Mechanisms of DNA Repair
Transcript of Part 2: Details of DNA Repair in Budding Yeast
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.