Exam 2 Microbiology

Eukaryotic Meiosis

- During Meiosis I, sister chromatids are linked by cohesions. Homologous chromosomes are linked by homologous recombination - homologous recombination is most prevalent in meiosis 1. In the diploid cell, there are 2 sets of homologous chromosomes. During S phase, the homologous chromosomes align. 2. The chromosomes replicate to generate sister chromatids 3. The sister sister chromatids are linked by cohesion. What aligns the chromosomes is homologous recombination. Homologous recombination can entangle two duplex DNA. 4. Homologous recombination generates crossovers and chromosomal exchange. 5. During Meiosis I the chromosomes will have different parts of the other chromosomes integrated into them. This generates genetic diversity 6. In the second round of meiosis, meiosis II, there is separation of the sister chromatids at the centromere.

RecA recombinase

- E. coli only has one RecA recombinase - Conserved from bacteria to humans - Promote strand invasion - Filaments - ATP is hydrolyzed - helps strand invasion to the other template strand 1. RecA binds to single stranded DNA 2. It strats by binding to the DNA by nucleation which is very slow. 3. It begins to fill in the single stranded DNA very fast with many RecA compelxes. It extends from the 5' to 3' while disassembling the opposite side. 4. Rec A forms a filament on the single stranded DNA and Rec A is required for the filament to grow

RecA Mediated Strand Exchange in Vitro

- Experiment that shows that RecA is able to bind to many types of single stranded DNA. - even circular DNA binds by RecA - In the experiment, scientists added double stranded DNA, linear DNA and circular DNA. Also added to purify the RecA ATP. RecA binds to the single stranded circular DNA initiating strand invasion. The branch will go through branch migration, displacing one of the strands, creating circular double stranded DNA - For the first 5 experiments, they added dysfunctional ATP. - In the presence of ATP a new species of DNA is formed in response to the duplex circular DNA. You also see the disappearance of the double stranded, linear DNA. - this experiment shows that RecA is sufficient enough to catalyze the recombination reaction

The lox/Cre System in P1 Phage Step 1

- Lox/Cre converts the linear phage DNA into a circular one-upon infection 1. The p1 phage contains 2 loxP sites which are aligned in the same direction. Once injected into the host, the two recombination sites align and generate a circular DNA - Why circular DNA? The circular DNA can go through the rolling circle replication

RuvAB and RuvC

- Ruv AB catalyzes branch migration in bacterial cells, using ATP. There are thousands of base pairs in a few seconds. Helps in the branch migration process. RuvB is related to helicases so they circle around the DNA. In hydrolyzing the ATP, it allows the branch migration. Hybrid DNA is generated by branch migration. - RuvC: resolvase, it resolves the holiday intermediate after the branch migration by generating a cut

Describe the 5' to 3' Exonuclease activity of Pol I

- The Pol I can cut the DNA from 5' to 3', the same direction as DNA synthesis - this is important to remove the RNA primer on the Okazaki fragment. There are tons of RNA primers on the lagging strand because each Okazaki needs an RNA primer. These need to be removed because they are not needed in the duplex DNA. This job is carried out by the Pol 1 enzyme 1. The Pol I enzyme has an extra domain, which is the 5' to 3' exonuclease which removes the RNA primer. 2. The polymerase is right behind the 5' to 3' exonuclease and starts to synthesize and full the gap. It extends the 3' end of the daughter DNA. 3. There is a nick in between Okazaki fragments because the Pol is not able to repair the nick left over. It is then translated, it moves, after the reaction. This nick translation is enabled by simultaneous 5' to 3' exonuclease activity as well as 5' to 3' DNA polymerase activity.

The Open and Closed Shapes of Pol I

- The Pol I enzyme has two shapes: opened and closed - accurate base pairing is enabled by the shape of the active site in the closed form - the mismatch will not fit into the active site, the insertion site. If there is a mismatch, it will be very bulky and the insertion site will not be able to accommodate it. - the fingers will be able to bring in the correct nucleotide, and the fingers will squeeze it to the palm. This squeezed form is known as the closed form. This is when the insertion site is very compact with limited space. So if there is a wrong base pair, it will not be incorporated into the DNA. - Worst case, a wrong base pair will be inserted into the DNA, it will then be chopped off by the 3'-5' exonuclease

Activities at the DNA Replication Fork:

- The beta sliding clamp: esures that Pol will not fall off. This is a ring that does not fall off - DnaB helicase: separates the two strands of DNA using the energy from NTP hydrolysis. Usually ATP, it is like a motor. - topoisomerase: untwists the duplex DNA ahead of the fork. Helps release the tension. - primase: binds helicase to synthesize 10-13 nt RNA primer, RNA pol. The primase binds to the helicase - Pol I and RNaseH: removes the RNA primer and seals the gap using polymerase action and replaces the RNA primer with DNA. RNaseH removes the primer but cannot polymerize the DNA - Ligase: seals the nick of the duplex DNA. This joins the two Okazaki fragments - SSB: prevents the secondary structure and degradation off ssDNA. Once the DNA is unwrapped, it is single stranded. These single strands can be targets of degradation, you also do not want secondary structure forming on the single DNA strands. The SSB prevents this by binding to the single strands.

DNA Repair Overview

- The key to DNA chemistry is fidelity - water and oxygen can be dangerous to DNA - the DNA is constantly under stress - an army of repair enzymes are evolved in bacteria and eukaryotes to constantly repair lesions - use the complementary strand to repair lesions, there is always a template strand - cost tremendous energy, but well worthy for cell survival. The cell is doing all it can do to prevent mutations in DNA

The Fundamental Chemistry: Phosphoryl Transfer Reaction

- These reactions use the Phosphoryl Transfer Reaction: DNA/RNA polymerase nucleases, topoisomerases, intron splicing, recombinases, transposases... - isoenergetic: delta G = 0 1. there is double stranded DNA. There is molecule X with a hydroxyl group which performs a nucleophile can attack to the phosphate. This breaks the PO bond, releases the upstream strand. 2. The nucleophile is now connected to the rest of the DNA strand. This will generate a 3' OH group which can do further nucleophilic attacks of the phosphate group. - the X can be the 3' hydroxyl group of double stranded DNA. It can also be tyrosine or serine residue which also contain a hydroxyl group. It can also be a water molecule which would be hydrolysis. - only if delta G is above 0 does the reaction require energy

Solar Radiation Generates Pyrimidine Dimers

- UV light creates pyrimidine dimers, a double bon cross links and forms sigma bonds. A cyclobutane ring can be created, linking two bases together - it can also result in the 6,4 linkages of the two thymine. This is when one pi bond becomes a sigma bond. This cross links the pyrimidines, C or G, this creaks a kink in the backbone of DNA - the kink leads to the DNA strand breaking

Nucleotide Excision Repair (NER) - Eukaryotes

- XPA-G: xeroderma pigmentosum. This is caused by mutations in the neuro system and is a neuro disease. This is a disease that makes a person very sensitive to UV light radiation. - Transcription coupled repair (TRC): damage recognized by RNA Pol - Very similar to bacteria Thymine dimer repair: 1. XPC proteins pushes open the double stranded DNA at the lesion site 2. It recruits XPB and XPD proteins. 3. The endonuclease XPF and XPG as well as the single stranded binding protein RPA is recruited to the lesion site. The XPF and the XPG will generate a nick on both sides of the damaged strand - this is different from the bacterial system because the bacterial system only has the UVRC exonuclease that generates both nicks. In humans, this function is carried out by two separate proteins 4. The 24-32 damaged portion of nucleotides is removed which generates a gap. 5. The gap is filled by the helicase, polymerase and ligase - this results in fixed, accurate DNA Another way of repairing bulky nucleotides in eukaryotes is: 1. Transcription coupled repair. This is very similar to the XP system except the original recognition of the damaged site is by the RNA polymerase instead of by the XPC protein. This only happens at the actively transcribed regions of the DNA. - in the human genome only 1% of the genome is actively transcribing protein coding genes. 2. The RNA polymerase detects the lesions in the DNA. Once the lesion is detected, it will pause on the template and recruit the XPB and the XPD proteins. 3. It then goes through the same repair process as before.

Base alkylation

- adding an alkyl group - bases are so electron-rich that they are nucleophiles -alkyl groups often have a positively charged C that lacks an electron - very bulky aromatic hydrocarbons can create covalent bonds to the bases

Beta Clamp Recycling

- beta clamps are left behind on the lagging strand - Once the Pol III is done, it dissociates, leaving the beta clamp. Before unloading and recycling the beta clamp has to.... 1. Recruit Pol I to removed the RNA primer. The Pol I then dissociates. 2. Then it recruits the ligase to seal the nick. The ligase the dissociates. 3. The delta subunit of the clamp loader is like a key and opens up the beta clamp. The beta clamp is then removed by the delta subunit.

Screening for carcinogens - Ames Test

- chemical structures that modify the DNA structure are known as carcinogens - mutate one gene in salmonella known as histidine synthase so that is cannot synthesize its own histidine which is an essential amino acid - the mutated salmonella can then only grow on a medium that has histidine - if you put mutated salmonella without the histidine synthase on a plate with no histidine, the salmonella will die. You do occasionally get some colonies because they overcome the gene loss to get the histidine gene back. The reacquire a functional copy of the histidine synthase. - first they spray the culture of the salmonella onto the selected plate. The salmonella has no histidine synthase and therefore will not grow on a plate unless it has histidine - soak a filter paper disk onto the potential mutagen that you are testing and put it in the center of the plate. Overtime the mutagen will diffuse over the plate, forming a gradient. You leave the salmonella grow. There will be some salmonella that restore the gene and stay alive. - there will be a circle of colonies formed around the disk at medium concentration of the mutagen. At very high conc. of the mutagen, it is too toxic. - if it is not a mutagen, you will not see the ring structure formed - this is the first-line test of carcinogens

Introduction to site-specific recombination and transposition

- discovered that the different colors of the maize seeds are random. Raised theory of jumping genes where some of the genetic elements can move freely through the genome - this was before the double stranded DNA structure was discovered and it was against what people thought because the structure and replication needed to be very accurate, very conservative - "jumping genes" - probably among the most ancient genetic entities 1. site-specific recombination requires short specific sequences. Does require a specific sequence and only limited recombination can happen. It can only happen between two specific sequences 2. transposable elements inert/delete randomly, genetic parasites. There is no specificity and you cannot predict where they will go - linked to evolution of viruses, introns and immune system

DNA Pol I enzyme

- found in e. coli, used to understand structure of DNA polymerase 1. Pol 1 binds to the growing strand and the template strand. The Pol 1 has two nucleotide binding sites, one is the insertion site and the other is the post insertion site. The post insertion site has the last synthesized base pairs. The insertion site has the next position in the template. 2. The incoming dNTP will bind to the insertion site. In this case it inserts A into the site. They base pair and a phosphodiester bond forms. 3. The Pol 1 will translocate ad slide to the right 1 position. The newly synthesized ATP pair will be docked to the post insertion site. 4. The next nucleotide, G in this case, comes in, ready for the next round of polymerization.

Homologous Recombination Overview

- homologous recombination: genetic exchanges at identical or near identical DNA sequences - homologous recombination is a DNA repair process at the double stranded break (DSB), called recombinational DNA repair. Highly accurate - DSB is mostly lethal. It can occur during DNA replication or UV light radiation - Bacterial and eukaryotic cells encounter tons of lesions within a day. Every second we are having thousands of lesions and some result in a double stranded DNA break. - homologous recombination generates genetic diversity in diploids. This is during the meiosis process.

Double-strand Breaks are Repaired by Recombination

- require a homologous duplex DNA - Use 3' ends because it can serve as the primer - recombinase - this happens between a double stranded break and another template wit near identical nucleotide sequence 1. one strand has a double stranded break and it has a template strand with near identical sequence. The first step is the end processing. The 5' to 3' exonuclease will chop off the top strand 5exposing the 3' ending which is a single stranded DNA overhang. This happens on both ends of the double stranded break. Generating a 3' single stranded DNA 2. The 3' end will invade the template DNA and displace the other reverse complimentary strand, base pairing with the template DNA. The displaced strand will generate a D-loop 3. this is followed by the second strand invasion. 4. The 3'OH is used as a primer to extend the DNA by DNA polymerase. - this is known as the Holliday model of DNA recombination

bidriectional

- when DNA replication starts there is a replication bubble. - the DNA replication happens on both strands in both directions - there are two DNA replication forks

Site-Specific Recombination and Transposition

Lecture 4 - Notes Below

Summary of Repair

Mismatch repair base excision repair photoreactivation nucleotide excision repair translesion synthesis

CQ: Which one of the following does NOT usually occur when a replication fork reaches a break in one strand of DNA? a. The fork collapses and a double-stranded break occurs b. Broken DNA ends are processed to create 5' overhangs c. Overhangs of broken ends invade their non-broken, double stranded DNA complement d. Invasion of the broken strand into its complement is mediated by recombinase

b. Broken DNA ends are processed to create 5' overhangs

QC: Reverse transcriptase: a. is used by HIV as well as some transposons b. can use either HIV as well as some transposons c. can have RNAse activity d. polymerizes in the same direction as DNA polymerase e. all of teh above

e. all of the above

The parent and daughter DNA strands

the parent strand goes from 5' to 3' and the other parent strand runs antiparallel 3' to 5' - at the replication fork there are two daughter strands being synthesized. One daughter strand is the leading strand synthesizing from 3' to 5'. The lagging strand also synthesizes 3' to 5'. There are two strands (Okazaki fragments) which can only synthesize 3' to 5'. - Discontinuous fragments are created and started from RNA primers. These RNA primers are 10-13 nucleotides - these RNAs are created by primase and are hybridized to the single stranded DNA - DNA polymerase then comes and synthesizes the DNA from the 5' to the 3' direction - it is called the lagging strand because that strand is pretty slow compared to the other strand due to the need to seal the gaps and synthesize more than one fragment - this type of synthesis is known as semi discontinuous

What does helicase in the replisome do?

unzips the DNA

Mechanisms of Recombinases

- Active site Tyr/Ser. This is because these have an OH group which can do the phosphoryl transfer reaction - they usually exist as tetramers. Tetramer: light gray: active subunits, dark gray: inactive subunits - Holiday intermediates - no need of ATP Example: 1. The two duplex DNAs are held together by the recombination of the inverted repeats. The light subunits are active. the tryosine residue will cleave open one strand of the two duplexes. 2. The cleaved end will now combine and be position near the blue strand. 3. The same tyrosine residue will catalyze the reverse reaction to join the two strands together 4. The two active site tyrosines will rotate away from the cutting and ligation site 5. The enzyme will isomerize to now have the other two subunits active (dark gray). The two tyrosines in the gray subunits will cut the other two strand open and religate them with a different strand

RecBCD Complex Sets the Stage for the Recombinase RecA

- Before the recombinase RecA can act on the 3' overhang needs to be prepared - RecBCD sets the stage for RecA to act - RecBCD: composed of a helicase and nuclease. It prepares the free 3' end and loads RecA at the 3' end - Chi sequence: a 8 nucleotide sequence, 1009 chi sequences on the E. coli genome - The complex has helicase activity to unwind the DNA, it also as exonuclease activity at the RecB domain. Rec C has a pin that helps keep the two strands of DNA separate. 1. The helicase unwinds the DNA at the expenditure of the ATP 2. the RecB (a nuclease) digests both strands of DNA. Not just the 5' to 3' strand but also the other strand. The helicase continues forward until the chi site is reached. 3. Once the chi site is reached the RecC will bind to it, stopping RecB from digesting the 3' end. T 4. The RecB can still digest the other strand, displacing the 3' overhand as a loop 5. RecBCD complex recuits the RecA protein to bind to the single stranded 3' overhang so it can start the recombinational process.

Recombination Mediated Mating Type Switch in Yeast

- MAT locus is active - HMLalpha and HMRa loci are silent. They are each about 700 base pairs long - this is an example of ow homologous recombination can contribute to phenotypic change - yeast can switch between different mating types, usually yeast are haploids and they usually do not go through meiosis unless they are stressed. - each haploid yeast cell has a mating type and they are either a or alpha - when stressed, the a haploid will mate with the alpha haploid generating a diploid. This diploid will then replicate and produce 4 spores. This helps the yeast go through the stress of the environment 1. The haploid is able to change the mating type from a to alpha depending on the environment. It does this through homologous recombination. On the yeast chromosome 3 there are three loci contributing to the yeast mating type. One is HMLalpha, one is MATa and the last one is HMRa. Usually the HML and HMR loci are not expressed. Only the MATa is usually active. Now the yeast is a type yeas. First the HO nuclease demonstrates a double stranded break resulting in gene conversion 2. The double stranded break will use the HML alpha to repair, going through the homologous recombination process. This results in the yeast converting its mating type from a to alpha 3. It can convert back to the a type through the OH nuclease by using the HMRa locus as the template during the DSDR - 700 base pairs is all the yeast needs to determine its mating type

Large Mutations Form Abnormal Chromosomes

- Within a single chromosome or with multiple chromosomes - across multiple chromosomes you can have a section of one chromosome be inserted into another chromosome - you can have a translocation where two chromosomes each exchange a section - on a single chromosome a section can be deleted or the same section can be deleted or a section can be deleted. Evolutionary Consequence: A chimpanzee has 25 different chromosomes while humans have 24. Because human chromosome 2 is fused from 2 ancestral chromosomes

Homing Endonuclease

- encoded by the intron - generate DSB - recombination leads to intron propagation - in eukaryotes there are many introns in the genome and some of those can propagate themselves to other places. One way they can do this is by homologous recombination catalyzed by this enzyme called homing endonuclease 1. Suppose two duplex DNAs are homologous and are on two homologous chromosomes. The gene X' has an intron to start and Gene X does not have an intron to start with. The homing endonuclease is transcribed from the intron region. (In this case the intron produces a functional protein) 2. The homing endonuclease goes over to the homologous gene and generates a double stranded DNA break at the same spot where it will introduce another intron 3. The double stranded DNA break will be repaired by its homologous chromosome which is X'. 4. After homologous recombination, gene X now has the same intron as gene X'

Three Polymerases Work Together at the Fork

- for the leading strand, you need one polymerase III. The lagging strand is very slow so there needs to be two polymerase III's working together. 1. The leading strand has one polymerase III which keeps going. The lagging strand starts with one polymerase III as well. It extends the primer and forms the loop. Another polymerase III subunits is "parked". The clamp will be waiting to load the second polymerase III on the lagging strand. It will use ATP to bring over the beta clamp. 2. Once the beta clamp is loaded, the primase interacts with the helicase to synthesize a new primer at the front end of the fork. 3. While the first polymerase III keeps extending the Okazaki fragment, the clamp will be loaded to the primed place of the lagging strand near the clamp loader. 4. Now there are two loops. The first polymerase III is still continuing. The second polymerase III is loaded onto the primed lagging strand with the beta clamp. While the first polymerase III is finishing up the Okazaki fragment, the second polymerase III has started polymerizing the second Okazaki fragment. At this point there are two Polymerase III's on the lagging strand and one on the leading strand. 5. When the first polymerase III finishes the Okazaki fragment, the Pol III will release the lagging strand, leaving the beta clamp. The beat clamp is still on the Okazaki fragment. This generates a problem because the beat clamp needs to be recycled

DSB During Meiosis is Generated by Spo 11

- homologous recombination requires a double stranded DNA break - this is a purposeful double stranded break 1. it is catalyzed by an enzyme called spo11. This protein has two tyrosines at its active sites. These tyrosine are the phenyl side chains off the protein. The OH acts like a nucleophile to attack the phosphodiester bond of the target DNA. This generates the double stranded DNA break. The Spo11 proteins is now covalently linked to one strand of DNA - the chemistry: the deprotonated tyrosine residue acts as a nucleophile to attack the phosphate group of the phosphodiester bond. This breaks the OP linkage. This generates a free 3' OH and the protein is covalently linked to the end of the broken strand 2. The next step is end processing by a couple enzymes. This processes the 3' end in order to expose the 5' OH. This is done by the protein complex Mre11-Rad50-Xrs2. This protein processes the strand attached to Spo11. It chews backwards. 3. The free 3' end is generated and ready for homologous recombination. 4. The free 3' end is processed by the endonuclease Sae2 and. 5. Sgs1 (which is a helicase) and Dna2/Exo1 (exonuclease) further enlarge the the 3' ending strand - Eukaryotic recombinases Rad51 or Dmc are specific to meiosis and are loaded to the 3' end - after this process, it is followed by normal homologous recombination

Repair After Lesion Bypass

- homologous repair can also happen after lesion bypass - a lesion bypass is when the replication fork skips over a lesion and continues on

Indels Caused by Template Slippage

- in humans its likely to have repeated sequences such as glutamine - sometimes slippage occurs in the tempate where there is an insertion of an extra fewnew nucleotides such as an extra CAG. This results in an extra glutamine in teh final protein product - this is not detrimental as a frameshift but adding or deleting the amino acid. Adding or deleting a few amino acids usually does not cause diseases - if the poly glutamine sequence is too long or too short it can cause genetic disease known as triplet expansion diseases. These can include Huntington's Disease, Fragile X syndrome, et.

Telomere Protection in Mammals

- in mammals the telomere repeat is TTAGGG. The enzymes that bind to and protect the telomere sequence is known as the sheltering complex. The 3' end is looped over by a POT1 protein which creates a T-loop. The 3' end is protected by looping up which discourages DNA degradation of the telomere. - increasing telomere length in mice leads to cancer

The Protein Structure of the Pol I Enzyme

- it looks like your right hand, it has a palm and three fingers - the replication fork is in the center of the palm. - in the middle of the palm are the two insertion sites - the fingers are responsible for bringing in the dNTP's - the thumb holds the duplex DNA in place - the 3' to 5' exonuclease activity is located slightly away from the polymerase active site, 10-20 angstroms - when a mismatch happens, the daughter strand will be bent toward the 3' to 5' exonuclease active sit in order to be hydrolyzed - the cleaved daughter strand will then go back to the insertion site in order to continue the polymerase reaction

Site-specific Recombination leads to inversion, deletion or insertion

- leads to inversion, insertions or deletions Ex: the two FRT sites are inverted 1. in the middle is the A,B and C genes. 2. the recombinase comes and binds to the two FRT sites and align them so they are parallel 3. This leads to the exchange of strands. 4. After recombination, the 3 genes flip their sequence to C, B and A - inversion happens when the two site-specific recombination sites are inverted. Ex: the two FRT sites are in the same sequence leading to deletion and insertion 1. The two recognition sites will align 2. after recombination, there is a deletion that results. This generates a circular DNA and a short stranded DNA - The reverse reaction can also happen which is the insertion. 1. the circular DNA can align with the binding site 2. This results in the insertion of the three genes

Mitotic Recombination

- limited to S and G2 phase - happens between sister chromatids, these sister chromatids only exist between the S and G2 phase - The recombinase is RAD51 in eukaryotes and is similar to RecA in bacteria which helps the 3' strand invasion process - Rad52, BRCA2 are like RecFOR, they are mediators. 1. There is a broken strand with a 3' end 2. There is invasion of the strand 3. This is followed by polymerization A. 4. This can be resolved by SDSA. This causes it to directly dissociate from the other sister chromatid 5.It synthesizes back leading to a non-crossover product B. 4. This can also be resolved by a double stranded break (DSBR). 5. This repair process can lead to double holiday intermediates. It can then go through a dissolution which generates a non-crossover and or a resolution which can generate a crossover.

The lox/Cre System in P1 Phage

- lysogenic pathway: prophage - lytic pathway: lysis, bacteria dies - bacterial transduction: the DNA transfer between bacteria which is mediated by a phage. The virus also chops up the host genome, so once the phase is assembled, some of the host genome will be inside the virus. When these viruses invade and infect another host cell, it will introduce the DNA from the previous e. coli to a new e. coli. - in biotechnology, scientists use the above process for research - this system was originally found in the P1 phage - lox is a specific site with 13 base pairs and Cre is a recombinase for the lox site. 1. A phage is a virus for bacteria. The phage attacks the E. coli cell, injecting its genome. 2. The genome circularizes and the genome next has two pathways that it can potentially go through: A. 3. It can go through the lysogenic pathway where the plasmid is inserted into the genome of the e. coli as a prophage and it continues living B. 3. Or it can go through lytic pathway where the circular DNA starts to replicate and make multiple copies. This produces proteins that assemble the phage and is followed by lysis. This results in the e. coli cell dying while the phage is replicated and then released into the environment

Base deamination by nitrous acid

- nitrous acid can be generated by food additives such as sodium nitrate cytosine to uracil, adenine to hypoxanthine

The hix/Hin System in Salmoneela typhimurium

- phase variation: two flagellins: FljB or FljC - the is a more complex example of site-specific recombination - Salmonella lives in guts and is a pathogenic bacteria - our immune system recognizes the flagellins and triggers an immune response - the bacteria doesnt want to be detected and therefore has two types of flagellins. Flagellin is a protein that makes up the flagella. Our immune system will not know which flagellin to target. This phenotype is known as phase variation which is enabled by the hix/Hin system. 1. On the genome of Salmonella, there are two operons, there is a promoter followed by the Hin. Hin is a recombinase for this system. Another operon has a promoter and then three genes, fljB, fljA and flC. The fljB and the fljC are the two flagellin genes making two flagella. The fljA is a repressor for the fljC gene. There are also two hix recombination sites flanking the Hin recombinase. They are inverted which will generate an inversion if it works. Usually when nothing happens in this combination, fljA and fljB will be expressed. fljB will go on and make the flagellin while fljA will repress the fljC from expressing any flagellin. 2. Once the hin is transcribed and made into a recombinase, it will make the two hix sites recombine, causing the hin gene to be inverted, Once inverted it also causes the promoter of the flj and the fljC to become inverted. Once the promoters are inverted there will no longer be an fljA or fljB proteins. 3. Without the fljA repressor, the fljC will be expressed. This is how the cell can flip between the flagellins.

oxidative damage

- reactive oxygen species: H2O2, OH-, O2- - bases are so electron-rich that radicals love to get an electron from them - oxidative species have free electrons known as radicals that are very reactive cytosine to 5-Hydroxycytosine thymine to thymine glycol adenine to 8-oxoadenine guanine to 8-oxoguanine - 8-oxoG pairs with A leading to a GC to AT mutation

Recombinational DNA Repair

- recombination as a DNA repair process - enzymatic machines in bacterial recombinational DNA repair (RecBCD, RecA, RecFOR, RuvABC) - homologous recombination in eukaryotes. Aids in the strand invasion process - nonhomologous end joining

Spontaneous hydrolysis by water: deamination

- spontaneous hydrolysis by water leads to deamination of the bases - C to U is the most common mutation. Left untreated a C-G pair will become an A -T pair. This could be a reason that there is a T instead of a U in DNA. - the oxygen with the lone pair of electrons from the water molecule is able to attack the carbon of the cytosine leading to the breakaway of the primary amine. The ammonia is released and is substituted by an oxygen, forming uracil - C to U is most common because there is water everywhere - all of the bases have an amine group so they all go through spontaneous deamination. Cytosine to uracil adenine to hypoxanthine guanine to xanthine 5-methylcytosie to thymine - 5-methylcytosine is linking to gene expression - hypoxanthine will pair to C and not a T

Direct Repair - Photolyase

- sunlight helps bacteria to recover from UV radiation - present in bacteria, archaea, eukaryotes, but not humans 1. The chromophore in DNA photolyase absorbs a photon of visible light and passes the excitation energy to FADH- 2. FADH- donates an electron to catalyze bond rearrangements in the pyrimidine dimer. The unpaired electron is able to attack the pyrimidine dimer. 3. Electron rearrangement restores pyrimidines, and the electron is transferred back to FADH to regenerate FADH-. This restores monomeric pyrimidines - this can happen with pyrimidine dimers - the enzyme photolyase uses energy from the sunlight to repair the pyrimidine dimer - photolyase has two cofactors, one is chromophore. This is a small molecule that absorbs light. The second factor is a redox factor, FAD. - there is no longer a cyclobutene ring.

Nucleotide Excision Repair (NER) - E. coli

- targets large bulky lesions - Exinuclease: two lesions are made on the same strand of DNA. This is the spatial activity of the UVr complexes - E. coli: Uvr-D - Uvr = UV resistant - NER is catalyzed by the Uvr enzymes. - present in e. coli and eukaryotes Ex: Repair of the thymine dimer: 1. Initiated by the UvrB protein which pushes open the double stranded DNA at the lesion site. It also recruits two UvrA proteins. It creates a bubble-like protein structure at the lesion site 2. The two UvrA dissociate and the complex recruits the UvrC protein. This is a specific spatial endonuclease which creates two lesions on the same strand of DNA. It creates a 3' and a 5' nick. The strand which has the lesion now has two nicks, one on both sides. 3. UvrC leaves and the UvrD protein comes in and takes away the UvrB protein as well as the 12-13 nucleotides generated by the UvrC. This portion of the DNA is removed, generating a gap. 4. This gap is then repaired by the Pol I and the ligase

The lox/Cre System Step 2

- the circular phage genomes enables rolling-circle replication, producing many copies of the genome. 2. The circular DNA gets nicked on one strand and the 3' OH will prime the DNA synthesis. The orange part is newly synthesized DNA. On the other strand it starts lagging strand synthesis with the Okazaki fragments. The leading strand continues while the lagging strand also continues. This produces linear DNA with many of the circular DNA's content. Once a long, linear DNA strand is generated, the loxP will become active again circularize to create a genome

The Architecture of E. coli Origin of Replication

- the e. coli origin of replication is also known as the Ori C, it is 245 base pairs long - it contains two regions: 1. DnaA protein binding sites, this region contains 4 long repeats (R1-R4). 2. DNA unwinding element: A=T rich with 13-mer repeats. This element is easy to open

Gene Conversion

- the homologous regions are not exactly the same - Remember, no crossovers do not mean no consequence because the region is different from the original strand. Any genes involved in this region can be different, the strands are hybridized. - during the next round of replication, if there is a gene change then it will be replicated to the daughter strand. - gene conversion can occur whether homologous recombination results in crossover or no crossover. - The homologous chromosomes are not exactly the same! There are still some small differences!! there is always a hybrid part! - Where the resolase cuts determines if there will be a cut or not - they are homologous chromosomes but heterogenous chromosomes - homologous recombination can occur between identical strands or near identical strands that may have an amino acid change - the couple of amino acid difference can result in a nonfunctional protein

repair After Lesion Bypass - Gap Repair

- the invasion DNA does not have a 3' end - mostly non-crossover 1. there is a gap at one arm of the replication fork. The strand invasion happens from the opposite daughter DNA. 2. This is not invasion of the 3' end. This generates two crosses which then goes through branch migration. This enlarges the base pairing region so that the growing strand can base pair. 3. the gap is filled in by replication producing a holiday intermediate 4. The holiday intermediate is resolved from resolase and ligase. There is no cross over and replication can resume

Replication Fork in Motion: the Trombone Model

- the leading strand and lagging strand have different replication speed, how does the replication fork ensure that the two strands are synthesized simultaneously? - the leading strand is a lot faster than the lagging strand because it is continuous. The lagging strand has to have many Okazaki fragments, it has to have the RNA primer chopped off and it needs to resynthesize DNA to fill the gap. - To synthesize simultaneously, a loop is created on the lagging strand. The is know as the trombone model. It is generated, enlarged and then released over and over. 1. The leading strand keeps going. As the fork moves forward, the helicase and the primase are associated so that the primase is able to synthesize a primer for the Okazaki fragments. 2. The Pol III core enzyme will be extending a primer ahead of the primer that has just been synthesized. While the Pol III goes, the loop will be generated. The loop is between the helicase and the Pol III core enzyme on the lagging strand. 3.As the Pol III synthesizes DNA, the loop becomes larger. 4. Once the Okazaki fragment is completed, the Pol III will dissociate from the lagging strand, leaving the beta clamp behind. 5. The clamp loader will load a second beta clamp onto the newly primed lagging strand. 6. The Pol III will then come and extend the newly synthesized primer for another Ozaki fragment. - the reason for the loop is because the lagging strand is slower than the leading strand - the Okazaki fragment length in E. coli is about 1-2 kilobases whereas in eukaryotes, it is 100 to 200 base pairs. The Okazaki fragments in eukaryotes are therefore much smaller. - a loop is formed every 2-3 seconds

Metal Ions are Important for the Activity of the Polymerase

- the magnesium ion is involved in DNA polymerase catalysis - the double-positively charged magnesium ion interacts with aspartic acid, this is the amino acid residue at the active site. Aspartic acid is negatively charged so it interacts with the magnesium ion. The aspartic acid stabilizes the two magnesium ions in the two active sites. - aspartic acid is negatively charged. This negative charge is attracted to the positive magnesium ion - there are two magnesium ions at the reaction center of DNA polymerase - these magnesium ions are important because they are interacting with the 3' hydroxyl group of the extending DNA and also brings close the alpha phosphate. It interacts with the oxygen of the alpha phosphate - the magnesium ion holds everything in place so that the oxygen of the hydroxyl group can attack the phosphate of the alpha phosphate group. - these groups need to be held close together in order for them to interact with each other, this job is carried out by the magnesium ion - there is another phosphate that holds the beta and gamma phosphate groups in place. It positions them correctly so that the nucleophilic attack on the alpha phosphate can happen. This is for the polymerase set. The magnesium ions are very important for the 3' to 5' exonuclease activity - the goal of the magnesium ins is to break up the incorrect base pair - the phosphodiester bond needs to be hydrolyzed, meaning that the water will come in and break the bond. In this case, the water is in the form of a hydroxide which will attack the phosphate of the phosphodiester bond. - the dNTP is released, the goal of the second magnesium ion is to bring over the nucleophile, the hydroxide ion. Without the nucleophile, the hydroxide will not be accessible in the active site.

E. Coli Pol III Holoenzyme

- the main function is to synthesize DNA Consists of three parts: 1. 3 Pol III core: yellow structure 2. 3 sliding clamps: ring structure interacting with the Pol III core 3. 1 Clamp loader: looks like a jar with 5 fingers. This loads the clamp ring onto the DNA.

DNA-Damaging Agents in Chemotherapy

- the mechanism of cancer cells is that they divide way faster than somatic cells. If we can disrupt the DNA replication process, we can kill the cancer cell and not the somatic cells - Bleomycin is isolated from bacteria and can bind to the double helix, stopping the replication fork - Doxorubicin is similar and is another cancer drug - it will also attack the normal cells, but somatic cells do not replicate. In mammalian cells that are mature, they do not replicate even after damage but tumors grow due to rapid replication.

Base Excision Repair (BER)

- the most prevalent method - bacteria, eukaryotes including mammals - AP site: apurinic or apyrimidinic site - repairs uracil, methyladenine, 8-oxoG, etc. Bacteria BER: 1. remove the wrong base by glycosylase. It hydrolyzes the n-glycosyl bond between the ribose and the base, generating an AP site 2. The AP endonuclease cleaves the phosphodiester backbone over the AP site, exposing the free 3' hydroxyl and the 5' phosphate group. 3. The Pol I polymerase has 5' to 3' exonuclease activity. those fills the gap after the exonuclease chops off the bases. 4. Ligase is necessary to repair the nick in the double stranded DNA Eukaryotic BER system: 1. 1. remove the wrong base by glycosylase. It hydrolyzes the n-glycosyl bond between the ribose and the base, generating an AP site 2. The AP endonuclease cleaves the phosphodiester backbone over the AP site, exposing the free 3' hydroxyl and the 5' phosphate group. A. 3. Recruits a polymerase and starts the 5' to 3' synthesis. There is no polymerase that has a 5' to 3' exonuclease. This generates a flap 4. A flap endonuclease cuts the flap 5. The final nick is repaired with a ligase. This generates a long patch of newly synthesized nucleotides. B. 3. Poly beta is a specialized polymerase that can repair the DNA more precisely. It is capable removing the 5' phosphate and the ribose. 4.It is then capable of filling in one nucleotide 5. The final nick is repaired with a ligase. This generates a short patch of newly synthesized nucleotides.

Direct Repair: O6-Methylguanine Demethylation

- the normal guanine pairs with cytosine, the G-C pair - once guanine is methylated at the 6 position, the methylated guanine will pair with the T. If left unfixed, the GC pair with be converted into an AT pair in the next generation - methyltransferase completely demethylates the methylguanine. The methyltransferase contains a cystine residue with a thiol group. The thiol group transfers the methyl group to itself. The enzyme itself gets methylated and it is then targeted for degradation. - each reaction costs an entire enzyme - the enzyme is made from transcription and translation and usually enzymes catalyze multiple reactions

Describe the 3' to 5' Proofreading Exonuclease of Pol I:

- the polymerase is able to detect and remove errors - there is about one accuracy every 10^6 to 10^8 base pairs - this occurs in the opposite direction of DNA synthesis 1. The polymerase mispairs dc with dt. 2. The polymerase repositions the mispaired 3' terminus into the 3' to 5' exonuclease site - the mispaired base is not away from the active sites of the polymerase 3. The exonuclease hydrolyzes the mispaired dC. 4. The 3' terminus repositions back to the polymerase site 5. The polymerase incorporates the correct nucleotide, dA

Translesion synthesis (TLS)

- the previous repair systems are very accurate and do not generate new mutations - TLS Pol lacks proofreading and can cause new mutations at the damaged site of the DNA - When the cell is super stressed due to a lesion at the replication fork, either the cell will die or scramble and resume the replication fork even with the risk of adding a new mutation - this is essential because once a replication fork collapses, the cell fails to divide and the cell will die - E. coli: Pol V, Pol IV, Pol II - Human has at least 10 TLS - Human Pol n put two As opposite of the thymine dimer, quite accurate 1. The Pol III detects a lesion and dissociates from the growing strand of DNA 2. TLS Pol is recruited. It base pairs and puts "something" at the lesion site. 3. The Pol III then resumes and the TLS Pol dissociates - this bypasses the lesion - How does the TLS Pol tolerate weird matches? The crystal structure of the TLS has a huge hole in the middle of the active site. It is crystalized with a Benzo-alpha-pyrene DBE adduct which fits into the active site. This compound is huge and the active site of teh enzyme is able to fit huge bulky bases.

Point mutations

- the smallest mutation - Types of point mutations: 1. Transition mutation: Involves the change from a purine to a purine or pyrimidine to pyrimidine Transversion mutation: a purine to a pyrimidine or a pyrimidine to a purine Consequences of point mutations: - if the point mutation occurs in the coding sequence this could lead to an amino acid change: 1. silent: no change in the amino acid or protein 2. missense: the amino acid is changed based on the point mutation 3. nonsense: an amino acid is switched, forming a stop codon

Replication Termination in E. coli:

- the termination site is halfway around the chromosome from the oriC - Ter sites: These are specific repeats genome that are termination sites and are therefore directional. On the chromosomes of e. coli there are multiple ter sites, but they go in different directions. The ter sites do not stop the replication fork if it is going in the same direction. The replication fork will continue until it encounters a ter site that is going in the opposite direction. - Tus protein: bind to Ter sites and physically block DnaB helicase - if one replication fork goes faster, it will be blocked by the first ter site that it encounters and wait for the other replication fork to complete

Overview of E. Coli DNA Replisome

- there are many enzymes working on the DNA replication fork - The Pol III is bound to a beta sliding clamp which is a ring surrounding the DNA and a clamp holder. This complex is known as the Pol III holoenzyme - There is a DNAB helicase which binds to one strand of DNA to unzip it. It breaks the hydrogen bonds between the bases and is placed in front of the fork in order to drive the fork forward. - There is primase to synthesize the RNA primer for the Okazaki fragments as well as for the leading strand - The Pol III holoenzyme, the DnaB helicase and the Primase form a complex at the replication fork called the Replisome - DNA topoisomerase is located at the duplex DNA. It releases the tension of the DNA strands ahead of the replication fork

Indels - Small Insertion and Deletions

- these are slightly bigger than point mutation - 1-6 nucleotide mutations - these mutations might cause a frame shift if the insertion or deletion are not in the fold of 3 Ex: a insertion or deletion of 1 or 2 in the coding sequence can lead to a frame shift - a frameshift almost always leads to a nonfunctional protein

Recombinational DNA Repair at a Collapsed Fork

- this happens if the repair has been initiated but stops, resulting in a collapsed fork The Goal of homologous recombination here is to reattach the broken arm to the forks. - Invasion strand is not extended by the polymerase and there is no synthesis of new nucleotides, but branch migration happens 1. A lesion results in a double stranded break. There is 5' end processing to expose the 5' 3' overhang 2. This is followed by 3' invasion of the other arm of the previous replication fork 3. This is followed by branch migration. This is where there is a cross over and the branch goes to the left. This generates more base pairing annealing. This enlarges the annealed places 4. The holiday intermediate is resolves in a non-cross over matter, this repairs the replication fork so that replication can continue

Single Strand not Processed by RecBCD...

- this happens, for example, when the gap repair process is just a gap so that the RecBCD cannot bind and recruit RecA - this requires the RecFOR complex. 1. the gap in the dulex DNA is protected by SSB proteins. SSB displaces RecA. Rec A cannot bind to any single-stranded region. 2. The RecFOR complex will interact with the SSB protein and recruit RecA over. 3. RecA is then able to replace SSB and fill the single stranded DNA place - RecFOR does not catalyze and instead helps to recruit another protein to the DNA. A protein that helps other proteins to bind to the DNA is a mediator. - happens when the 3' overhang is not necessary for homologous recombination

How does the clamp loader work?

- this is a clamp that binds the beta clamp a the bottom of the 5 subunit clamp loading complex. - without ATP the beta clamp will not bind at the bottom of the clamp loader - only when the ATP is bound to the subunits, it will load the beat clamp - once the beta clamp is loaded onto the 5 subunit complex, the beat clamp pops open. The DNA is already primed and is then winded through the clamp loader. This is followed by ATP hydrolysis to eject the clamp. The beta complex then closes.

The FRT/Flp System In Yeast 2u Plasmid

- this is a site-specific recombination system - FRT is the recombination site - Flp is the recombinase - you just need these two elements to do site-specific recombination - The two forks go in the same direction - more than two copies of the chromosome are created in one replication cycle - yeast has two kinds of plasmids: 2u(micron plasmid and Cen plasmid. The cen plasmid is a low copy plasmid. The 2 micron plasmid is a high copy plasmid because it uses the FRT/Flp system to many many copies in one round 1. Start with a 2u plasmid. Once the replication starts it generates a replication bubble and two replication forks. One FRT site is located closer to the origin than the other one, so one FRT site will get replicated. 2. This FRT site recombinates with the other FRT site which flips this region of the chromosome. This means that after the recombination, it is inverted. 3. Once it is inverted, previously the replication fork that was going in two opposite directions will go in the same direction. this cause replication to keep going and not stop. 4. The replication forks keep moving in the same direction around the plasmid. This generates a bid, oddly shaped plasmid. The FRT site is replicated multiple times. 5. Inversion happens again which will delete the regions between the inversion site and cut the DNA into multiple, monomeric plasmids. - in one cycle, this generates multiple genomes

Recombinational DNA Repair at a Stalled Fork

- this is when the fork stalls and go backwards, this requires homologous recombination in order to repair the damage 1. A fork stalls and goes backwards. The lesion is base pairing with the new template. The newly synthesized daughter DNAs base pair with each other. This allows the cell to repair the lesion before the fork resumes. 2. the daughter strands that base pair with one another are digested by the nuclease and the replication restarts Or... 2. If the lesion is not repaired after a long time, the daughter strand of one will use the other daughter strand as a template and extend over. This generates a four stranded holiday intermediate. 3. This allows branch migration to migrate to the right and bypass the lesion. The replication can then resume

The Model of the Eukaryotic Replication Fork:

- this model is very similar to that of prokaryotes The replisome progression complex (RPC): - Helicase: the Mcm2-7 which is the ring. The difference between the prokaryotic and the eukaryotic helicases is that the eukaryotic helicase is on the leading strand while the prokaryotic helicase is on the lagging strand. The eukaryotic helicase has more partners. such as GINS, Cdc45 protein. The proteins: Mcm2-7, Cdc45 and GINS compose the MCM complex. The functions of GINS and Cdc45 are that they are regulatory proteins of the helicase because they are targets of cyclin and kinase. This means that eukaryotes can divide anytime they want, unlike prokaryotes. These two proteins make sure that the reaction only happens during the S phase. - Primase: known as Pol alpha - Polymerase: It has different polymerases on leading strands. On the leading strand is Pol epsilon and on the lagging strand is Pol delta. - Sliding clamp: called PCNA, this has the same ring structure as the beta clamp, just a different name - Clamp Loader: structurally very similar to the prokaryotic one, but it is called RFC (replication factor C). This is not in complex with the Pol holoenzyme and the helicase. - RPA: this is a single stranded binding protein that is similar to SSB. - CMG complex: the helicase complex containing the MCM2-7 helicase, GINS and Cdc45 regulatory proteins. - RPC (replisome progressing complex): this is similar to the replisome in prokaryotes. This contains the CMG helicase complex, the polymerases as well as the primer.

Nonhomologous End Joining (NHEJ)

- this process does not require homologous recombination - unique to eukaryotes - happens when there is no homologous duplex DNA,. This might happen in somatic cells during the G1 and G0 phase when cells are not actively dividing. Here there are no sister chromatids or homologous chromosomes. - NHEJ increases with genomic complexity by introducing mutations - can result in deletions of a few nucleotides - it directly stitches the broken strands together which usually results in insertions or deletions of a few nucleotides 1. First there is a double stranded break that needs to be repaired. This recruits the Ku70-Ku80 complex to bind to and protect the ends. 2. This further recruits the DNA-PKcs and Artemis. the DNA-PKcs is a protein kinase that will phosphorylate Artemis. Artemis is a nuclease and once phosphorylated it will start chewing back both of the double stranded DNA which will widen the double stranded break region. 3. It then recruits DNA polymerase to fill the gap 4. The two strands anneal in an imperfect way 5. This is followed by ligation to fill in the gap. The double stranded DNA break is now repaired!!! - this is not an accurate way of repairing the double stranded DNA break

Site-Specific Recombination Sites: loxP and FRT

- two inverted repeats - an asymmetric core - recombinase binding and cleavage site - the most common site-specific recombination sites are loxP and FRT - At the loxP site there is a specific sequence ATAACTTCGTATA and this is one loxP site. There is also an inverse repeat of the same sequence. This is a very small region and is only 13 bps. These two inverted repeats are enough for recombination. Between the two inverted repeats is the core sequence wich is assymetric. It can be short or long. The recombinase binds to the short inverted repeats and will clease at the end end of them. - loxP and FRT are very similar except the binding site sequences are different

Mismatch Repair (MMR)

1. DNA polymerase misincorporates a nucleotide, creating a mismatch. The newly synthesized GATC site is hemimethylated. 2. DNA recruits MutS and MutL. MutS binds the mismatch and forms a complex with MutL 3. MutS-MutL scans DNA bidirectionally, forming a loop. It tries to bind a specific sequence. The palindrome sequence GATC. Once the MutS and MutL finds this sequence, the GATC sequence is always methylated on the template strand. This tells the cell which is the template strand. 4. MutS and MutL recruit MutH which is an endonuclease that cleaves the daughter strand. - Mismatches: a switch in base that is still a normal base. 5. Helicase II and Pol I exonuclease unwind and degrade the newly replicated DNA strand past the mismatch 6. Pol LLL fills the gap 7. Ligase seals the DNA - the parent DNA is always methylated but the newly synthesized DNA is not methylated at first. This is how the cell determines which is the mutated strand. - Dam methylase: this catalyzes DNA methylation in bacteria. - the GATC site is spaced out on average every one Kb - this costs alot of dNTP

Initiation of Replication in E. coli

1. DnaA oligomerizes on binding the origin and wraps the DNA around the oligomer. The DnaA protein also binds to ATP. DnaA is associated with the protein Hu. HU facilitates open complex formation. This happens by twisting and coiling which forces it to open. 2. DnaC loads DnaB helicase onto each strand, forming the prepriming complex. DnaC is a loader protein. DnaC hydrolyzes ATP to load the DnaB helicase onto the replication bubble. Once the bubble opens, it generates two replication forks leading to bidirectional replication of the lagging strand in e. coli. 3. DnaB helicase expands the replication bubble. It recruits the primase which forms RNA primers at each replication fork. The primed template recruits the beta clamp. This clamp prevents the polymerase from falling off the DNA during synthesis. 4. RNA primers direct clamp loading and assembly of one Pol III holoenzyme complex to the template. The holoenzyme complexes advance replication to the opposite DNAB helicase. This starts replication. There are two holoenzymes added to each fork as it goes in both directions. 5. Continued DnaB helicase activity allows lagging stand priming the Okazaki fragment synthesis

Site-specific Recombination in Biotechnology

1. It can be used to generate gene transcription. For example, you have a promoter, a stop and a gene. The stop codon is flanked by recombination sites. Once the recombinase is introduced, the stop codon will be deleted out. This is followed by active transcription of the gene. It can be used to induce gene expression. 2. It can also be used to insert a gene. On the chromosome there is one recombination site and you introduce another recombination site to a gene. The gene will be inserted into the genome 3. In another instance you might want to substitute one gene for another. Say you have the original genome and allele A is flanked by the two recombination sites and you introduce a plasmid with a different allele, with allele A'. It has the same recombination site and therefore it will recombine and substitute allele A for allele A'. This is how you engineer the genome using site-specific recombination. - site-specific recombination is preferred because it is predictable. It only happens at the site where the recombination site is located. It is also very simple in that you only need the recombinase and the recombination site. Examples: - lox/Cre in transgenic mice generation. All of this type of mice is generated this way. - FRT/Flp in lamda red recombination to knockout E. coli genes. Classic way to insert a gene into am e. coli - attB/P: Gateway Cloning

There are Two Paths for Completing Double Stranded Break Repair

1. Synthesis-dependent strand annealing (SDSA) 2. Double-strand break repair (DSBR) - crossover - non-crossover Above are the two results - in bacteria the homologous recombination occurs at the replication fork - at the replication fork there are two identical sequences at the two arms. This is why bacteria use homologous recombination during DNA replication * recombination is possible at replication forks because of the identical sequences of the two arms

Replication Termination in Eukaryotes: loss of DNA at the ends of linear chromosomes

1. the lagging strand generates Okazaki fragments which have RNA primers. These RNA primers can be repaired by certain polymerases to fill in the DNA 2. The problem is the first RNA primer, since this is not primed and there is no DNA sequence in front, once the RNA is degraded, it can not be filled 3. The daughter strand has a gap and in the second generation replication, this daughter strand will be used as a template which will result in shortened duplex DNA, with missing information

What Mechanisms catalyze tranposition?

1. transposase 2. integrase 3. reverse transcriptase

How does the DNA polymerase reaction work?

1. you have duplex DNA with one strand as the template. The bottom strand is the template while the top strand is the growing strand. - as you can see the next unpaired nucleotide is a thymine 2. A is brought by dNTP which has three phosphates. The first phosphate is alpha, then beat, then gamma. 3. The 3' end of the last nucleotide does a nucleophilic attack to the alpha phosphate. The electrons of the phosphate are pulled towards the oxygen. This breaks the P-O bond, forming a new phosphodiester bond between the thymine and the adenine 4. The pyrophosphate is kicked out, the beta and gamma phosphates.

Compare the origin of replication between bacteria and eukaryotes.

Bacteria only have one origin of replication while eukaryotes have many. This is because the chromosomes of eukaryotes are very large in comparison in with just one ORC, replication would take a very long time - this is why have many ORCs in eukaryotes is important

CCQ: What are the major differences between helicases, topoisomerases, and ligases used in DNA replication?

Helicases and topoisomerases unwind DNA - helicases in the prereplication process, while topoisomerases relieve supercoil stress ahead of the replication fork - and ligases seal DNA that has been broken or synthesized as Okazaki fragments

Is Pol I the Major DNA Polymerase in E. Coli?

No! Pol III is! - in e. coli there are 5 DNA polymerases - Pol I is the most abundant polymerase in e. coli. This was the first polymerase of e. coli to be studied. - the Pol I is mainly used for its 5' to 3' removal function enabled by the 5' to 3' exonuclease. Only the Pol I has this activity - Pol III is the chromosome replicating polymerase - Pol II,IV and V are involved in DNA repair as well as translation synthesis - Pol IV and V do not have the 5' to 3' exonuclease activity. They are not versatile - If the cell is stressed, replication stops - if a polymerase adds a mutation, replication can continue and the cell will survive - deleting Pol I is not lethal, it is used to remove RNA primers - nick translation - Pol II and III are error-prone and therefore expressed during stress - the Pol III does not have the 5' to 3' exonuclease activity

Chapter 11 - DNA Replication

Notes Below

Lecture 2 - DNA Mutation and Repair

Notes Below

Lecture 3 - Recombination DNA Repair and Homologous Recombination

Notes Below

Initiation of Replication in Eukaryotes

Origin: There are multiple origins. ARS (yeast - autonomously replicating sequence) is the origin of replication. Has conserved A and B1-3 region which are repeated sequences. These are conserved in all eukaryotes ORC: This is the origin of replication complex that comes in and binds to the origin. It also recruits the helicase Mcm2-7 as well as other proteins involved in the cell cycle included Cdc6 and Cdt1. Prereplication complex: The complex happens at the G1 phase and consists of OR, Mcm2-7, Cdc6 and Cdt.. During the G1 phase the replication has not started yet so it parks there. Once the cell enters the S phase, the cyclin kinase in the cell increases. The cyclin kinases such as CDK and DDK phosphorylate the ORC complex. They leave the origin of replication. This releases the Mcm helicase which starts to create the replication bubble and recruits the rest of the replisome complex. The Pol enzymes, the clamp loader, and other components.

Recombination Leads to Dimeric Chromosomes in Bacteria

Recombination that went wrong in bacteria, dimeric chromosome: 1. Bacteria genome undergoes recombination. One fork is normal and the other is undergoing recombinational DNA repair. This creates a holiday intermediate 2. a. If resolution happens by cleaving the X site, then it will not generate a cross over. Replication will continue as normal. The end result is two daughter, normal DNA chromosomes b. However if the replication process goes through the Y path then it will generate a cross over which will repair the replication fork as well as generate a dimeric genome 3. b. this dimeric genome is resolved by monomers by the XerCD system

Holiday Intermidiate

SDSA Pathway: the strand dissociates and anneals to the original strand. The template strand goes back to its original form and the broken strand extends and anneals to its own broken strand. This completes the replication but generates a patch of DNA that is different from the original sequence. (slightly different) DSBR Pathway: this involves Holiday intermediates. The branched intermediate will complete replication ligation. - the Holiday intermediate is the four strands of DNA entangled - there are two ways to resolve the holiday intermediate: crossover and non crossover Crossover: the broken strand will cross over with the template strand. The template strand becomes attached to the broken strand. Non-Crossover: the template strand and the broken strand do not attach together - To resolve the holiday intermediates, you use the enzyme resolyase. These are endonucleases that cut open the holiday intermediates. If the cuts are at the X positions, it will generate non-crossover products. Only the middle portion is newly synthesized here. If the resolase cuts at the X site and the Y site, it will generate a crossover resulting in the template and the broken strand being attached. This results in larger segments of DNA exchange on the chromosome.

Telomerase ad Telomerase Solve Termination Problem in Eukaryotes

Telomere: TTGGGG repeats in mammals. There is a spatial polymerase that is like a clock counting down the number of cell divisions Telomerase (TERT): - a cool reverse transcriptase: RNA to DNA. It uses RNA as a template to synthesize DNA. It carries its own RNA template - Carries its own RNA template - this can extend the telomer lengths Telomere binding proteins: protect the imperfect ends from degradation and recombination in telomerase was discovered by Elizabeth Blackburn and graduate student Carol Greider - they used Tetrahymena as their experimental organism because it is known for never dying. It is called the immortal algae. - In human somatic cells we do not have telomerases which extend the telomere. We are born with a certain telomere length and with each cell cycle it becomes shorter and shorter. This correlates with many age-related diseases. How the telomerase works: 1. The telomerase carries the RNA template strand. 9 of the couple hundred RNA nucleotides can base pair with the TTGGGG telomere sequence of the algae. It is 1.5 fold of the TTGGGG motif. 2. The telomerase is able to extend the 3' end based on the RNA template. 3. The enzyme will translocate to bind to the newly synthesized TTG and then extend again 6 nucleotides. 4. With continuation of this process, you get a telomere that is elongated. 5. The single-stranded telomere needs to be converted to double stranded DNA for stability. This is accomplished by primase which binds to the end sequence which is then filled in by polymerase. 6. The RNA primer is chopped off, leaving some nucleotides unpaired. 7. The unpaired nucleotides are then protected by the telomere binding protein which protects the end of the telomere. The 3' single stranded DNA is a great target for degradation or homologous recombination.

Possible Effects of a Lesion at a Replication Fork

What are the three things that can happen when there is a lesion at the DNA 1. translesion synthesis: translesion synthase that bypasses the lesion 2. fork stalls: there is a lesion in the leading strand causing the fork to stall. This leads to fork regression 3. repair initiated: there is a gap in one of the strands but the gap is not sealed causing the fork to collapse resulting in a double stranded break in the daughter strand 4. Lesion bypass: this is where the replication fork jumps over the lesion and leaves a gap, restarting the priming process and leaving the gap. This is the same for the leading and lagging strands. This will then go through a gap repair process - 2 through 4 rely on homologous recombination which is a repair mechanism that is very accurate

CQ: Humans have at least 10 trans-lesion synthesis (TLS) polymerases, even though these polymerases all lack proofreading activity and are thus highly prone to errors. This could be due to any of the following, Except: a. it is vital that cells keep growing, especially once they have committed to division and have started replicating their DNA b. some DNA has to divide faster than other DNA and this DNA tends to be less sensitive to mutation c. given the need for DNA replication to complete once it has begun, many TLS polymerases have evolve to deal with the numerous types of lesions that might arise d. lack of proofreading is the "biological price" a species must pay to overcome replication blocks-allowing a few mutant cells to survive

b. Some DNA has to divide faster than other DNA and this DNA tends to be less sensitive to mutation

CQ: Which of the following is NOT true about gene conversion? a. gene conversion does not generate crossovers b. gene conversion does not occur in yeast nor in most other eukaryotes c. gene conversion can be programmed to respond to environmental cues d. gene conversion can be a byproduct of recombination during meiosis or following DSB repair

b. gene conversion does not occur in yeast nor in most other eukaryotes

QC: Despite the fact that they often make up a sizeable percentage of a genome, few transposable elements are actively moving in genomes. This is most likely because: a. such movements take too long and are unlikely to have time to occur during the cell cycle. b. such movements have been selected against as it would risk the destruction and loss of the transposons c. such movements have been selected against as it would damage the host genome. d. all of the above

c. such movements have been selected against as it would damage the host genome.

What is the most important thing about DNA replication?

fidelity

N-beta-glycosyl bond hydrolysis-depurination

hydrolysis occurring between the ribose and the base -being hydrolyzed by water lets a base leave leading to an abasic site

What factors can lead to a DNA mutation?

hydrolysis, oxidation and alkylation - water, oxygen, UV light, X ray, polycyclic aromatic hydrocarbons

Semiconservative

method of replication that implies that each new strand of DNA is half original and half new - daughter DNA needs to be the same copy as the pervious - the DNA duplex splits into two strands and the newly synthesized DNA has one old and one new strand

the lox/Cre system and the FRT/Flp system help[ the DNA to...

replicate and produce more than one copy per cycle

What is fidelity?

the genetic information is passed down from one generation to the next generation without a mistake - this is why DNA is a duplex, because it needs a template.

All DNA replication happens from an elements known as

the origin of replication