L12-HOMOLOGOUS RECOMBINATION AT THE MOLECULAR LEVEL

Beginning of NCO/CO pathway

*1. Spo11 makes a DSB (double-strand break)* in DNA. In both DNA strands, Spo11 breaks one of the phosphodiesther bonds in the DNA backbone by covalently binding to the 5'-end and releasing a free 3'-end. Spo11 is the protein that initiates homologous recombination in meiosis. Spo11 is expressed only in the cells that are going through meiosis. It is tightly repressed in somatic cells: somatic cells don't want to induce recombination! Recombination can be DAMAGING in somatic cells, because it can lead to LOH (loss of heterozygosity) that results in expression of recessive traits in some cells of a heterozygous individual). Spo11 binds DNA randomly, NOT sequence-specifically. Thus, DSBs are randomly distributed in the genome: locations where they occur in one cell are different from locations where they occur in another cell. But, there are places in the genome, where Spo11 binds better: those are recombination hotspots (very loosely packed regions of chromosomes); there are also places in the genome where Spo11 doesn't bind very well: those are recombination coldspots (very tightly packed regions of chromosomes). *2. Resection of 5'-ends to make 3'-overhangs*. Resection of the ends involves two enzymes. First, *ENDOnuclease* must cleave a short end piece of DNA with Spo11 covalently attached to it. Then, a free DNA end is formed, and *EXOnuclease* can come in and resect the 5'- end further. Both endonuclease and exonuclease are enzymes that "cut" DNA (cut = hydrolyze phosphodiester bond). Endonuclease cuts inside the chain (endo = inside); it doesn't need a free end to make a cut. Exonuclease cuts nucleotides one by one from the end (exo = outside); it won't cut DNA if there is no free end present. The difference between endo- and exo- nucleases is based on their interaction with DNA: endonuclease attaches to the continuous chain of DNA; exonuclease "bites off" nucleotides from the end one by one. You should understand the distinction between endonuclease and exonuclease. *3. Invasion of one 3'-overhang into the homologous (non-sister) chromatid, which results in formation of D-loop.* -Dmc1-coated single-stranded DNA tail searches the entire nucleus for the one and only homologous sequence. This is how in most organisms (including yeast and humans), homologous chromosomes find each other and eventually pair to form bivalents in prophase of meiosis I. -RPA binds single-stranded DNA and stabilizes it (prevents formation of secondary structures, such as hairpins). Outside of meiosis, RPA in eukaryotes has the same function as SSB in prokaryotes: it is involved in DNA replication (RPA = replication protein A). -You should know the NAMES and ROLES in homologous recombination of the following proteins: Spo11, endonuclease, exonuclease, Dmc1, RPA, DNA polymerase, DNA ligase, mismatch repair system (see further slides). While other proteins are also involved in coordinating the incredibly complicated process of homologous recombination, you don't need to know their names for the purpose of this class. *4. Extension of the invading 3'-overhang*. DNA polymerase extends the invading strand using homologous chromosome as the template: DNA synthesis replaces the DNA degraded during formation of the 3' tails. Since the template is the "red" molecule, "red" sequence will be added to the "blue" sequence. Remember that red and blue colors represent slightly different DNA sequences due to homologous chromosomes having different alleles for the same genes.

Homologous recombination at a molecular level

Purpose: 1. Genetic diversity 2. Formation of bivalents in MI (--> segregation of homologs) Homologous recombination in meiosis serves two purposes: 1). *To create new combinations of alleles for linked genes via crossing over (when two homologous chromosomes exchange portions of DNA)*. Creating new combinations of alleles is beneficial for a population's survival: it increases the genetic diversity of a population, thereby increasing the chances that some individuals in the population have the right combination of alleles that would allow them to survive and reproduce in a changing environment. Recall that new combinations of alleles arise via two mechanisms in meiosis: (1) independent assortment of non-homologous chromosomes, and (2) crossing over between homologous chromosomes. 2). *To ensure proper segregation of homologous chromosomes during meiosis I*. Recombination between homologous chromosomes is what brings the two homologous chromosomes together during meiosis; it is what allows them to pair and form a bivalent. Formation of bivalent is necessary for proper segregation of homologs in meiosis I. Without recombination, bivalents don't form and therefore mis-segregation (nondisjunction) of chromosomes can occur. Homologous recombination occurs in meiosis I, when each chromosome consists of two sister chromatids. Sister chromatids are products of replication of one chromosome and therefore are identical. Homologous chromosomes carry the same genes, but can carry different alleles for those genes. After crossing over occurs, two chromatids are recombinant (those that engaged in crossing over) and two are parental (those that did not engage in crossing over).

End result: crossing over of the flanking loci, with or without gene conversion at the middle locus

*11. Correction of the heteroduplex by the mismatch repair enzymes.* -Similarly to the NCO pathway, correction of heteroduplex occurs randomly: mismatch repair system either corrects the "red" sequence towards the "blue" sequence or the other way around, with equal probability. *End result*: crossing over (with or without gene conversion in the middle). -Again, note, that we are only looking at a small region in the chromosomes. In our drawings, we are zooming in on the area where recombination is occurring (around gene B). -Notice that there are two regions of heteroduplex in each of the recombinant chromatids. What if in a given chromatid, one heteroduplex gets corrected towards the "red" sequence and the other one towards the "blue" sequence? Will it produce allele B, allele b or a third, new allele? The answer depends on what is the difference between alleles B and b. Alleles B and b are two versions of the same gene. In the simplest scenario, there is one base pair difference between B and b: for example, allele B codes for a normal, functional protein B and allele b has one mutation that changes a normal codon into a premature stop codon, and therefore it does not produce functional protein B. There is one base pair difference between alleles B and b: everything else besides that one base pair is identical. Now, think about the two heteroduplexes: if this one base pair that is different will end up in one of the heteroduplexes, it will matter how that heteroduplex will get corrected. The other heteroduplex won't have any mismatches, so technically it won't even be a heteroduplex: it will be a perfectly complementary double-stranded DNA. This is the most likely scenario. -If there are TWO differences in the DNA sequences of alleles B and b, and one of them is in one heteroduplex, but another one is on the other heteroduplex, then it is possible to produce a new allele, which will be a hybrid allele between B and b. This is a very unlikely scenario though.

*NCO (noncrossover) pathway* -SDSA (synthesis-dependent strand annealing)

*5. Return of the extended 3'-end back to its original chromatid.* -This is the point where the interference signal acts and says: "We already have a CO, we don't need another one, so you go back." -The invading strand leaves the homolog that it invaded and anneals back to its old complementary strand. "Annealing" means the same as renaturation: when one DNA strand associates with a complementary strand to form double helix. *6. DNA synthesis to fill in the gap.* DNA polymerase synthesizes the missing piece. Because the template is "blue," the "blue" sequence will be synthesized. *7. Ligation.* DNA ligase joins the ends, restoring strand continuity.

Heteroduplex is corrected by a mismatch pair of enzymes

*8. Correction of the heteroduplex by the mismatch repair enzymes.* -*Heteroduplex* is the region in DNA, where the sequences of the two strands don't match perfectly: in our example, one strand is "blue" and the other one is "red." In reality, it means that the sequences of the two homologous chromosomes are slightly different (one has allele A and the other one has allele a). The heteroduplex shown here contains one *mismatch*. -*Mismatch repair system* is a group of proteins in the cell that corrects mismatches in DNA. Here, the mismatch occurred, because the "red" sequence was added to the "blue" 3'-end using the "red" homolog as the template, and then it went back and annealed with the "blue" sequence. So, the mismatch is just a consequence of the two homologs having slightly different sequences. The mismatch repair system would detect this mismatch (because it interrupts the perfect shape of the double helix) and make a random call: with equal probability, it will either correct the "blue" strand to become "red," or it will correct the "red" strand to become "blue." The probabilities of getting either allele are equal, because the two strands are "equal" - one strand is not better than the other strand in any way, so mismatch repair proteins choose randomly which strand to keep and which one to correct. -(Mismatches can also occur in DNA during replication, when DNA polymerase makes a mistake and fails to correct it by proofreading. In this case, correction is not random; it is directional: mismatch repair always corrects the mismatch towards the old strand, thus fixing the error. We will talk about it later in the semester). -Correction of heteroduplex produces either "blue" (a) or "red" (A) allele with equal probability.

End result of noncrossover

-*End result of the NCO (non-crossover) pathway*: Depending on how the heteroduplex has been corrected, there will either be no genetically visible consequences of recombination, or gene conversion. -Gene conversion is deviation from 2:2 segregation of alleles in meiosis. In this case, one recombination event that resulted into a noncrossover, produced *3A : 1a* segregation of alleles. -Mendel's first Law (the Law of segregation) says: "Each individual possesses two alleles encoding a characteristic, which segregate during gamete formation *in equal proportions*. Two gametes, one from each parent, unite at random at fertilization." Gene conversion breaks Mendel's first Law: the alleles don't segregate in equal proportions! -The NCO pathway is *non-reciprocal*: only one chromatid gets broken by Spo11. The other chromatid is unchanged: it is only used as template for DNA synthesis. -Question: Can you explain why this logic produced 3A : 1a as opposed to 3a : 1A ? What would you change to produce 3a : 1A ? --> Answer: To produce 3a : 1A, the DSB must occur in the "red" chromatid (with the A allele), and the heteroduplex must be corrected towards the "blue" sequence (the a allele).

Tetrad analysis in S. cerevisiae (budding yeast)

-A lot of what we discussed today was discovered in the budding yeast (aka baker's yeast) S. cerevisiae, because yeast is a fantastic model organism for genetics. The two main reasons are (1) yeast can stably exist either as a haploid organism or as a diploid organism and (2) after meiosis, the four haploid spores remain together in a tetrad, which allows us to see what happened in that particular meiosis. Tetrad analysis*: • Take two haploid yeast of known genotypes and mate together to make a diploid. • A population of diploid yeast cells of identical genotype goes through meiosis. • Each cell forms a tetrad = four haploid spores that remain together in a special sac. • We can carefully take tetrads, one by one, and dissect each tetrad under the microscope, freeing the spores from the sac using a tiny needle. • We can place each haploid spore separately on the surface of agar in a Petri dish, laying 4 spores from one tetrad one below the other. • Each haploid spore will divide by mitosis and produce a colony of haploid cells, so we will get 4 colonies that represent 4 spores in that tetrad. • We can examine phenotype of each colony, and since the cells are haploid, we will immediately know their genotype. • By examining one tetrad, we know exactly what happened in meiosis in that one diploid cell. • By examining multiple tetrads, we know what happened in meiosis in each of those diploid cells. -This technique is more powerful that population-averaged analysis, where we cannot tell which gametes came from one meiosis. -If no recombination events occurred in the region between A and C, all four spores in the tetrad will have parental combinations of alleles for these three genes. (Note that we are looking here at a SMALL REGION on the chromosome; even though no recombination events occurred in this particular region, there must have been a crossing over event somewhere else on that chromosome to ensure formation of bivalent and proper segregation in meiosis).

Detection of crossing over in yeast tetrads

-By analyzing the genotypes of the four haploid spores in a tetrad, we can infer what happened in that particular meiosis. -We can detect crossing over and infer its location. Here, on this slide, crossing over occurred either in the region between genes A and B (top) or in the region between genes B and C. -There is no gene conversion here (all alleles segregate 2:2). -We can detect gene conversion as a deviation from the 2:2 segregation of alleles. -Gene conversion can occur without crossing over (in the NCO pathway of homologous recombination = top) or with crossing over (in the CO pathway of homologous recombination = bottom). -Notice the *AbC* spore in the top scenario (the third one from the top): if we only saw this one spore, without seeing the rest of the spores from the same tetrad, we would have probably said that this is a double-crossover event. But crossing over is always a RECIPROCAL exchange, so if this was a double-crossover, we would also expect to find s spore with the aBc genotype in the same tetrad - but we don't! Instead, we see that the AbC chromatid appeared during NON-RECIPROCAL transfer of genetic information. This is how we know that this AbC chromatid resulted from heteroduplex correction during a noncrossover event, rather then from a double crossover. This is why tetrad analysis in yeast is such a powerful technique: we don't just see individual gametes: we see all four product of one meiosis, so we get to know the context of how a particular chromatid was generated!

... with or without gene conversion of the middle locus

-Gene conversion can occur in the CO pathway. -We are looking at four chromatids here. The top chromatid and the bottom chromatid are non-recombinant, so the top chromatid always has the B allele and the bottom chromatid always has the b allele. The two "middle" chromatids engaged in crossing over (see that one chromatid begins as red and ends as blue, and the other chromatid begins as blue and ends with red - this is the result of crossing over). -Depending on how heteroduplex is corrected in those two middle chromatids that engaged in crossing over, there might be different distribution of alleles B and b. -*Gene conversion is deviation from 2:2 segregation of alleles in meiosis*. Here, we can either end up with 3B : 1b, or with 3b : 1B depending on how correction of heteroduplex goes in the two recombinant chromatids. (We can also end up without gene conversion, 2B : 2b). -When we talk about 3 loci (3 genes: A, B, C), we refer to them as the *middle locus* and the *flanking loci*.

NCO/CO possible outcomes

-Homologous recombination is a process of repairing a DSB (double-strand break) in DNA using homologous DNA sequence (found on the homologous chromosome). A particular double-strand break can be repaired either by a noncrossover pathway, or by a crossover pathway. Both are considered to be homologous recombination events. Before, we used to think that homologous recombination = crossover. Now, we know that homologous recombination = crossover or noncrossover. -Both pathways can create recombinant combination of alleles.

Logic: many DSBs --> CO and NCO pathways

-In meiosis, the cell intentionally starts homologous recombination by making DSBs (double-strand breaks) in DNA, because CO (crossing over) is necessary in meiosis I. DSBs occur randomly, throughout the genome. *The cell makes MORE DSBs than it needs to make COs* - probably, because it is better to make more than not enough. Therefore, not all DSBs become COs. Some DSBs develop into COs and the rest develops into something called NCO (non-crossover). We will discuss both pathways of homologous recombination in meiosis: CO and NCO pathways. Both pathways begin with a DSB, but the result is different. -Who decides if a particular DSB will develop into a CO or into a NCO? The exact answer to this question is still unknown, but there is a hypothesis that makes a lot of sense. Recall that when we talked about meiosis, we discussed how each bivalent must have at least one CO (this is called obligatory CO) - otherwise, bivalent won't form. Also, recall that when we discussed three-point cross, we talked about *crossover interference*. Crossover interference is observed when occurrence of a CO in one region of the chromosome inhibits formation of another CO nearby. This means that the DSBs nearby develop into NCO (because once they were made, they must be repaired). The hypothesis proposes that the *first* DSB that successfully establishes the contact with the homolog becomes designated into CO by default. CO designation sends an inhibitory bidirectional signal, thus making other DSBs on the same chromosome develop into NCO. The inhibitory signal fades over distance, so DSBs located far away have a chance to develop into CO. The exact nature of the signal is unknown. The beauty of this hypothesis is that obligatory CO and crossover interference are intrinsically connected: one CO *must* occur per each bivalent (obligatory CO) and the remaining DSBs on that chromosome will develop into NCO by way of interference signal. -Up until now, we were using the terms recombination and crossing over interchangeably. But, they are not the same. -*Homologous recombination* is the process of repairing a DSB (double-strand break in DNA) using homologous DNA sequence (found on the homologous chromosome). Sometimes, it results in a crossing over event (CO); sometimes, it doesn't (=NCO), but in both cases, we consider the process to be homologous recombination. In CO, two chromatids "swap" pieces. In NCO, sometimes we can see non-reciprocal transfer of information between chromatids (like a "patch" of the red sequence within the blue chromosome shown here); other times, we don't see any consequences in the NCO pathway. Today, will talk about molecular mechanisms of these two pathways of homologous recombination in meiosis: the CO and the NCO pathways. -The term "recombination" has a genetic meaning and a molecular meaning. From a *genetic perspective*, recombinant means "not parental." For example, even if genes A and B are on two different chromosomes, in a cross AA BB x aa bb --> Aa Bb, parental combinations are AB and ab, and recombinant combinations are Ab and aB. From a *molecular perspective*, homologous recombination occurs when two DNA molecules engage in an act of repairing a double-strand break using homologous sequence as the template (even though there might be no observable genetic consequences, as we will see in this lecture). -Note that in the next slides, we will only be looking at *TWO* chromatids (those that are involved in homologous recombination). *Remember, that there are four chromatids in a bivalent, we are just not showing those two that are not involved in the recombination event that we are considering*. -Homologous chromosomes are colored in red/blue: homologous chromosomes contain identical genes, but may have different alleles for those genes. Different alleles means slightly different DNA sequences. Thus, red and blue colors represent slightly different DNA sequences. -We will examine in detail what is happening at the very location of the recombination event. Keep in mind that we are looking at A TINY REGION on the chromosome.

Post-meiotic segregation confirmed the existence of heteroduplex DNA

-Left tetrad = normal, Mendelian 2:2 segregation of alleles = normal. -Middle tetrad = 3:1 segregation of alleles = gene conversion = rare. (Note that gene conversion is shown here as part of the NCO pathway; it can also occur during the CO pathway.) -Right tetrad = post-meiotic segregation = super rare! -Post-meiotic segregation is a result of uncorrected heteroduplex in one of the haploid cells. Postmeiotic segregation is a very rare event: normally, heteroduplex is corrected before the cells proceeds with meiosis. Very rarely, heteroduplex remains uncorrected and the cell goes through meiosis. In this case, one spore (haploid gamete) will inherit a chromosome with a region of heteroduplex in it. After the first round of replication, one daughter cell will have one allele and another daughter cell will have another allele. Therefore, half of the colony will have one phenotype and half will have the other phenotype.

*CO (crossover) pathway* occurs via double Holiday junction

-Let's go back to step #4, after which the NCO and CO pathways separate (extension of the invading 3'-end). *5. Annealing of the second 3'-overhang to the displaced strand in the D-loop.* *6. Extension of the second 3'-overhang.* DNA polymerase extends the second 3'-end. Once again, because the template for DNA synthesis is "red," we are extending the "blue" strand with the "red" sequence. *7. Ligation.* DNA ligase joins the ends, thus restoring the continuity of all strands. Black dotted line denotes the process of ligation (there is no new DNA added; red and blue strands are brought together and covalently joined). *8. Branch migration.* The two Holliday junctions are pushed away from the center, towards the sides. *9. Resolution of the double Holliday junction (4 cuts). * -The two Holliday junctions are always resolved (cut) in a "perpendicular" way, such that *each of the four DNA strands is cut once.* To ensure that a double Holliday junction is always cut in a "perpendicular" way, cutting (and re-ligation) of all four strands occurs simultaneously, in coordination. This can be done in two alternative ways (top and bottom fogures); the outcome will be the same. Black arrows show positions of cuts: each of the four strands is cut once. -This is done to make sure that the outcome of the double Holliday junction is always a crossover. It is important, because once a particular DSB has "committed" to become a CO, it MUST become a CO. Otherwise, there is a risk of not having any CO events per given pair of homologous chromosomes, which would be bad, since the bivalent wouldn't form in the absence of CO. -Imagine an alternative scenario: if the two Holliday junction were both cut "horizontally," such that the two "inner" strands would get two cuts each, and the two "outer" strands would get zero cuts. The outcome would be a noncrossover (the LEFT RED end of the chromosome would stay with the RIGHT RED end of the chromosome; the BLUE LEFT end of the chromosome would stay with the BLUE RIGHT end of the chromosome). *10. Ligation.* DNA ligase joins the ends, thus restoring the continuity of all strands. -After each strand is cut, strand are rejoined by the DNA ligase IN THE NEW WAY. -Look at the left Holliday junction in the top figure: the red strand and the blue strand each get a cut (shown by black arrows). Then, the cut strands are rejoined in a new way: the red with the blue, and the blue with the red (look at the result of resolving the left Holliday junction in the bottom figure). Same exact logic applies to the right Holliday junction: black arrows show positions of cuts; after cutting, red strand is joined with blue strand, and blue strand is joined with red strand (look at the result of resolving the right Holliday junction in the bottom figure). -Look at the bottom figure: this is a CROSSOVER: --> top chromatid: left side = red; right side = blue; --> bottom chromatid: left side = blue; right side = red. -Remember, we are looking here at a very small region in a very long chromosome: in the top chromatid, everything towards the left will be red and everything towards the right will be blue; in the bottom chromatid, everything towards the left will be blue and everything towards the right will be red. Areas in the middle of both chromatids contain heteroduplex DNA, which must be corrected.

Two outcomes are possible

1. *NCO (noncrossover)* --> SDSA (synthesis-dependent strand annealing) 2. *CO (crossover)* --> DHJ (double Holiday junction) Up to this point, both pathways progress identically. Next, CO and NCO pathways proceed differently. -We will talk about the NCO pathway first, because it is simpler. -Then, we will talk about the CO pathway, which is more complicated.