.24 whereas those double Holliday junctions that remain from this process can be acted upon by Holliday junction
00:56:01.12 resolvases and then they can end up as being crossovers.
00:56:05.06 So we have these two processes the crossover generating process which comes from the double Holliday junctions and then this antagonistic
00:56:17.01 process which removes the double Holliday junctions and ensures that many of the molecules will be present in this non crossover form.
00:56:26.22 So what happens in the absence of the BLMs helicase when this process doesn't occur?
00:56:33.15 Then in fact everything is going to be kept in this configuration
00:56:37.23 and the prediction is that there should be many more crossovers arising during this process and indeed
00:56:43.13 that's the case if one looks at bromodeoxyuridine labeled chromosomes in the way that I showed you before, in the absence of the BLMs helicase
00:56:54.20 what you see here is the fact that there's an astonishingly large increase in these harlequin
00:57:01.22 chromosomes where there are multiple sister chromatid
00:57:05.09 exchanges which are happening for every replicated molecule and that's occurring so far as we know because the BLMs helicase cannot dissolve
00:57:18.02 double Holliday junctions and argues that many of the events that we're seeing here are not simply the result of break induced
00:57:24.28 replication processes but may also be involved in double Holliday junctions caused by this gene
00:57:32.12 conversion mechanism. This idea that the BLMs helicase
00:57:40.23 and topoisomerase III take apart these double Holliday junction molecules can be seen even in a simple organism such as saccharomyces cerevisiae
00:57:50.11 as I'm going to talk about more in the subsequent lecture, we have the ability to create specific double strand breaks at individual places
00:58:00.15 on chromosomes in yeast by expressing a site specific enzyme that cleaves the DNA - a so called endonuclease
00:58:08.02 - and so a double strand break is made at some place on the chromosome and we provide here
00:58:14.27 on a different chromosome the homologous sequences that are needed
00:58:18.25 to repair this double strand break and most of the time when this double strand break is repaired, it occurs without a crossing over.
00:58:26.22 The crossover product which is shown here can be seen because we have a set of restriction enzymes which are differently placed in the top molecule
00:58:38.28 and the bottom molecule so that if there is a crossover now you have novel restriction fragments. There's actually a second crossover product
00:58:47.21 up here but it's so close to the parent sized molecule you really can't see it but here you can see that there's a very small amount of crossing
00:58:55.26 over in the normal process of this repair. If we take away the yeast equivalent of BLMs protein, which is called Sgs1, there is a significant
00:59:08.05 increase in the amount of this crossing over. If we take away the topoisomerase, we get exactly the same result
00:59:14.21 , and we take away both of them and it's the same. So in saccharomyces as well as in human cells this BLMs helicase plays a role in dissolving these double
00:59:25.29 Holliday junctions and forcing more of the molecules to be resolved as crossovers. I've shown you examples of a number of different mechanisms
00:59:34.27 of homologous recombination as ways to repair double strand breaks. Most of these double strand breaks have arisen during the process
00:59:44.13 of replication itself but I don't want to leave you with the impression that there aren't other sources of double strand breaks.
00:59:52.16 Double strand breaks do arise from things like ionizing radiation, they arise from the failed actions of topoisomerase II like molecules.
01:00:04.02 Topoisomerase II molecules break both strands of DNA and then rewind the DNA and then put them
01:00:11.03 back together sometimes that fails and if you have a drug
01:00:14.10 such as etoposide which is again another anti cancer agent you generate lots of these kinds of double strand breaks
01:00:22.06 from those kinds of sources and there are these different mechanisms that can be used to repair these double strand breaks.
01:00:30.12 In the following lecture I'm going to talk specifically about how we know how gene conversion
01:00:39.01 takes place in much more molecular detail than I've outlined
01:00:42.09 here and give you a better sense of how we know all of these steps actually occur in the living cell but I hope I've given you at least an overall
01:00:52.12 picture for the multiple ways in which breaks can be repaired in cells and the consequences of that kind of repair and leave you with the fact
01:01:01.20 that when some or all of these mechanisms are not functioning properly there are still other mechanisms which are less precise
01:01:10.17 that leads to the translocations and inversions and deletions that one sees in the cancer cells
01:01:18.12 that I started talking about. Thanks very much for listening.

Part 2: Details of DNA Repair in Budding Yeast

We were unable to retrieve the video

a00:00:06.01 Hello. I'm Jim Haber. I'm a professor at Brandeis University and director of the Rosenstiel basic
00:00:11.13 medical sciences center. In a previous video I talked more generally about mechanisms
00:00:18.13 of double strand break repair and how cells need to do this essentially every cell division in order to keep their chromosome from becoming rearranged.
00:00:29.00 In the talk I wanna give now I want to extend this to look in much more molecular detail at what happens in the repair of one specific
00:00:40.04 double strand break. Again this work is done with the model organism saccharomyces cerevisiae, budding yeast,
00:00:48.19 and again I'll say that the repair mechanisms that we're studying are very
00:00:55.01 strongly evolutionarily conserved and therefore have something to do with what
00:00:59.20 we know about human chromosomes and how they're repaired. In the last video,
00:01:07.10 I introduced two different mechanisms that can be used to repair double strand breaks.
00:01:14.15 One of these mechanisms which is called break induced replication is a way in which one end of the double strand break
00:01:22.17 invades homologous sequence, sets up a so-called displacement or D-loop, and then fills in that D-loop with both the leading and lagging
00:01:32.26 strand DNA polymerases, allowing the unidirectional replication fork to proceed down the DNA and this proves to be an important
00:01:43.04 mechanism to restart broken DNA replication forks and also apparently in the maintenance of telomeres of both yeast cells and human cancer cells
00:01:55.16 in those situations where the enzyme telomerase has not been reactivated. The second mechanism is a gene conversion mechanism
00:02:06.06 and in particular I talked about a mechanism called the double Holliday junction gene conversion mechanism and here the ends of the
00:02:16.19 double strand break are both processed in the same way as would happen in break induced replication by exonucleases that chew away the DNA,
00:02:26.15 but now both ends of the double strand break can invade into the donor locus, the two 3' ends of the single stranded regions
00:02:37.13 can initiate new DNA synthesis so you get a blue line here which is new DNA synthesis both here and again here and the new DNA synthesis
00:02:49.09 effectively allows the formation of an intermediate structure containing not one but two Holliday junctions. These symmetric structures can be acted upon
00:03:02.12 by resolvases or unwound by another pair of enzymes called a helicase and a topoisomerase to yield either non-crossover
00:03:12.07 outcomes or crossover outcomes associated with the repair of this double strand break. Now I'm going to complicate matters by telling you that although
00:03:24.10 this is an important mechanism of gene conversion and is apparently the major mechanism of gene conversion in the
00:03:33.16 repair of double strand breaks generated in cells undergoing meiosis
00:03:37.11 it is not the only mechanism of gene conversion in somatic cells or in meiotic cells either.
00:03:44.13 There is a second related mechanism of double strand break repair which is called synthesis dependent strand annealing
00:03:52.08 and this mechanism shares many of the features of these two previous mechanisms that I've generated but has its own special features.
00:04:00.08 Again, the DNA is first chewed away by exonucleases to generate single stranded regions that will attract the Rad51 recombinase protein
00:04:14.02 and allow the formation of a displacement loop in exactly the same ways that we looked at before, although here, for reasons that are not
00:04:24.01 entirely obvious to us, it is more likely that one or the other end invade rather than both ends invading simultaneously and now this end
00:04:35.08 that's invaded sets up new DNA replication that will start to copy out
00:04:41.24 the sequences that are in the region that has to be repaired, but the DNA synthesis
00:04:48.05 which is carried out in this process is different from what happens under normal DNA replication. In normal DNA replication,
00:04:56.21 we expect to see semiconservative replication that is to say the new DNA is associated with its template and stays associated with its template
00:05:08.29 but in synthesis dependent strand annealing it appears that the newly synthesized DNA is unwound from the template the way normally we think RNA
00:05:19.25 is unwound from DNA during transcription and this unwound DNA eventually will then anneal with the other end of this process
00:05:27.09 to produce now two newly synthesized strands and the repair of this double strand break and what's distinctive about this mechanism
00:05:38.07 relative to break induced replication is that essentially all of the outcomes will be without crossing over. So, this mechanism of synthesis
00:05:50.17 dependent strand annealing is a more conservative mechanism in somatic cells because it prevents crossings over that can lead to
00:06:00.24 loss of heterozygosity and preserves the original genetic information on both sides of the double strand break. Now the way we have come to know
00:06:14.01 a lot of the details of this process is by studying in budding yeast one particular example of a programmed double strand break
00:06:23.29 which leads to a particular outcome that we can study in some detail. This is known as mating type switching in budding yeast and I need
00:06:34.29 to present you with the features of this system. There is one locus on one of the sixteen chromosomes of saccharomyces known as the
00:06:46.15 mating type locus and it turns out that mating type can exist in two sexes. It can either be called mating type a or mating type alpha
00:06:55.22 and what is distinctive of these two alleles of this one gene is that they are encoded by completely different DNA sequences.
00:07:04.18 Here illustrated in red are the sequences that control the a mating type and the a1
00:07:10.19 gene product of this region is involved in regulating the cells a mating type.
00:07:16.15 In contrast matalpha cells have a pair of open reading frames which are in blue sequences, completely different from the red ones,
00:07:24.24 and encode two open reading frames that together allow cells to have the identity of being alpha cells. So this is a really remarkable
00:07:34.08 system because it turns out that in nature cells are capable of switching from a to alpha or from alpha back to a as often as every cell division
00:07:45.14 and this is not like any mutation process that we normally think about where you change one basepair from an A to a T or back.
00:07:53.22 Here we are replacing whole chunks of DNA with DNA that must have come from somewhere else in the cell and indeed that's the case because
00:08:03.24 on the same chromosome as the mating type locus there are in fact two donor sequences, one called HML which normally carries the blue
00:08:14.21 or alpha sequences, and a second donor locus which is called HMR and HMR normally carries the red or a sequences. In some of the experiments
00:08:27.06 I'll talk about we change the identity of these sequences from red to blue and the mechanisms that I'm talking about really don't have
00:08:34.22 much to do with whether we're using red sequences or blue sequences. In order for this system to work there has to be a way of stimulating
00:08:44.09 the replacement of these red sequences by blue sequences or vice versa. That is carried out by a site specific enzyme called HO endonuclease.
00:08:55.13 The HO stands for a process called homothallism which is the underlying process that these cells are going through which is a process
00:09:03.09 by which a cell can start with one mating type and produce cells of the opposite mating type and this is something that actually
00:09:10.05 happens in a number of different fungi, although the precise mechanism by which it happens is not the same in all of these organisms.
00:09:18.26 In the organism that we're studying, saccharomyces cerevisiae, there is an enzyme HO which will cleave in the MAT locus - in fact just at the junction
00:09:29.27 between the red sequence and the sequences that are the same in a and alpha and this chromosome break made by this enzyme
00:09:39.05 leads to the replacement of the red sequences by the blue sequences and those blue sequences
00:09:44.24 have to be copied from somewhere else so as I say there is a donor.
00:09:49.15 The donor is interesting in its own right. The donor is not expressed even though the donor has exactly the same sequences that are
00:10:03.28 expressed at the MAT locus, in the position of the donor, surrounded by a sequence called 'E' and another sequence called 'I' for the
00:10:13.20 imaginative names 'Essential' and 'Important' these sequences interact
00:10:19.23 with a histone deacetylase known as Sir2 which is able to remove the acetylated
00:10:30.16 residues on the N-terminal tails of histone H3 and H4 and lead to the positioning of nucleosomes across this region
00:10:40.06 in a way that silences the expression either of HML or of HMR so that these regions,
00:10:49.01 although they have intact genes, are not expressed. They are heterochromatic.
00:10:56.04 On top of that there is a particular nucleosome which is sitting about here which prevents the HO endonuclease
00:11:06.10 from cleaving this site. It's exactly the same site that HO endounclease can cleave here and HMR
00:11:13.13 has a site which is exactly the same as what will be cleaved
00:11:16.13 at the MAT locus but they aren't cleaved because there are nucleosomes sitting on these sites that prevents them from being both
00:11:25.26 transcribed and cleaved by the nuclease. That means that these two heterochromatic regions are templates for repairing the double strand break
00:11:33.25 but themselves are not expressed as genes. One of the mysteries of this process and something that we really just don't
00:11:40.25 understand in any detail is if HO endonuclease cannot cut this site,
00:11:47.09 how can the ends of the double strand break invade the same region, pry open the DNA,
00:11:52.29 and start copying DNA across this region? Because that's what must happen in order to repair the red sequences with blue sequences,
00:12:00.20 and I'll say a little bit more about that as we go along. The way in which we have been able to study this process in tremendous
00:12:09.16 detail came from an advance made by Ira Herskowitz's lab who placed the HO gene under the control of an inducible promoter.
00:12:19.03 So by simply adding the sugar galactose to the medium we can turn on the expression
00:12:25.09 of the HO endonuclease, it turns out that the control is extremely
00:12:29.10 tight so that there is essentially no HO endonuclease produced in the absence of galactose, but as soon as we turn on galactose
00:12:36.24 now the HO endonuclease can be expressed, and it's expressed sufficiently so that virtually every cell
00:12:43.00 in the population suffers the same double strand break
00:12:45.27 at the same time and that allowed us then to follow at the DNA level what was happening to the DNA during the synchronous
00:12:54.12 replacement of, in this case, a sequences by alpha sequences. Some twenty years ago my lab was first able to follow what was happening
00:13:06.22 during this process of double strand break repair. Shown here is a southern blot. The idea is that we're looking at a fragment of DNA
00:13:15.24 which comes from the MAT locus and probe with a probe just at the end of the MAT sequences. We start out with sequences that are in MATa
00:13:24.18 and then we turn on the HO endonuclease and in twenty minutes essentially all of the MATa sequences are cut into a somewhat smaller
00:13:33.11 restriction fragment by the action of the HO endonuclease which is shown here and then in what is an astonishingly long time the cells replace
00:13:44.04 the a sequences by alpha sequences and so in this way using southern blots we can easily see the beginning and the end of this
00:13:55.05 process and as I will show you, lots of intermediate steps in the process. The first take home message from doing this kind of an analysis
00:14:04.08 is that this repair process is surprisingly slow. Yeast cells have a division time under the conditions we're using of less than three hours
00:14:14.16 - two and a half hours - and yet it takes at least an hour for the cells from the time they make a double strand break to the time we see the repair product
00:14:23.14 and just to remind you we can easily see that the repair product is different from the starting product because the restriction sites in the blue
00:14:31.01 sequences are different and so we see a different fragment. So what we would like to understand in detail is where are the slow steps
00:14:42.10 in this process and it turns out it's not one slow step, it's many slow steps. Let's just look at the beginning of this process in more detail.
00:14:54.05 We went through this before in the first lecture that I talked about and briefly at the beginning. After there's a double strand break
00:15:01.19 made then there are these two exonucleases that can chew away the DNA to generate single strands of DNA. The single strands of DNA
00:15:10.19 are going to platform for the formation of the Rad51 filament, and the Rad51 filament is now going to search space in order to find sequences
00:15:21.10 which are identical with which it can catalyze the first step of the repair
00:15:26.08 process which is the exchange of basepairs between the single stranded DNA
00:15:30.15 at the MAT locus and the same region in the donor locus, in this case HML. I remind you that HML is in this case 200 kilobases
00:15:41.09 away on the same chromosome, so this search is considerably more difficult than searching for a sister chromatid which is going to be very
00:15:49.11 very close nearby during replication. Now this broken molecule must search over an entire chromosome or in some cases if we fool around
00:15:58.11 with the system, searching other chromosomes to find these sequences and to generate this first step in the repair process. So how do we know
00:16:11.05 these things that I just said? Well I'm going to show you how we know these things in this system. In 1990 Charles White
00:16:20.00 who was postdoctoral fellow in my lab worked out the fact that we could see the first of the intermediates of this process which is the action of this
00:16:29.24 exonuclease which is now moving down the DNA and if we ran the DNA on a southern blot but under denaturing conditions so that the Watson strands
00:16:40.16 and the Crick strands all migrated as single molecules, what Charles showed was the fact that as the exonuclease proceeded
00:16:48.18 down the DNA it would go past some restriction site, in this case the StyI
00:16:53.09 site, and when this site is single stranded it cannot be cleaved.
00:16:57.17 The enzyme requires double stranded DNA for cleavage and so at that point
00:17:02.08 the StyI fragment that is generated would be cleaved here and if you probed
00:17:09.08 this with sequences which are specific for this strand, the strand that ends 3', because it is the 5' strand that is being chewed away,
00:17:19.10 we could see the appearance of what looks like a partial digestion product. If you've run southern blots and you don't add enough enzyme
00:17:26.09 you would see what appears to be partial digestions because you wouldn't have cut this site, you'd have
00:17:31.23 cut the next site. That's exactly what we're seeing here
00:17:34.01 but it's not a partial digestion product, it's because this site cannot be cleaved.
00:17:39.00 Now we get a piece of DNA that's this long and as the exonuclease continues past this site, now this site cannot
00:17:48.11 be cleaved and we get a fragment this long and indeed
00:17:50.26 you can see another fragment very faintly here which is the next fragment in this resection process. Charles worked out
00:18:00.07 the fact that when there was a double strand break there was extensive 5' to 3' resection on this molecule. On the way to generating the product
00:18:12.15 which again goes from a to alpha in this particular setup. This represented the first recognition that the way that the recombination process
00:18:25.24 started was by this resection. You can actually make this assay more informative by removing
00:18:35.14 those two donors. Now when we make a double strand break,
00:18:38.03 the resection process is going to go on and on and on but instead of then engaging the donor and turning everything off
00:18:46.09 and doing the repair there's no way to do the repair and it turns out that this is a very unintelligent resection machinery. It just keeps on chewing.
00:18:54.14 From these kinds of measurements we can actually say the appearance of these partial digestion products if we measured
00:19:02.16 when they appeared, we learned another of the slow features of this recombination process: the resection occurs at about one nucleotide
00:19:12.02 per second which is about four kilobases an hour. This is surprisingly slow in terms of what you might imagine happening in the test tube.
00:19:21.24 Maybe it's slow because the nucleases going down the DNA not only have to chew away the DNA, but there are nucleosomes in the way
00:19:28.27 and those nucleosomes have to be removed and so forth. We don't really know why this process is quite so slow. But what we have
00:19:36.14 been able to learn and this has been mostly the work of other laboratories: Gregory, Ira, Lorraine Symington, and Stephen Jackson's labs
00:19:45.18 that there are at least two major exonucleases which are responsible for this digestion. One of them is an enzyme called Exo1 which also plays
00:19:58.28 roles in other repair processes such as nucleotide excision repair and the other is a complicated machine involving a helicase which is unwinding
00:20:10.02 DNA and an endonuclease which is clipping off the little tails of DNA which contains a protein called Sgs1 which I have mentioned previously
00:20:20.21 in the earlier video is a homolog of the Bloom's syndrome protein Bloom which plays many other roles in recombination besides the role
00:20:30.16 that it's playing here. So what's illustrated here is the fact that you can easily see the appearance of these partial digestion products in the wildtype
00:20:38.26 cell but if you remove both Exo1 and remove the Sgs1 helicase protein now there is essentially no resection and the broken DNA
00:20:51.16 basically stays in space; it stays for as long as the experiment is carried out. We now know something about which enzymes
00:20:58.27 are carrying out this process of resection. Once the resection has taken place the goal
00:21:08.00 of the next step is the formation of this Rad51 filament. In vitro
00:21:16.01 experiments - experiments carried out using these proteins in the test tube combined with some genetic identification of some other
00:21:23.28 components that are needed for recombination suggested that in order to form this Rad51 filament it was probably preceded by the formation
00:21:34.03 of a different filament which contained the RPA or ssDNA binding protein that first binds to this region. Among other things, single strand DNA
00:21:45.11 binding protein prevents single strand DNA from self pairing and forming little complicated secondary structures and so it stretches
00:21:54.09 out the DNA but this RPA can only then itself be removed from the single stranded DNA by the action of a set of what are called mediator
00:22:02.23 proteins, one of which is called Rad52 and another pair of proteins are called Rad55 and Rad57. The latter of these proteins, Rad55 and Rad57
00:22:15.04 , actually have some resemblance to Rad51 protein but they can't do Rad51's job. They're called Rad51 paralog proteins and they play
00:22:26.00 a fundamentally important role in this process of getting rid of the single stranded binding protein and replacing it with the growing filament
00:22:36.05 of Rad51. These proteins are very important in this process and again we know this from test tube work but I'm going to show
00:22:46.19 you that now we know this by looking inside living cells. I'll also make a digression just to say that one of the curiosities about evolution
00:22:56.26 is that the yeast cells that we study do not have another protein which is called BRCA2. This is one of the two major familially inherited defects
00:23:11.00 in breast cancer and it turns out that the BRCA2 protein plays a role very similar to Rad52 in helping to displace RPA and to put Rad51
00:23:23.18 in its place. It turns out saccharomyces just doesn't have this protein and for a long time we thought maybe this was a protein that was only
00:23:33.07 evolved late in evolution and would only be found in worms or flies or in mice or people but in fact some other yeasts - a yeast called ustilago
00:23:47.24 which for those of you know is also known as corn smut and is actually a Mexican delicacy - that organism has a BRCA2 homolog, so we now
00:24:01.12 think that saccharomyces has just lost this function and has kept Rad52 as the only way of doing this kind of repair. Rad52
00:24:11.19 and these paralog proteins are now going to facilitate the formation of this filament. We can see that this is happening in vivo by taking
00:24:23.04 advantage of a technique called Chromatin Immunoprecipitation. For those of you who haven't been exposed to this before, it's a remarkable
00:24:33.05 way in which one can see what parts of DNA are bound by a particular protein and the basic steps in chromatin immunoprecipitation
00:24:40.27 are illustrated here. The proteins are bound wherever they happen to be bound on DNA at any given time. If one adds formaldehyde to the
00:24:50.05 cells it crosslinks proteins to each other and crosslinks proteins to DNA. Then the chromatin - proteins and DNA combined - is sonicated
00:25:02.13 and shattered into relatively small pieces and if one then uses an antibody against whatever protein you happen to be interested in, in this case
00:25:11.22 Rad51, then you can purify all the DNA that is associated with that particular protein. If you have a different antibody you can purify
00:25:22.28 all the DNA that was associated with this protein or that protein but here we're interested in Rad51 and so we can pull down all the DNA
00:25:31.17 that is purified with this antibody and then it turns out almost like magic by simply
00:25:38.24 raising the temperature one can reverse these formaldehyde induced
00:25:43.05 crosslinks and then get rid of the protein and end up with a pure solution of all the DNA molecules that were initially associated
00:25:52.23 with, in this case, Rad51. By doing this, we can then purify and ask to what sequences does Rad51 bind at any point in this repair process?
00:26:04.23 We use polymerase chain reaction to amplify and confirm the presence of any individual set of sequences. That's what we did here.
00:26:14.17 Here's the southern blot again - the one I showed you before - that takes this very long time before we get a product and here is chromatin
00:26:25.07 immunoprecipitation results. The first result I'm showing you is not Rad51 but RPA. We think RPA binds before Rad51
00:26:35.11 and indeed as soon as we see the double strand break at this position, we see an enrichment
00:26:42.08 in the amount of DNA that is at the MAT locus which is brought down
00:26:46.17 by the anti-RPA antibody and again we're using a PCR amplification to ask whether MAT sequences are being enriched and indeed
00:26:58.28 as soon as we see the break we see the appearance of RPA at the break site.
00:27:03.07 In contrast, and surprisingly, it takes ten minutes before we see the appearance
00:27:10.17 of Rad51. Here's the Rad51 ChIP and again we've immunoprecipitated using the anti-Rad51 antibody and we're identifying
00:27:19.10 the region by using a PCR analysis of this region and we see that unlike RPA which appears as soon as the break is there, it turns out
00:27:28.20 that it takes about ten minutes before we see the appearance of Rad51. That's again something we had no anticipation of from just thinking
00:27:38.07 about this from biochemical terms where all of these things happen very very rapidly in the test tube. In the living cell from which we've extracted
00:27:47.15 this information there's a real lag between the time that it can do one of these steps and the time that it can do the next step.
00:27:53.20 We don't yet know why. The other thing that we know by doing the same kind of analysis is that there's essentially no Rad51 loading at the MAT
00:28:06.28 locus. Let me just go back and say now we're looking at this reaction so we're asking when does Rad51 load onto the MAT DNA?
00:28:15.23 What happens if we do this in the absence of that Rad52 protein and the answer is nothing
00:28:21.12 happens because you absolutely need Rad52 in order to displace
00:28:25.27 RPA and to put Rad51 onto this single stranded molecule. In contrast, if we knock out one of the paralog proteins, Rad55,
00:28:34.24 what we find is that eventually we can load the Rad51 protein onto the DNA but unlike what happens in wildtype cells where this happens within
00:28:44.26 an hour, it takes a long long time before it can do this in the absence of the Rad55 paralog. Rad55 paralog is important, Rad52 protein
00:28:56.26 is absolutely necessary for the formulation of this intermediate and in both of these cases the cells fail to carry out the recombination reaction
00:29:06.20 because they don't have a functional Rad51 filament. This same requirement is found in mammalian cells. This is not work that we've done.
00:29:16.11 Or I should say in vertebrate cells. This is again the remarkable system of DT40 chicken cells which carry out homologous recombination
00:29:24.24 at a very high level and it has been extensively studied primarily in the laboratory of Shinichi Takeda where they can create knockouts
00:29:33.22 of the genes in the same way that we create knockouts very easily in saccharomyces cerevisiae so it's possible to now compare yeast
00:29:42.13 results and vertebrate results using this particular system. This is not by knocking genes down by siRNA, but really ablating them by complete
00:29:55.00 deletions. Again what one sees here are in the different assay is the fact that you don't have Rad51 filament formation in the absence
00:30:05.05 of either BRCA2 deficiency or in the absence of one of the five paralogs that vertebrate cells
00:30:12.19 have that are related to the Rad55 and Rad57 proteins. This one is called XRCC2.
00:30:18.15 So if you irradiate cells with gamma radiation in wildtype cells after a while one sees these bright spots.
00:30:29.06 Those are staining with an antibody against the Rad51 protein. These foci represent places
00:30:34.15 where a filament has formed by assembling Rad51 at the sites of double strand breaks and Rad51 simply fails to form
00:30:43.18 in the absence of BRCA2 and in the absence of the paralogs. Very much like we could see by ChIP, it can be seen
00:30:52.03 in vertebrate cells using this somewhat more remote technique because they're not looking directly at the DNA
00:30:58.29 or an individual set of DNA sequences. That's taken us to this point where we have a filament. The next thing we
00:31:09.00 want to know is: how does that filament carry out the search to find homologous sequences and to carry out this
00:31:16.13 point of strand invasion? The first question is how long does it take? It turns out that the same chromatin
00:31:24.02 immunoprecipitation technique that I showed you can be used to figure out when Rad51 binds to these sequences
00:31:32.17 can actually be used to figure out when the Rad51 filament finds the donor sequences and that's basically just illustrated
00:31:40.16 here. Once you crosslink Rad51 protein to DNA, you can see when you pull down the MAT sequences, but when
00:31:49.12 the strand invasion process happens Rad51 at least for a time should be associated not only with the single stranded
00:31:56.13 DNA, but with the donor sequence. This is the double stranded DNA and the single stranded DNA
00:32:01.19 coming together in the filament in order to do the strand exchange process and so if you crosslink this intermediate
00:32:08.26 with formaldehyde and do exactly the same enrichment with the antibody you should bring down not only the MAT
00:32:15.27 sequences but the donor sequences as well. And indeed that turns out to be the case. If you do the same experiment
00:32:24.27 and now use a probe not for MAT but a set of PCR primers that are specific for the donor sequences,
00:32:33.20 the donor sequences do not become associated with Rad51 until considerably later than the donor sequences which were bound
00:32:44.26 at MAT. That makes sense - it takes real time for this search to take place - for the Rad51 filament to search all over this chromosome to find a tiny
00:32:55.01 little region it's in fact only about a 320 basepair region that has homology to the end of this double strand break.
00:33:02.28 The whole chromosome turns out to be almost 320 thousand basepairs and we're looking for 320 of those basepairs
00:33:14.13 as the site of homology and it takes real time for this search to take place. As it says here, if we were to put these donor
00:33:23.28 sequences on a different chromosome, as we have done, this search takes even longer because it turns out to be easier
00:33:31.17 to search within a chromosome than to search between chromosomes and of course any chemistry teacher you
00:33:40.03 ever had told you that an intramolecular reaction was faster than a bimolecular reaction but to think about that at the
00:33:48.04 level of a whole chromosome is kind of amazing. That is to say an intrachromosomal interaction 200 kilobases away
00:33:55.24 is much more efficient than searching between different DNA molecules. I wish we knew more about why that was.
00:34:05.14 We were progressing - we've gotten to the point where there's strand invasion - the next step in this process is that there
00:34:15.04 has to be new DNA synthesis and new DNA synthesis is going to be initiated from the end of this single strand here
00:34:23.24 adding DNA polymerase and now copying across this region and as I mentioned earlier this is different from normal DNA
00:34:31.10 replication because we think that the newly synthesized strand is being extruded and is only at the end of this
00:34:38.15 extrusion process that the second end is going to be able to carry out synthesis now in the opposite direction to fill in all of this.
00:34:47.20 During this process, there also has to be the clipping off of these red sequences and this turns out to be done
00:34:56.00 by a set of proteins including the Rad1 and Rad10 proteins whose normal job is to clip DNA during the repair of photodimers.
00:35:07.29 Here recruited to clip off these nonhomologous tails and it turns out that this also is recognized by two proteins
00:35:17.12 that normally have a role in recognizing mismatched DNA called MSH2 and MSH3 which have a role in mismatch repair
00:35:25.16 and they also play a role in doing this clipping but none of the other proteins of nucleotide excision repair
00:35:32.10 and and none of the other proteins of the mismatch repair are involved in this process so this is a new little machine that is assembled
00:35:39.14 out of components of two other important DNA repair processes and that allows the removal of these red sequences.
00:35:47.19 The next thing we want to focus on if you will is the initiation of new DNA synthesis. How do we see that?
00:35:58.29 Well, we see that by using another really nice assay that was developed also by Charles White in 1990 and that is a
00:36:08.05 polymerase chain reaction, so what Charles figured out is that if you put one primer in HML just to the left of the shared
00:36:17.26 sequences in essence the alpha sequences that are going to be added to this in this repair process and another PCR primer
00:36:27.29 which is here at the MAT locus these two primers initially are 200 kilobases apart and so no amplification but once
00:36:39.15 strand invasion has taken place and you allow fifty basepairs of new DNA synthesis from the 3' end, now there's a covalent
00:36:49.04 piece of DNA which joins this primer and that primer and you get a PCR product that you can assay. We can measure now,
00:36:58.19 here by chromatin immunoprecipitation, here the synaptic set by chromatin immunoprecipitation, and then by this very
00:37:07.03 clever PCR assay we can see the appearance of this new product and what I'm going to show you is basically that
00:37:17.20 both of these are in and of themselves slow steps. The search, refined by taking more points, is not a half an hour
00:37:26.07 as I first showed you but about 15 minutes. But it takes then another 15 or 20 minutes before we see the appearance
00:37:32.29 of new DNA synthesis, and why it should take so long to assemble the DNA repair machinery, the replication machinery
00:37:41.14 at this end is another of the things that we still need to be studying This is unpublished work from a graduate student
00:37:50.28 in my lab named Wade Hicks. Basically, Wade just took lots more points to sort of carry out this assay and it repeats
00:37:59.22 what I just said: we can see the kinetics of the Rad51 protein at the broken end of the DNA molecule at MAT,
00:38:08.07 we can see the appearance - and it turns out it's about 15 minutes later - of the Rad51 protein at the donor sequence HML,
00:38:17.08 and then we have to wait another long time before we see the beginning of new DNA synthesis by that PCR
00:38:23.16 reaction that I just showed you. These are all slow steps in this process and we haven't even gotten to the end because,
00:38:31.18 if I just go back, this step here is the step to begin this process but it takes it turns out another 15 or more minutes
00:38:42.16 maybe longer before we get to the very end of this process and we've completed all the copying and cleaning
00:38:48.01 up steps to go from MATa to MATalpha. I don't really have a lot to say about this approach but I just want to show you
00:38:59.28 another way in which it is now possible to carry out these kinds of analyses and that is actually watching the process
00:39:07.15 of mating type switching by watching. Here we've borrowed the technology that was first developed in John Sedat's
00:39:17.04 lab by Wallace Marshall and Aaron Straight and Andrew Belmont in which one can tag individual locations on chromosomes
00:39:28.21 and one does that by inserting arrays of the Lac operator sequences. This is the operator for the Lactose operon
00:39:37.01 of bacteria - e. coli - and then binding to it a fusion protein between the Lac repressor protein which is fused to green
00:39:45.20 fluorescent protein and that will create a green mark at some point in the cell. Here right next to the donor HML.
00:39:53.15 By the same kind of approach, one can put a different set of sequences - the operator for the tetracycline inducible system
00:40:02.18 - and have the Tet repressor bound to it now fused in this case to red fluorescent protein so we have two spots
00:40:10.00 one red and one green and the question is what do they do when you turn on the HO endonuclease? This uninformative
00:40:19.20 little movie is just simply saying we can actually do what I want to call single molecule measurements. This is one
00:40:28.01 chromosome in one cell watching how these sequences come together, stay together, and come apart during the process of repair.
00:40:35.19 The data that I'll show you which is on this slide actually were done earlier in fixed cells - because we're not
00:40:44.03 finished with the live cell analysis - using two green spots but the principle is exactly the same. The first thing is I told
00:40:52.08 you that MAT is going to recombine with HML to replace a sequences by alpha and one question would be 'are those
00:40:59.13 sequences already together before the repair process takes place?' and the answer is no because in almost all the cells
00:41:06.24 we see two spots. What this graph shows is what fraction of the cells have only one spot and we see one spot when the
00:41:14.19 two green spots come together and they're no longer resolvable because they're too close together and so only
00:41:20.28 a very small fraction of cells have one spot at the beginning but as soon as we turn on the HO endonuclease the fraction
00:41:28.16 of cells with one spot goes up and up and up. That's reflective of the fact that they're coming together
00:41:33.13 to do this repair process. They stay together for some period of time necessary to do the repair and then they come
00:41:40.13 apart and so we can see now essentially this whole process in either fixed cells and now in living cells as they're doing this process
00:41:51.17 unperturbed by any of these other mechanisms. We're not using chromatin immunoprecipitation, we're not using
00:41:57.12 PCR, we're just watching and yet it's telling us a lot about this process. One thing it has told us is the function of
00:42:08.02 another protein that is unable to carry out repair. This is another of the so called Rad genes. All of these Rad genes were
00:42:17.01 identified initially because they're sensitive to x-rays. So Rad52, Rad51, and Rad55, and Rad57, and others were all
00:42:28.11 identified because the cells failed to properly repair double strand breaks made by ionizing radiation by x-rays and one of those
00:42:35.19 proteins was Rad54. Rad54 also cannot carry out the completion of mating type switching or other double
00:42:44.21 strand break induced events. But its defect is different from Rad55, or Rad57, or Rad52 because it loads the Rad51
00:42:54.13 proteins fine - there's no defect in the loading of Rad51 onto the ends of the double strand break - and it can do this
00:43:02.11 strand invasion process fine which we can see here because it can put these spots together - the two green spots together -
00:43:09.29 no problem but it can't finish the process. And in fact, if we analyze this using polymerase chain reaction of the
00:43:18.17 sort I showed you before, the reason that Rad54 is defective in this process is it cannot carry out the next step in the process
00:43:26.02 which is that 50 basepairs of new DNA synthesis that we analyzed by polymerase chain reaction. So Rad54
00:43:36.14 can apparently carry out something like a normal strand invasion process but it doesn't do something in the strand invasion
00:43:45.00 process to allow the end of the DNA to be available for now new DNA synthesis and this has been one of those
00:43:52.21 things we figured out by combining these very different techniques: chromatin immunoprecipitation and this visual light
00:44:00.20 microscopy to sort of see where the Rad54 protein really has a defect. The next things I'm going to tell you are things
00:44:13.04 which we're really still in the process of understanding and so I'm going to show you more ideas about what we're trying
00:44:21.11 to learn than final conclusions because we're not done with these kinds of approaches. But one of the things
00:44:28.07 that really fascinates us is the problem that I mentioned at the very beginning: how is it that the HO endonuclease
00:44:35.15 cannot cleave the donor sequences? Apparently because of the presence of this nucleosome which is sitting in this region
00:44:42.27 and I'll just say that Kirsten Weiss in Bob Simpson's lab some years ago showed by using micrococcal nuclease
00:44:51.15 and other techniques she could map the exact position of all of the nucleosomes across the silent HML region and so there's
00:44:58.25 this whole series of very highly positioned nucleosomes across this region and one of them sits where HO ought to be able
00:45:06.24 to cut. So the question is: how can Rad51 filament engage exactly the same
00:45:13.18 region in order to initiate strand invasion and new DNA synthesis? And the take home message is we don't know.
00:45:21.13 Here's how we are trying to figure it out. We can analyze how well these nucleosomes are positioned by cleaving
00:45:30.24 the DNA with micrococcal nuclease, which will only cleave in the spacer regions between nucleosomes, and if a nucleosome
00:45:40.23 has not moved and is in one position then a pair of PCR primers that goes around the nucleosome will amplify a short
00:45:49.26 region of DNA that we can see. But if that nucleosome has been pushed out of the way, then we ought to see that that
00:45:57.21 region is no longer protected and that signal should drop. The aforementioned Wade Hicks who had been doing the
00:46:04.26 experiments that I showed you previously has been trying to do these experiments and he sees something but not
00:46:11.12 a lot and that is that as strand invasion is taking place and this is the Rad51 ChIP showing that these sequences
00:46:19.08 have come together with the donor there is a really small change in the protection of the nucleosome at the place where
00:46:28.27 the strand invasion is taking place. These experiments were done under conditions where the process could
00:46:35.01 continue and the cells could go ahead and finish the repair process and maybe if we could stop the process here before
00:46:42.19 new DNA replication had taken place we would be able to see a more profound change in the nucleosome structure
00:46:49.12 and that's a question that we'll address in the future. Since I just mentioned the fact that it would be nice to stop the process
00:47:02.11 by blocking DNA replication, I should tell you a little bit about what we know about how DNA replication
00:47:08.11 is really taking place in this region. It's clearly different for normal DNA replication in that the newly synthesized DNA is being extruded
00:47:18.05 and then being used to copy the other strand and that predicts that all the newly synthesized DNA
00:47:23.19 should be in the MAT locus and the donor should remain unaltered and we believe that this is the case. First let me
00:47:31.21 show you what we know about what proteins are required for the DNA replication in this special circumstance.
00:47:38.18 We know that DNA replication normally involves the clamp protein called PCNA which is necessary for holding the DNA
00:47:49.01 replication machinery on the template and that turns out to be true we need it in this situation as well. Here is a cold sensitive
00:47:57.11 mutation of PCNA and if we block the cells so that they cannot do normal DNA replication at low temperature
00:48:06.04 then instead of - this is also done at low temperature - normally we would see the appearance of this product
00:48:12.23 which is now MATa switching to MATalpha. But in the cells that lack this functional PCNA protein at the low temperature,
00:48:20.26 there isn't any synthesis, ergo, you need PCNA for this process to take place. It looks as if you need either
00:48:30.18 pol epsilon or pol delta to carry out this process. These are two of the major three polymerases in saccharomyces and I won't
00:48:39.15 show you the data but you don't need the lagging strand polymerase pol alpha or its associated RNA dependent
00:48:47.05 primase step but you need apparently either pol epsilon or pol delta because if we do the same kind of experiment
00:48:55.25 here using high temperature sensitive mutations, if we have a temperature sensitive mutation of pol epsilon or a temperature
00:49:03.23 sensitive mutation of pol delta and we look at these at their nonpermissive temperature what we find is that
00:49:10.19 in both cases we do get product though it is somewhat slower and somewhat reduced and this at least raises the possibility
00:49:19.15 that either of these polymerases can work. It doesn't quite tell us whether one of these does 98% of the time
00:49:26.23 the job, but when we knock it out the other one can pick up the slack, or whether they really are just sharing the process.
00:49:33.03 We still don't know for sure what the answer to that is. We've looked at a lot of other components of DNA replication.
00:49:42.06 One of the ones we've looked at is a protein complex called Dpb11-Sld2-Sld3. This is a complex necessary
00:49:51.29 for the loading of the helicase proteins and the setting up of a normal DNA replication fork at an origin
00:49:59.02 of replication and we have also looked at a kinase called Cdc7 which again is needed for all normal DNA replication.
00:50:09.23 So the question is: do you need Sld2/3 and Dpb11 - do you need Cdc7 - for normal MAT switching, which is not the same as regular
00:50:21.04 replication? To do this, we used a very clever trick by Kevan Shokat at UCSF which is to create what is known as an analog
00:50:32.21 sensitive allele of Cdc7. The Cdc7 analog sensitive allele has been mutated in such a way that its normal
00:50:43.10 ATP binding site - it's a kinase, it uses ATP to phosphorylate proteins - that active site has been enlarged
00:50:50.18 in a way that will accommodate an inhibitor and no other kinase in the cell has an enlarged site like this so the inhibitor
00:50:58.15 specifically inhibits Cdc7 without inhibiting any of the other kinases in the cell. That means we can specifically
00:51:06.09 block the action of Cdc7. If you look here, we have blocked the action of Cdc7, these cells cannot do normal DNA replication
00:51:15.27 because they need Cdc7 for that purpose, but they have no problem carrying out MAT switching.
00:51:21.09 So that means Cdc7 is not required for MAT switching even though it is required for normal DNA replication. Now, we take those cells
00:51:30.28 which are blocked prior to initiating DNA replication and we ask: do they require Dpb11? The answer is, apparently,
00:51:40.20 yes they do because in the absence of other DNA replication - because Cdc7 has been blocked
00:51:46.21 - if you knock out Dpb11 there's no product. This argues to us that Dpb11 is required in the process of normal MAT switching
00:51:57.23 as well as for normal DNA replication when Cdc7 is knocked out. We've been doing other experiments to try and get
00:52:05.18 a list of all the proteins that are needed for replication under these conditions and although I won't show you the data
00:52:13.04 for this, it turns out that the helicase that is necessary for normal DNA replication - which involves the MCM proteins,
00:52:21.22 MCM2, 3, 4, 5, 6, and 7 - also is not required for doing the little patch of DNA synthesis that we see in mating type switching.
00:52:31.22 That would suggest that we are looking at an unusual process which uses a subset of the DNA repair
00:52:45.08 machinery to carry out this process and again argues that all the newly synthesized DNA should be in the recipient
00:52:52.14 locus so we've tried to address that question as well and then that's the question: does MAT switching really proceed by
00:53:00.22 this SDSA mechanism? The key thing is really asking the question: is the DNA synthesis in this process conservative
00:53:10.05 or semiconservative? In semiconservative replication, the kind that we think about in normal DNA replication, the
00:53:21.07 newly synthesized DNA is going to always be associated with the old strand from its template that it was copied from, but in this synthesis dependent
00:53:32.10 strand annealing mechanism, all the newly synthesized DNA is going to end up in the recipient
00:53:38.16 locus. And so the question is how do you tell the difference between these two? To do that we basically
00:53:44.05 followed a technique that was developed in 1958 and is sometimes called the most beautiful experiment done
00:53:51.05 in molecular biology by Meselson and Stahl in which they demonstrated that DNA replication was in fact semiconservative
00:53:59.03 and just to remind you of this classic experiment, they grew the cells in heavy medium until all the DNA
00:54:07.16 was heavy labeled with heavy nitrogen. They then shifted the cells to normal N14 containing medium and then they separated
00:54:16.13 the DNA by its density in a gradient of cesium chloride and they were able to show that starting with heavy
00:54:25.29 DNA that after one round of DNA replication, that all the DNA was uniformly of one sort, it was heavy-light,
00:54:34.19 that is to say one strand was still heavy and the newly synthesized strand was light and then if that went through another round
00:54:42.25 of DNA replication you'd end up with a population of light-light DNA and an equal amount
00:54:49.15 of heavy-light DNA from the first subsequent round of replication.
00:54:53.18 So this density technique allows one to tell whether or not one has semiconservative
00:54:59.11 replication or do you in a single jump go from heavy-heavy to light-light? And so we carried
00:55:06.20 out that experiment - Gregory Ira - who was a postdoctoral fellow in the lab and now has a lab at Baylor university modified the MAT
00:55:16.12 locus here for technical reasons we had to make a somewhat larger piece of DNA which we put in place
00:55:23.21 of the normal donor sequence - A sequences - and so when this switches into MAT we can identify those sequences
00:55:32.04 and the question is: when you get switching, cells grown in heavy medium, transferred to light medium, and then
00:55:39.10 asked to go through the switching process, do they generate light-light DNA or heavy-light DNA? Heavy-light DNA
00:55:50.18 would be the mark of semiconservative synthesis. We carried out this experiment. What we get is that DNA is almost
00:56:01.21 exclusively light-light which means that it went from heavy to completely light in a single switching event which is what you
00:56:09.10 would expect if this was happening by SDSA and not what you'd expect if it was happening by a so-called
00:56:15.19 semiconservative replication mechanism. The sequences are slightly heavier than light-light because part of the sequences
00:56:23.25 in the fragment that we analyzed don't have to be replaced during the repair process and they stay heavy
00:56:29.15 and so we don't get exactly light DNA but if it were heavy-light it would have been unequivocally seen in the middle
00:56:38.09 which is it not. That tells us that all the newly synthesized DNA is in the recipient locus. The last set of questions I would
00:56:50.02 like to address with you have to do with how accurate is the DNA replication process? This isn't the normal DNA
00:56:56.18 replication process and it isn't even clear that the normal machinery that monitors errors in replication is going
00:57:04.22 to be working. Does mismatch repair work in this context? How does the cell know, for example in this situation, is this
00:57:13.25 an old strand and this a new strand? How does the cell define old and new strands which is necessary for accurate
00:57:20.08 mismatch repair in a situation where you do not have a complete replication fork.
00:57:25.02 To address these questions we wanted to ask how often during this process of switching do we see mutations arising during this process?
00:57:35.00 To do this, and again this is work which is unpublished but may soon be published, we had to trick up the mating type system so that we could
00:57:47.28 follow these events more precisely. To do that we replaced the sequences of the donor locus - HMR - with a copy of a uracil3 gene
00:57:59.04 - a biosynthetic gene - which was derived not from saccharomyces but from a related yeast called kluyveromyces and it turns out as
00:58:06.26 I'll show you this was a piece of incredibly good luck. Maybe we did this because we were a bit lazy because somebody already had
00:58:13.18 these sequences but if we hadn't used them we wouldn't see what I'm going
00:58:17.12 to tell you we can now see. The nice thing about this arrangement is that
00:58:23.08 these sequences are not expressed because they're still subject to the same silencing mechanism that the donors normally have and
00:58:31.26 so the cells are initially uracil minus. Then we turn on HO and the cells switch and when they switch they bring the uracil 3 gene sequences
00:58:41.23 into the MAT locus but now there's no silencing and these cells are ura plus, but of course that meant we could look for mutants
00:58:49.29 that didn't put in ura plus here. They turn out to be resistant to a drug called 5-flurourotic acid and so we can select for uracil minus cells that had
00:59:02.04 arisen during the switch and confirm that these mutants were new - that they didn't preexist in this template - we could do another trick
00:59:13.08 and it turned out that simply by adding nicotinamid to the medium, nicotinamid is an inhibitor of the Sir2 histone deacetylase that I
00:59:22.03 mentioned before and when you do that the silent mating locus becomes not silent and under those circumstances now these sequences as well
00:59:31.24 as these sequences are going to be transcribed and it turns out that these cells become uracil plus again because the donor is not
00:59:39.03 altered it's just the copy of the uracil gene that came in the MAT is altered. It turns out when we do this what we see is about a thousand
00:59:50.16 fold increase in the level of mutations relative to those same sequences undergoing mutagenesis just by replication and I'll just summarize
01:00:00.24 for you what we found so far and I'm just going to then show you a real quick example. The thousand fold increase in the mutation
01:00:10.16 rate is mismatch repair independent, the mismatch repair machinery doesn't seem to know how to work in this context, and it turns out to be both
01:00:19.22 dependent on pol delta or pol epsilon. I mentioned before we would like to know whether pol epsilon mostly worked or pol delta mostly
01:00:28.02 worked and interestingly what we found is that both of them are working even in this context because we have mutations that are error prone
01:00:35.26 and both error prone mutations of pol epsilon or pol delta change the mutation rate in this system. The most important thing I think is
01:00:45.02 that the mutations that we see have a kind of a signature. They're different qualitatively from the mutations that we see in normal DNA
01:00:53.10 replication and their hallmarks are three, or one depending on how you look at it. They're all examples of what we call template jumping or
01:01:04.19 template switching, the simplest one of which is minus one frame shifts in short runs of the same sequence. Then there are more extreme versions
01:01:15.02 which are called quasi-palindrome mutations which I'll show you very briefly, and finally there are astonishing jumps between
01:01:21.28 templates on different chromosomes. Those are the events we never would have seen if we had done this experiments in a different way.
01:01:28.18 You don't need to pay any attention to the details of this. The things above the line are basepair substitutions, so for example here's a C that
01:01:39.13 changes to a T and they did it twice in two independent experiments. Some of these are nonsense mutations so you have what would
01:01:49.00 be a U now a UAG termination of this protein in its translation. The ones below the lines are deletions or frame shifts and there are no increases
01:02:01.28 but you have lots and lots of cases where three Ts becomes two Ts or in this case down here four Cs becomes three Cs. This is a hot
01:02:12.27 spot for some reason because this happened nine times after the first fifty something events that we looked at. There are some other
01:02:20.17 classic examples of - sorry, just to reemphasize if you look in spontaneous mutations, only half of them in homonucleotide runs and they're both
01:02:32.28 frameshifts plus and frameshifts minus. All of these are minus and they essentially all occur with one exception in these homonucleotide runs.
01:02:42.12 There are also more classic frameshift examples in which apparently the DNA polymerase came along, got to this point,
01:02:52.13 fell off, and re-initiates copying at exactly the same sequence but it leaves a 65 basepair deletion.
01:03:00.10 Here's a simple case where there's the same sequence twice: ATC . . . ATC and you end up with one copy of ATC.
01:03:09.14 All of these represent cases where the polymerase essentially
01:03:12.25 has to get out of register and then re-establish register and make some mutations. Then there are some more complicated mutations
01:03:22.15 which involve both basepair substitutions and changes either additions or deletions of the sequence and it turns out that these all can be
01:03:31.13 accounted for by what are known as quasi-palindrome events. They were first seen by Lynn Ripley and have been seen in other cases by
01:03:40.09 Jeffrey Strathern's lab but here what's happening is the newly synthesizing piece of DNA appears to fall off, it self anneals, or else it starts copying
01:03:56.00 from the displaced strand and we can't tell which is which. But in any case, it starts to copy off itself and then it reverses this and
01:04:07.02 goes back to the thing it ought to have been doing and in the course of this it leads to simultaneously basepair substitutions and insertions
01:04:15.10 in the DNA sequence, all of which are templated. They're templated by these quasi-palindromes and again these are very very rare in
01:04:24.04 spontaneous mutations. And then finally as we analyze these sequences we discovered that sometimes the sequences at MAT had sequences
01:04:34.25 here in red that don't come from the Kluyveromyces gene at all. It turns out that they come from the uracil3 gene but it's the uracil3 gene that
01:04:44.06 Saccharomyces normally has and it's located on a completely different chromosome, on chromosome five, and not at HMR locus
01:04:52.21 on chromosome 3. This represents a case where the template - the newly synthesized DNA - is going to dissociate from the template, fall off,
01:05:02.04 and then find a partner which it turns out is only 73% identical at the DNA sequence level, and start copying out sequences from this foreign
01:05:11.01 location, and having done that, it's now going to have to dissociate a second time and go back to the locus to finish
01:05:19.24 because that's the only way we're going to recover this event at MAT. These are really remarkable and unexpected events in which there is dramatic
01:05:27.15 jumping of this newly synthesizing product from one chromosome to another in order to pick up sequences in order to finally finish the
01:05:37.20 DNA repair event and is certainly nothing we have ever seen in spontaneous events. The reason that these events are of some
01:05:47.02 interest is that recently Jim Lupski's lab at Baylor has been very interested in what are called copy number variations in various human diseases
01:05:56.22 and these copy number variations appear to involve cases where the DNA replication machinery stops copying, as it ought to be, along
01:06:05.11 this way, falls off, then goes back and copies some sequences, falls off again, copies some more sequences, falls off again, and copies some
01:06:14.20 other sequences which are a considerable distance from each other. All of the junctions involved in these events have a very small
01:06:23.26 number of basepairs - microhomology - and that's exactly what we saw in the case that I showed you in the MAT case where we're getting
01:06:32.05 again jumps from Kluyveromyces uracil3 to saccharomyces uracil3 using very small numbers of basepairs at the junction in order to jump back
01:06:42.04 and forth and so we're very interested to understand how these events happen because we think we have a model system now to study
01:06:51.03 these microhomology mediated events whether they're break induced replication as Luskpi's lab thinks, or whether they're SDSA events
01:07:00.09 as we imagine they might be is something we'll have to sort out but we now can use this Saccharomyces system as a way to look at these
01:07:08.10 kinds of surprising template jumps that appear to occur during the process of repair. Again just to summarize that part,
01:07:18.12 what we see is that there's a very elevated rate of mutation associated with the DNA repair and that these DNA repair events have a kind of
01:07:27.07 novel signature which tells us that the process of DNA repair is much less processive than the normal process of DNA replication. Finally, I'm
01:07:40.13 going to say a couple of words about how we can extend this system out of the MAT context to ask questions in a broader context and the
01:07:50.16 first thing that we did is simply to take the donor sequences away and use another copy of MAT but on another chromosome as the donor.
01:08:00.23 So now there's no silencing anymore. This doesn't involve HML or HMR it's just another copy of MAT and as I had mentioned in the previous
01:08:09.29 lecture there happened to be a series of restriction site differences between the donor and the recipient so that if repair happens
01:08:18.12 between the break made at MAT and the donor on a different chromosome and if those are then associated with a crossover we will get a novel pair
01:08:27.27 of restriction fragments, one of which is shown here, which tells us that some of the time crossovers are being produced during this
01:08:37.04 process of double strand break repair. As had been shown or inferred from mammalian cells, if you take away the Bloom's protein
01:08:50.25 which is called Sgs1 or Topoisomerase 3 then one should be able to prevent double Holliday junctions from being so called dissolved and there ought
01:09:02.27 to be an increase in crossover products and indeed that's what we see if we take away Sgs1 or Top3 or the double mutant there's a significant
01:09:13.28 increase in these crossover products. What that means, we think, is that Sgs1 and Top3 does some of the time involve a repair mechanism
01:09:25.07 which is not SDSA but the other mechanism that I mentioned, the double Holliday junction repair mechanism because that mechanism
01:09:32.19 can lead to crossovers and those crossovers are defeated if helicase Sgs1 and its associated topoisomerase are dissolving those intermediates
01:09:45.11 by unwinding them so that they are normally found as non-crossover products. If we take away Sgs1 or if we take away Top3
01:09:54.25 now what we see is that there's an increase in the crossovers and that tells us that at least some of the time repair is not happening by SDSA
01:10:04.11 but by some kind of double Holliday junction mechanism. That's what I just showed you but we can go a little further because there
01:10:16.00 are other helicases besides Sgs1 that play a role in repair and so we've looked at all of these again a lot of this work was done either by
01:10:27.27 Greg Ira when he was in my lab or subsequently when he was in his lab at Baylor. Here what we see is that there are two other helicases
01:10:37.28 which also change the ratio of crossovers to non-crossovers. A helicase known as Mph1 if we knock it out also has the effect of dramatically
01:10:48.12 increasing the level of crossovers and if we make a double mutant with Sgs1 and Mph1 the crossover level actually goes up even higher.
01:10:57.17 Then there's another of these proteins called Srs2 which also appears to change the ratio between crossovers and non-crossovers but in fact
01:11:08.09 with Srs2 what we discovered is it's kind of a misleading conclusion because Srs2 actually doesn't increase the number of products
01:11:19.08 that are actually crossovers but it destroys in ways that we don't understand - the absence of Srs2 destroys the ability to recover
01:11:28.07 many of the non-crossover products from the cell. So the total efficiency of repair goes way down the fraction of those events which are associated
01:11:37.27 with crossovers stays constant and therefore there's apparently an increase in crossing over but it's really for a different reason.
01:11:45.05 If we take all of these conclusions together we can start to get a picture of what these three helicases are doing and give us a much better
01:11:55.20 picture really of what's happening when the cell suffers a double strand break. What we think now is that there's a double strand break,
01:12:02.23 it starts to be acted upon by the exonucleases and even by Rad51 and it starts to do some kind of strand exchange process but there
01:12:13.23 has to be a decision: is it going to go down the double Holliday junction pathway, or is it going to go down this SDSA pathway.
01:12:23.11 It appears that Mph1 plays some critical role in deciding that this will be an unstable
01:12:30.13 D-loop where you're going to unwind the newly synthesized DNA
01:12:34.19 as it goes and that this is going to drive things into this pathway. When you take away Mph1 things go down this pathway and crossovers increase.
01:12:45.11 If you knock out Sgs1 things are already going down this pathway but now they can't be resolved as non-crossovers and
01:12:53.05 crossovers increase even more but if you knock away Srs2 the cells are committed to this pathway - most of them - those that go down
01:13:03.18 this pathway are going to still lead to crossovers but now the guys
01:13:07.07 that start to go down this pathway mostly just die and we don't know why they die.
01:13:12.11 Somewhere along this process they failed to complete the process and we don't recover them as repair products
01:13:21.10 but we don't know yet how that happens. Lastly, there's a competition between
01:13:32.06 two different gene conversion mechanisms, between the double Holliday junction
01:13:39.10 mechanism and SDSA now I'm going to tell you that there's finally a competition between gene conversion mechanisms
01:13:46.04 and break induced replication. Really all of these things are happening in the cell at the same time and the cell needs to have a way of deciding
01:13:54.02 which of these things it ought to be doing at any given moment.
01:13:56.28 So the difference between gene conversion and break induced replication clearly is that
01:14:02.16 in this case both ends engage the same template and do a little patch
01:14:06.10 of new DNA synthesis whereas in break induced replication
01:14:09.28 one end engages and starts to synthesize and the other end for whatever reason doesn't get involved. We discovered,
01:14:19.26 and this will become important in how I can show you the difference, that these events, the events of break induced replication depend on the
01:14:29.08 nonessential pol delta subunit called Pol32. These events, the events in SDSA, do not. Suvi Jain who was a graduate student in the lab decided
01:14:40.12 to set up a test of how the cell really knows the difference between doing gene conversion and break induced replication and the basic guts
01:14:50.18 of this experiment are shown here. She set up a break inside a gene called Leu2 on chromosome five and she provided a template
01:15:00.26 for that as another Leu2 gene on chromosome three and so when the break is made it can use this as the template to do gene conversion and to repair
01:15:11.27 this break and that doesn't change the arrangement of all the markers in this chromosome. The extreme opposite version
01:15:19.19 of this is that after the break is made it can only find homology to one side of the break and so now the cell has no way to do gene conversion
01:15:29.24 it can only do break induced replication and as a consequence you copy all the sequences on the template from here all the way to the end
01:15:39.13 and you also lose all the sequences that were on the other end of the broken chromosome because there's no place for them to go and we
01:15:47.07 obviously set this up in a way where those sequences were dispensable. At the end of break induced replication you have two copies of these
01:15:55.06 sequences and no copy of these sequences whereas in gene conversion there's one copy of each and there's been a little patch
01:16:02.12 of new DNA synthesis. Suvi asked a very interesting question which is at what point does the strand invasion machinery lose sight of the fact
01:16:12.26 that the two ends are both trying to work on the same template. Here . . . is a case where the template ends are very close together
01:16:28.17 and so she started pushing the ends further and further apart by adding more DNA sequences
01:16:33.19 in between the LE and the U2 sequences and asked how would that effect the way in which these cells would repair?
01:16:41.05 The first thing she found out was that the viability kept going down
01:16:45.14 towards the level that we see with break induced replication as these sequences became further and further apart and the second thing
01:16:53.06 that she found was that increasingly the results of this repair were that the sequences
01:17:00.28 over here which are marked by hygromycin resistance are lost.
01:17:04.24 Those are the sequences we can lose and they are lost because the cells increasingly are going from doing gene conversion
01:17:11.24 repair towards doing break induced replication. When Suvi then looked at this more by doing kinetics by looking at when was new synthesis
01:17:22.24 starting and again this is one of those assays in which she's not looking at the completion of the process but only when do you see the first
01:17:31.06 50 nucleotides of new synthesis and so there's a PCR primer in this little space, there's a PCR primer on the other side of LE and once they
01:17:41.09 join together you have strand invasion and the beginning of new DNA synthesis, that's what's being measured at the bottom.
01:17:47.18 The results are amazingly striking . . . even when there's a 1.2kb gap between these sequences, the kinetics of this process are quite rapid
01:18:01.21 and very complete, but if you push the sequences further apart, the kinetics change completely. It's not merely that the efficiency goes down,
01:18:11.03 they are slow to start and they have a completely different kind of kinetics which suggests that they have changed their mechanism.
01:18:18.11 We know that this slow kinetic process is characteristic of break induced replication and to show this even further, Suvi knocked
01:18:27.01 out the Pol32 gene in these constructs and what you see is by the time you get to 2.5kb
01:18:33.26 or even 1.2kb, these events become increasingly and then finally
01:18:40.13 completely dependent on Pol32 which we take to be an indication that at this point the cells are mostly doing gene conversion using the SDSA
01:18:51.07 mechanism but as the gap between these ends gets further and further, there's some kind of loss of understanding
01:18:58.17 of the two ends relative to each other and each end then begins to behave as if it was the only end and each end starts to do break induced replication
01:19:07.22 which is a slow and Pol32 dependent process. That has led Suvi to the idea that there is something called a
01:19:16.27 Recombination Execution Checkpoint, that the cell actually has a mechanism to decide whether
01:19:23.17 or not the two ends of strand invasion are both happening on the same
01:19:28.15 template close enough together, and I haven't shown you this, in the proper orientation. She actually flipped around some sequences
01:19:36.08 and showed that they could invade but they were pointing in the wrong way. The cell doesn't know that this is appropriate for doing gene conversion.
01:19:45.20 When the ends are close together and engage the same template, everything goes relatively quickly. When they are far apart,
01:19:52.27 it's as if each end is completely independent. There's no communication between the two ends and the repair goes by this other mechanism.
01:20:02.11 We think this must have something to do with signaling by virtue of how you enlarge this invasion loop so that the
01:20:11.08 other end can be engaged from the same structure but we haven't proven that at this point.
01:20:16.11 Why is this important? We think this is important in order for the cell to avoid some
01:20:24.07 of those chromosome rearrangements that are characteristic of cells where bad things happen.
01:20:30.22 The idea would be that if you have a break inside a repeated sequence, each of these two ends independently could go looking
01:20:40.09 for a partner and if this end goes to this partner, but this end goes to this partner, they're going to start doing two break induced replication
01:20:50.01 events which would lead to lots of chromosomal instability and chromosomal rearrangements. The cell has in fact built in a checkpoint,
01:20:58.20 a way of slowing down this process to say "are you sure that it is impossible to do invasion so that both ends are on the same
01:21:06.06 template whereupon the rest of the event takes place in a relatively timely fashion?" This recombination checkpoint is a way of preventing
01:21:15.07 break induced replication when it's a bad solution to an otherwise soluble problem. But of course, if there isn't a solution to the problem,
01:21:24.08 doing break induced replication is better than doing nothing and that's what the cell then does. That's kind of a summary of where we are in
01:21:35.00 terms of knowing about one specific double strand break and how it's repaired. Breaks made by HO. But there are lots of questions,
01:21:42.27 some of which are illustrated here. We really don't understand how the searching by the Rad51 filament takes place. What happens
01:21:50.23 inside that little filament in order to do this strand exchange? How do they line up properly to find those sequences? For example, if the Rad51
01:22:01.20 filament bound to DNA and stayed tightly bound to the DNA and just scanned it up and down it would miss all the homologies that were
01:22:08.09 in the opposite orientation. It has to be able to get on, search, fail, fall off, get on in another orientation to try again, and we don't know anything
01:22:17.29 about how that process is really taking place. We really don't understand why it takes so long for the replication machinery
01:22:26.09 to assemble at the points where new DNA synthesis is going to take place. In gene conversion it takes twenty minutes. Apparently in break induced
01:22:35.23 replication it takes several hours and the question is "what's it doing during that time and why is it so slow?" We really don't know the
01:22:43.07 answer to that. We don't understand yet at all how strand invasion can take place in the context of positioned nucleosomes but obviously
01:22:53.18 it does and even more mysterious is to understand after DNA repair is over, how is chromatin re-established, how are nucleosomes
01:23:03.28 put back in the places where DNA repair has taken place? That's another of these mysteries that we really haven't been able to solve although
01:23:14.24 I will say we have a recent paper that's published that argues that the chromatin chaperones AsfI and CafI play an important role in
01:23:25.09 re-establishing this chromatin but the details we don't know. Then there's the question of if the replication doesn't require the MCM helicases,
01:23:35.23 how is the DNA being opened up at all for repair? There must be some other helicase but we don't know what it is. And finally, if all of
01:23:45.09 this is taking place while the cell is sending out alarms and saying "there's damage here" and stopping the cells from going through the cell cycle
01:23:54.02 which is the case, how does that signaling get turned off at the end of the process so that the cells can resume cell cycle progression
01:24:02.10 and we know that it's not simply enough to have completed pieces of DNA. There's a real mechanism that turns off the DNA damage
01:24:12.01 checkpoint but that will be a subject for another lecture and this lecture, I think, has come to an end. Thank you very much.

Videos in this Talk
  • Part 1: Mechanisms of DNA Repair by Recombination
    Part 1: Mechanisms of DNA Repair by Recombination
    Audience:
    • Student
    • Researcher
    • Educators of H. School / Intro Undergrad
    • Educators of Adv. Undergrad / Grad
  • Part 2: Details of DNA Repair in Budding Yeast
    Part 2: Details of DNA Repair in Budding Yeast
    Audience:
    • Researcher
    • Educators of Adv. Undergrad / Grad
Total Duration: 2:26:12
Recorded: October 2009
All Talks in Genetics and Gene Regulation

Talk Overview

Part 1. The most deleterious form of DNA damage is a double-strand break (DSB), which can arise from errors in DNA replication, from the failure of topoisomerases to complete their cycles of DNA cutting and rejoining, from mechanical stress and from the action of endonucleases that cleave DNA. Here we review mechanisms of DNA repair, specifically how DSBs can be repaired either by nonhomologous end-joining mechanisms or by several homologous recombination pathways including single-strand annealing, gene conversion and break-induced replication.

Part 2.  A specific set of examples of homologous recombination in budding yeast is presented based primarily on data from the Haber lab.  Here a site-specific HO endonuclease can be induced to create a single DSB in the genome and the kinetics of repair can be followed in real time.  The roles of key recombination proteins and the different types of outcomes provoked by this DSB are reviewed.

Playlist: DNA Replication and Repair

Related Resources

Part 1
Pâques, F., and Haber, J.E. (1999). Multiple pathways of recombination induced by double-strand breaks in Saccharomyces cerevisiae. Microbiol Mol Biol Rev 63, 349-404.

Haber, J.E. (2006). Transpositions and translocations induced by site-specific double-strand breaks in budding yeast. DNA repair 5, 998-1009.

Aylon, Y., and Kupiec, M. (2004). New insights into the mechanism of homologous recombination in yeast. Mutat Res 566, 231-248.

Krogh, B.O., and Symington, L.S. (2004). Recombination proteins in yeast. Annu Rev Genet 38, 233-271.

Weinstock, D.M., Richardson, C.A., Elliott, B., and Jasin, M. (2006). Modeling oncogenic translocations: distinct roles for double-strand break repair pathways in translocation formation in mammalian cells. DNA Repair (Amst) 5, 1065-1074.

Part 2
Ira, G., and Haber, J.E. (2002). Characterization of RAD51-independent break-induced replication that acts preferentially with short homologous sequences. Mol Cell Biol 22, 6384-6392.

Ira, G., Malkova, A., Liberi, G., Foiani, M., and Haber, J.E. (2003). Srs2 and Sgs1-Top3 suppress crossovers during double-strand break repair in yeast. Cell 115, 401-411.

Krogh, B.O., and Symington, L.S. (2004). Recombination proteins in yeast. Annu Rev Genet 38, 233-271.

Sugawara, N., Wang, X., and Haber, J.E. (2003). In vivo roles of Rad52, Rad54, and Rad55 proteins in Rad51-mediated recombination. Mol Cell 12, 209-219.

Wang, X., and Haber, J.E. (2004). Role of Saccharomyces single-stranded DNA-binding protein RPA in the strand invasion step of double-strand break repair. PLoS Biol 2, 104-111.

Wang, X., Ira, G., Tercero, J.A., Holmes, A.M., Diffley, J.F., and Haber, J.E. (2004). Role of DNA replication proteins in double-strand break-induced recombination in Saccharomyces cerevisiae. Mol Cell Biol 24, 6891-6899.

White, C.I., and Haber, J.E. (1990). Intermediates of recombination during mating type switching inSaccharomyces cerevisiae. EMBO J 9, 663-673.

Wolner, B., van Komen, S., Sung, P., and Peterson, C.L. (2003). Recruitment of the recombinational repair machinery to a DNA double-strand break in yeast. Mol Cell 12, 221-232.

Reader Interactions

Comments

  1. health news says

    I’m now not positive where you’re getting your information, but great
    topic. I needs to spend some time learning more or working out more.
    Thanks for magnificent information I was searching for
    this info for my mission.

    Reply

Leave a Reply

Your email address will not be published. Required fields are marked *

iBiology and iBiology Courses are part of the Science Communication Lab (SCL). Our mission remains the same, to connect people to science. However, our focus has shifted to producing and evaluating cinematic films for education and the public, which you can find on the Science Communication Lab website. For more information, please see this blog post!