Mechanisms of DNA Repair
Transcript of Part 1: Mechanisms of DNA Repair by Recombination
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.