Quiz 15 Study Guide: Ch 34 & 35- DNA replication and Repair Ch 34: DNA Replication ∙ Different possible models of DNA replication o Conservative- parent strand is retained fully o Semi-conservative- mix of parent and new strands ▪ Each strand oDon't forget about the age old question of tenseme
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f DNA acts as a template for synthesis of a new strand ▪ Daughter DNA contains one parental and one newly synthesized strand o Dispersive- pieces of parental DNA are retained and dispersed in new strands ∙ Meselson Stahl Experiment to test o Grew bacteria in heavy N15 and light N14 o Run through Cl gradient- heavy DNA will settle at the bottom o Then they took bacteria grown in heavy nitrogen and drew them in N14 and allowed DNA to replicate o After 1 rounds of replication: 1 heavy strand and one light strand o 2 rounds: 2 molecules of mixed heavy/light DNA and 2 molecules of light DNA o Conclusion: E. coli DNA replication is semi-conservative o He said you should be able to understand the graphical results on the following slide:∙ Replication involves initiation, elongation, and termination o E. coli is circular, double-stranded DNA o Replication begins at a unique site (origin) o Proceeds bidirectionally until the two replication forks meet at the termination site o Replisome- protein machinery for replication ▪ 1 replisome at each of the 2 replication forks ∙ DNA polymerases o Require a template o Need a 3’ OH as primer o Polymerization runs in the 5’ to 3’ direction o Requires all 4 activated precursors- dATP, dGTP, dCTP, and TTP- as well as Mg++ o Many polymerases can correct mistakes by removing mismatched nucleotides ∙ Polymerization is a Nucelotidyl-group-transfer-reaction ∙ 3’ OH of 3’ nucleotide makes nucleophilic attack on alpha-phosphate in NTP– releases PPi (pyrophosphate) which is hydrolyzed to catalyze phosphodiester-linkage formation ∙ E. coli has 5 DNA polymerases, but 3 are key to promoting formation of phosphodiester linkages joining units of the DNA backbone o DNA polymerase I: repairs DNA and participates in DNA synthesis of lagging strand by degrading & replacing RNA primer o DNA polymerase II: role in DNA repair o DNA polymerase III: major DNA replication enzyme, responsible for chain elongation DNA polymerase I o Moderately processive: adds about 20 residues then dissociates ▪ Processive=enzyme stays bound to DNA and polymerizes completely ▪ Distributive= enzyme continually binds and unbinds o E. coli DNA Polymerase I has 3 different active sites on a single polypeptide chain o Activities of DNA Polymerase I: ▪ 5’ ???? 3’ polymerase: adds dNTPs to a 3’ end ▪ 3’ ???? 5’ exonuclease: proof-reading, detects mistakes∙ Can correct mismatched base pairs by looking at H-bonds and analyzing shape ▪ 5’ ???? 3’ exonuclease: editing, degrades RNA primers o A protease can cleave off the active site for 5’ ???? 3’ exonuclease activity: that’s the small fragment ▪ Large fragment has the other 2 activities (3’????5’ exonuclease and polymerase) DNA Polymerase III is the replication enzyme in E. coli o Activities of III: ▪ 5’ → 3’ polymerase ▪ 3’ → 5’ exonuclease ▪ Lacks 5’ ???? 3’ exonuclease o Alpha subunit catalyzes activity, beta subunit clamps to DNA (making it processive) o Because it is highly processive due to b-subunits clamping to DNA, you need fewer polymerase molecules for complete replication, and allows rapid rate of replication ∙ Replication occurs in the 5’ to 3’ direction even though it looks like it grows 3’ to 5’ ∙ Okazaki fragments: lagging strand is synthesized in short pieces in opposite direction of fork mvmt, but still in the 5’ to 3’ direction. All DNA polymerases synthesize 5’ to 3’ o RNA primer is used to start the lagging strand ▪ Primosome is complex containing primase enzyme that synthesizes short pieces of RNA that are complimentary to one of the template DNA strands at the replication fork o DNA pol can’t begin without existing 3’ OH ∙ Leading strand is made in direction of fork, beginning at origin and ending at termination site∙ Replisome = primosome + DNA polymerase Lagging strand loops back through the replisome so the leading and lagging strand are synthesized in the same direction: “trombone slide model” ∙ Single stranded binding proteins bind to single stranded DNA to make sure lagging strand doesn’t form into a weird second structure ∙ Helicase breaks H bonds between base pairs so you can separate the strands o Requires ATP: hydrolyze it to provide energy to unwind DNA o This adds tension: supercoils o Topoisomerases then relieve supercoiling, but they are not part of the replisome ▪ Ex: topoisomerase II, gyrase, utilizes free energy of TP hydrolysis to add negative supercoils to DNA o IGNORE METHOD OF HELICASE IN BOOK ∙ Joining of Okazaki fragments by DNA pol I and DNA ligase o Polymerase I has 5’????3’exonuclease to digest and remove RNA primer. As it does so, it catalyzes polymerase to fill gaps with DNA o DNA ligase catalyzes phosphodiester bond, uses ATP for energy ∙ Errors o DNA pol III uses its 3????5’ exonuclease activity to remove mispaired nucleotide before polymerization continues ▪ Recognizes distortion in DNA caused by incorrect pairs: there will be kink in helix that enzyme can recognize ▪ Very low error rate: 10^-7 ▪ Lowest error rate of any enzyme but mistakes still do occur ∙ DNA replication in E coli requires DNA Pol I and III 1. DNA supercoiling is relieved ahead of & behind replication fork by topoisomerases (not part of replisome) 2. Replication fork is site of simultaneous unwinding (by helicase) and DNA synthesis (DNA polymerase III) + single stranded binding protein, (SSB tetramers); all part of replisome 3. Primase synthesizes the RNA primer (~10 nucleotides); also part of replisome 4. DNA Pol III synthesizes new DNA in 5’→3’ direction using parental strand as template 5. DNA Pol I removes RNA primer (5’→3’ exonuclease) and fills in gap (5’→3’ polymerase); not part of replisome 6. DNA ligase joins ends of daughter strands (i.e. closes nick); not part of replisome Initiation and Termination of DNA replication ∙ Proteins bind to terminating region and block DNA helicase so the replication fork cannot continue ∙ Replisome assembles at origin site (oriC) ∙ Initial assembly depends on local unwinding of the DNA caused by binding certain proteins ∙ DnaA is one initiation protein ∙ Terminator utilization substance (Tus) binds to the ter site ∙ Tus inhibits helicase activity and thus prevents replication forks continuing through this regionDNA replication is similar in eukaryotes and prokaryotes ∙ Differences o Replication is slower in eukaryotes than prokaryotes because histones slow down the replication fork o Eukaryotes have at least 15 DNA polymerases α) initiation δ) repair and recombination, ε) major DNA replication β) repair γ) mitochondrial DNA replication o Eukaryotic chromosomes are large linear, double-stranded DNA molecules o Replication is bidirectional and initiates at specific origins on each chromosome o Occurs during a specific phase of the cell cycle called “S” (G1-S-G2-M) o No specific termination sites o Topological constraints (chromatin) present o nucleosomes disassemble then reassemble after replication. New ones also have to be added since there is more DNA to form chromatin. o The “end problem” in eukaryotes is that you cannot completely replicate the ends because DNA polymerase requires a primer. Not all terminal gaps become filled because the RNA primer is removed but there is no way to replace it with DNA. ▪ Every time you replicate, the ends of your chromosomes become degraded because they are exposed to exonucleases ▪ Eventually you lose a gene that is essential and your cell dies ▪ The more you replicate, the shorter the ends get: molecular clock ▪ E. coli doesn’t have this problem since it’s circular ▪ At the end of chromosomes of all eukaryotes is telomere: non-essential junk DNA that is a repetitive sequence of AGGGTT. Synthesized by enzyme called telomerase. At some point, our body turns off telomerase, so we can’t continue to protect the ends of our chromosomes and they become more degraded over time. Our body wants a molecular clock ∙ Cancer cells turn telomerase back on to lengthen telomeres to continue replicating against the molecular clock ∙ Telomerase has a protein component that catalyzes elongation but and RNA portion that provides the template. It’s a protein-RNA complex Ch 35- Replication and Repair ∙ Replication- maintain genetic information from generation to generation o Rapid, accurate ∙ DNA repair mechanisms o DNA is the only cellular macromolecule that can be repaired because the cost to the organism of damaged DNA far outweigh the energy of repair ▪ Our bodies don’t repair proteins or RNA because we can just degrade them and make more o Errors in replication that are not corrected by proofreading need to be corrected o DNA damage also needs to be corrected that can occur from base modifications, nucleotide deletions, and x-link DNA strands ▪ Smoking and high oxygen environments damage our DNA. “Oxygen catastrophe” oxygen is poisonous, can form superoxide radicals, and it introduces oxidative stress to nucleic acids1. Mismatch-repair: errors in replication not corrected by proofreading a. The complex of proteins Mut S,L,H find and repair mismatched nucleotides b. Mut H is the endonuclease- it cuts the newly synthesized strand c. DNA polymerase III fills in the gap with the correct base pairs d. In E. coli, the S and L proteins can recognize which strand is the newly synthesized strand because the parent strand is methylated but the new strand is not yet modified ∙ DNA damage o Specific repair enzymes scan DNA to detect alterations o 3 types of repair: 1. Direct Repair- do not need to break phosphodiester backbone of DNA ▪ Photolyase enzyme recognizes thymine dimer, binds, and uses visible light to catalyze reversal of cross-link and breaking of dimer, restoring normal base pairing and repairing the DNA ∙ Photodimerization o Double-helical DNA is very sensitive to damage by UV light o Dimerization of adjacent pyrimidines in a DNA strand is common (e.g. thymines) o Replication cannot proceed in the presence of pyrimidine dimers (template strand is distorted) o Thymine dimers are repaired in all organisms 2. Base excision repair ▪ DNA glycosylase removes the faulty base, you’re left with an apurinic/apyrimidic site, AP endonuclease cuts backbone adjacent to missing base, deoxyribose phosphodiesterase removes the deoxyribose phosphate, Polymerase I adds correct nucleotide, DNA ligase seals ▪ DNA can be damaged by alkylation, methylation, deamination, bulky adducts (cigarette smoke) ▪ excision-repair pathways can repair many of these defects ▪ oxidizing guanine residues or deaminating bases lead to mismatched base pairs ▪ methyl group added to C residues regulates gene expression. When deamination occurs, the methyl-C becomes thymine. So a C:G base pair becomes T:A ▪ When C becomes deaminated, it becomes U. Because this occurs naturally all the time, your body couldn’t recognize if a U naturally present in DNA would be correct or a deaminated C. That is why uracil is not used in DNA. Methyl group on T is a tag that distinguishes T from deaminated C. 3. Nucleotide Excision Repair ▪ Not all damaged bases are recognized by DNA glycosylase ▪ Damage is recognized, complex of proteins called UvrABC endonuclease cleaves the backbone and excises about 12 nucleotides, DNA polymerase fills in correct bases, DNA ligase seals it off ∙ Why do you need mismatch repair and nucleotide excision repair if they are doing similar things? o Mismatch repair used when errors during replication occur that are not corrected o Nucleotide excision repair is used if damage happens after replication occurs (environment, UV light, oxygen) Quiz Review: ∙ Replisome is protein machinery responsible for replication, and there is 1 at each replication fork o Know components of replisome ▪ 2 DNA Polymerase III core complexes ▪ Primosome ∙ Primase ▪ Helicase- unwinds DNA ▪ Single stranded binding proteins o Topoisomerases (not part of replisome) ∙ DNA Polymerase III is the main replication machinery, whereas DNA Polymerase I is useful for joining Okazaki fragments o Know the steps of DNA Polymerase I and its activities ▪ 5’ ???? 3’ exonuclease removes RNA primer, at the same time, 5’ ????3’ polymerization activity synthesizes new DNA in the 5 ‘to 3’ direction ∙ Go over components of Polymerase III o See blue-green image of the subunits above^ o B-subunit clamps to DNA that makes the processive nature of the enzyme- allows enzyme to remain bound to DNA in replication ▪ Allows for less enzyme needed for complete replication and rapid replication rate ∙ Meselson Stahl experiment o Understand the experiment and how it distinguished between the 3 proposed models o Why the data they got explains the conclusion o Know and understand the orange and red slide about the experiment above ^ ▪ Seriously, know this. Pay attention to the chart on the slide that shows how the data is presented ∙ Replication o Requires existing 3’ OH to begin o What phosphate would you need to label if you wanted to label DNA? Alpha position because pyrophosphate is released but the alpha phosphate is attacked by the 3’ OH o If you didn’t have a 3’OH, replication could not take place. This is a chain terminator (ddNTP) aka 2’,3’-dideoxy analog ▪ Fred Sanger sequenced insulin in 1958 and sequenced DNA in 1980- won 2 Nobel Prizes ∙ He figured out if you add dideoxy, you will terminate the reaction o ddATP= reaction stops everywhere there’s an A o ddCTP = terminates every time there’s a C, etc. o ddGTP o ddTTP ▪ make a gel of this and that’s how you sequence DNA ▪ you still need some regular dA(or any letter)TP with the ddA(or any letter)TP or else all the products would end at the first A. This way, the chain is extended until all of the A’s are terminated, resulting in different sized fragments that all end in A.∙ AZT is a drug used to treat HIV o It’s a chain terminator that has a 3’ azido group (N3) instead of a 3’ OH. o It blocks polymerization of the virus, but does not terminate elongation of all polymerases. ▪ HIV reverse transcriptase does not have proofreading ability so it can’t make DNA from its RNA ▪ Most polymerases in our body can recognize this and proofread by cleaving it and moving on, but this process contributes to the drug’s side effects o AZT has the nitrogen added because the charge of N can get into your cell’s membranes easier (N increases lipophilic nature) o Mitochondrial polymerase can be somewhat affected because it has ancient polymerases that aren’t as efficient as removing this chain terminator???? leads to side effects o 100-fold greater affinity for viral reverse transcriptase than your own enzymes o AZT is not a cure- not eliminating virus. But you are reducing the virus’s ability to replicate so you minimize its growth. ∙ The end replication problem o At the end of eukaryotic chromosomes, when the RNA primer is removed there is a gap. So information is lost every time we replicate. To solve this problem, our bodies add telomeres to the end of our chromosomes, which are non-coding “junk” sequences. So we only lose “nonessential” DNA in replication ∙ Why is thymine used in DNA instead of uracil? o C naturally, spontaneously becomes deaminated to form uracil. Because that happens, the body needs to be able to recognize this damaged DNA. The enzyme uracil DNA glycosylase excises uracil from DNA when it sees it because it knows it does not belong. o RNA still has U. The C in RNA also get naturally converted to U. RNA kept U because the energy to repair RNA is not worth it. It doesn’t need to be corrected, you can just degrade it and make more. Potential errors don’t lead to lasting damage, RNA is short-lived. ∙ PCR Polymerase Chain Reaction o Kary Mullis won Nobel prize for this in 1993 o PCR: a repetitive method that yields a ~ 106 fold amplification of a specific DNA sequence. Can detect as little as one DNA molecule o Components: a pair of primers that hybridize with the flanking sequences of the target, all 4 deoxyribonucleoside triphosphates (dNTPs), and a heat-stable DNA polymerase o Steps: 1) Heat a double stranded piece of DNA to 95° to separate 2) Cool to 55° to allow hybridization of primers 3) Heat to 75° to allow Taq polymerase to elongate both primers. o Second cycle: do it again. Separate strands, hybridize primers, and amplify. o After 25 cycles, you amplify 106 fold o We use the enzyme Taq polymerase in PCR because it can withstand being heated. It comes from Thermus aquaticus bacteria that live in hot water environments, like in hot-water springs