Chapter 5 Essays

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Chapter 5 Essays
2015-02-05 23:48:22
Test One
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  1. Explain the discovery of OF?
    a.       Using tritium thymidine in a pulse-chase experiment, they noticed that the most recently replicated DNA of the pulse (just behind the replication fork) became radiolabeled. There was one large band and several smaller bands, indicative of fragments. 
  2. What is the first proofreading step? 
    It occurs before a new nucleotide is added to the growing chain. The correct nucleotide has a higher affinity for the moving polymerase than does the incorrect nucleotide, because the correct pairing is more energetically favorable. Also, after nucleotide binding, but before covalent addition of the nucleotide, the enzyme must undergo a conformational change in which its fingers tighten around the active site. This allows double checking of the exact base-pair geometry before catalytic addition
  3. 1)      What is the second error-correcting reaction? 
    a.       It is exonucleolytic proofreading, which takes place right after an incorrect nucleotide is covalently added to the growing chain. Because a 3’—OH end is required for elongation, those DNA moleucles with mismatched nucleotides are not effective as templates and cant be elongated. So, DNA polymerase corrects the mismatched pair viaa separate catalytic site. This 3’-to5’ proofreading exonuclease clips off any unpaired residues at the primer terminus, continuing until enough nucleotides have been removed to regenerate a coorectly base-paired 3’—OH. 
  4. 1)      Why can’t DNA replication occur in the 3’-to5’ direction? 
    a.       If there were additions in this direction, the growing 5’ end would provide the phosphates needed for covalent linkage; and, any mistakes can’t be hydrolyzed away because it will terminate DNA synthesis. Corrections can only occur in the 5’—to—3’ end. 
  5. 1)      Why is an erasable RNA primer preferred to a DNA primer? 
    a.       It is because the enzyme that starts chains anew can’t be efficient at self-correction. And since it is already marked as suspect, it can be efficiently removed and replaced. 
  6. 1)      What is used to prevent rapid dissociation of DNA polymerase on the leading strand? 
    a.       An accessory prtotein called a sliding clamp keeps the polymerase firmly on the DNA and releases it as soon as the polymerase runs into a double-stranded region of DNA. It forms a large ring around the DNA double helix. One side binds to the back of the DNA polymerase, and the whole ring slides freely along the DNA as the polymerase moves. The assembly of the clamp around the DNA requires ATP hydrolysis by a special protein complex, the clamp loader, which hydrolyzes ATP as it loads the clamp ont to a primer-template junction
  7. 1)      Though the enzymes involved in DNA replication are spoken about as if they were independent, what is really the case? 
    a.       In reality, most of the proteins are held together in a large and orderly multienzyme complex that rapidly synthesizes DNA. It remains stationary while the DNA is wound throughit. At the front of the fork, DNA helicase opens up the helix. Two DNA polymerase molecules work at the fork on each strand. This close association increases efficiency of replication. This is possible due to folding back of lagging strands. It also facilitates the loading of the polymerase clamp each time an OF is synthesized. 
  8. 1)      What was the experiment done for determination of eukaryotic origins of replication? 
    a.       Human cells were labeled with tritium thymidine so that the DNA synthesized became highly radioactive. They were then lysed and streaked on the surface of a glass slide coated with a photographic emulsion. Autoradiography reveals the pattern of labeled DNA. The replicated DNA can be detected in the LM as lines of silver grains. This was done via a pulse-chase experiment. In other cases, the pulse was done; then, the chase was done after getting rid of the radioactive thymidine. The results showed tapering off of the silver grains, showing that they were moving bidirectionally form the central replication origin. A replication fork stops only when encoutneing a replication fork moving in the opposite direction.  
  9. 1)      What did further experimentation show regarding replication origins? 
    • a.       They tend to be activated in clusters, called replication units.
    • b.      New replication units seem to be activated at different times during the cell cycle until all of the DNA is replicated.
    • c.        Within a replication unit, individual origins are spaced at intervals 30,000-250,000 nucleotide pairs from one another.
    • d.      Replication forks are formed in pairs and create a replication bubble as they move in opposite directions away from the common point of origin, stopping only when they collide head-on with a replication fork moving in the opposite direction. 
  10. 1)      When exactly does DNA replication occur? 
    a.       It occurs in the S phase of the cell cycle. It lasts for about 8 hours. By the end, each chromosome has been replicated to produce two complete copies, which remain joined until the M phase. 
  11. 1)      What is suggested by the 8 hour time length of the S phase? 
    a.       The replication origins are not all activated simultaneously and DNA in each replication unit is replicated during only a small part of the total S-phase interval. 
  12. 1)      What else can replication depend on? 
    a.       It can depend on chromatin structure. Heterochromatin is condensed and is replicated late in S phase. Euchromatin is less condensed and allows transcription. Regions of te genome whose chromatin is least condensed are replicated first. 
  13. 1)      Explain the experiment for determining the fact that replication origins are well-defined DNA sequences. 
    a.       Yeast cells were lacking the ability to produce histidine were placed on a medium that lacked histidine. A cicular plasmid containing histidine was introduced into the mutant yeast cells, allowing them to synthesize histidine. They also chopped up the genome and inserted it into yeast cells as well. The yeast cells that carried the plasmid containing an origin of replication were able to proliferate because they passed on the essential gene. Those that received a plasmid with no origin of replication were unable to pass on the histidine producing gene to their progeny. 
  14. 1)      What is the effect of deleting various replication origings on chromosome III, for example? 
    a.       Removing a few origins has little effect because replication forks that begin at neighboring origins of replication can continue into the regions that lack their own origins. The deletion of more replication origins results in the loss of the chromosome as the cell divides because it is replicated too slowly. 
  15. 1)      What does each DNA sequence that can serve as an origin of replication have? 
    • a.       A binding site for a large, multisubunit initiator protein called ORC, for origin recognition complex
    • b.      A stretch of DNA that is rich in As and Ts and therefore easy to unwind
    • c.       At least one binding site for proteins that help attract ORC to the origin DNA
  16. 1)      In eucaryotes, how is the process regulated to ensure that all the DNA is copied once and only once? 
    • a.       The answer lies in the way the ORC complex is activated and deactivated. The ORC-origin interaction persists throughout the entire cell cycle, dissociating only briefly immediately following replication of the origin DNA, and other protiens that bind to it regulate origin activity.
    •                                                               i.      These include the DNA helicase and two helicase loading proteins, Cdc6 and Cdt1, which are assembled onto an ORC-DNA complex to form a prereplicative complex at each origin during G1 phse.
    •                                                             ii.       The passage of a cell from the G1 to S phase is triggred by activation of the protein kinases (Cdks) that lead to dissociation of the helicase loading proteins (degrading phosphorylated Cdc6), activation of the helicase, unwinding of the origin DNA, and loading of the remaining replication proteins including DNA polymerases. The protein kinases that trigger DNA replication simultaneously prevent all assembly of new prereplicative complexes until the next M phase resents the entire cycle b phosphorylating ORC.
    •                                                           iii.      This provides a single window of opportunity for prereplicative complexes to form and a second window for them to be activated and subsequently disassembled. Each origin of replication can fire once and only once a cell cycle.  
  17. 1)      Aside from DNA replication, what does chromosome duplication require? 
    a.       It also requires the synthesis and assembly of new chromosomal proteins onto the DNA behind each replication fork. The cell requires a large amount of new histone protien, approximately equal in mass to the newly synthesized DNA, to make the new nucleosomes in each cell cycle. For this reason, most eukaryotic organisms possess multiple copies of the gene for each histone. 
  18. 1)      Explain this new nucleosome process.
    • a.       Histones are synthesized mainly in S phase, when the level of histone mRNA increases as a result of increased transcription and decreased mRNA degradation.
    • b.      The major histone mRNAs are degraded within minutes when DNA synthesis stops at the end of S phase.
    • c.       As the replication fork advances, it must pass through the parental nucleosomes. To replicate chromosomes efficiently in the cell, chromatin-remodeling proteins, which destabilize the DNA-histone interface, are required. Aided by such complexes, replication forks can transit even highly condensed heterochromatin efficiently.
    • d.      As the replication fork passes through chromatin, most of the old histones remain DNA0bound and are distributed to daughter DNA helices behind a replication fork. But, because new histones are needed, the histone octamer is broken into an H3-H4 tetramer and two H2A-H2B dimers.
    •                                                               i.      The H3-H4 tetramers remain associated with DNA and are distributed at random to one or the other daughter duplexes, but the H2A-H2B dimers are released from the DNA and are added later to complete the nucleosome
    • e.      Histone chaperones help with this method. 
  19. 1)      After histones are made, what must be done? 
    a.       There must me modifications to enable the DNA histones to resemble the parental histones. Each daughter chromosome is seeded with the memory of the paternal pattern of H3 and H4 modificaiton. Once nucleosome assembly is complete, the parental patterns of H3-H modification are reinforced through histone modification enzymes in reader-writer complexes that recognize the same type of modification they create
  20. 1)      Aside from telomerase, what else happens at the end of the replication? 
    a.       Because the 3’ DNA is longer than the 5’ end, the protruding end loops back and tucks its single-stranded terminus into the duplex DNA of the telomeric repeat sequence to form a t-loop
  21. 1)      What is the belief with telomerase? 
    a.       It is believed that telomere repeats provide each cell with a counting mechanism that helps prevent the unlimited proliferation of wayward cells in adult tissues. Some cells retain full telomerase activity; others are lowered. After ecll generations, a process called replicative cell senescence may cause some cells to inherit defective chromosomes and withdraw from the cell cycle. Telomerase activity may be linked to aging. This was tested on mice, the later generations of which had more defects and premature aging. 
  22. 1)      What was the experiment to determine telomere length is a measuring stick?
    a.       The experiment used the telomere at one end of a particular chromosome, artificially making it longer or shorter. After many cell divisions, the chromosome recovers, showing an average telomere length and a length distribution that is typical of the other chromosomes in the yeast cell. 
  23. 1)      What are some mutations that occur spontaneously every day? 
    a.       About 5000 purine bases (A and G) are lost because their N-glycosyl linkages to deoxyribose hydrolyze, called depurination. Similary, deamination of cytosine to uracil in DNA occurs at a rate of 100 bases per cell per day. 
  24. 1)      What can be done to fix these changes?
    a.       Using base excision replair or nucleotide excision repair. 
  25. Compare bse with nucleotide excision.
    a.       They both excise the damage and restore the original DNA sequence using undamaged strands as the template and sealing it with DNA ligase. 
  26. Explain base excision repair.
    • a.       Base exicision repair involves enzymes called DNA glycosylases, each of which recognizes specific types of altered bases in DNA and catalyzes its hydrolytic removal. A key step is an enzyme-mediated “flipping-out” of the altered nucleotide from the helix, which allows the DNA glycosylase to probe all faces of the base for damage. These enzymes travel along the DNA using base-flipping to evaluate the status of each base. Once damage is recognized, it removes the base from its sugar.
    •                                                               i.      The missing tooth created by DNA glycosylase is recognized by AP endonuclease, which, along wiith phosphodiesterase, cuts the phosphodiester backbone, after which the damage is removed and the damage is repaired. DNA polmyeriase adds new nucleotides and DNA ligase seals the nick. 
  27. Explain nucleotide excision repair.
    a.       Nucleotide excision repair can repair damage caused by any large change in the structure of the helix. Such bulky lesions include those created by the covalent reaction of DNA bases with large hydrocarbons. In this pathway, a large multienzyme complex scans th DNA for a distortion in the double helix, rather than for a specific base change. Once it finds a bulky lesion, it cleaves the phosphodiester backbone on both sides of the lesion, and a helicase peels away a single-strand oligonucleotide containing the lesion. DNA polymerase and ligase then repair it. 
  28. What is an additional mechanism that maximizes the effectiveness of their DNA repair enzymes?
    a.       They delay progression of the cell cycle until DNA repair is complete. The orderly progression of the cell cycle is maintained through the use of checkpoints that ensure the completion of one step before the next step can begin. The cycle stops if damage is detected. 
  29. 1)      What guides homologous recombination? 
    a.       DNA base-paring in that it takes place only between DNA duplexes that have extensive regions of sequence similarity as they base pair. 
  30. 1)      What must occur for the DNA hybridization critical for homologous recombination do? 
    • a.       The strand in the helix must steparate. It is formed when a replication fork encounters a DNA nick, falls apart, and an exonuclease degrades the 5’ end at the break, leaving an unpaired single strand at its 3’ end. The single stranded 3’ DNA end is cated on by proteins that allow it invade a homolgous DNA duplex, sch as RecA and Rad51.
    •                                                               i.      RecA binds tightly to single-stranded DNA forming a nucleoprotein filament. With more than one binding ite, it can hold the single strand and the double helix together, allowing it to catalyze the multistep DNA synapsis reaction that occurs between a DNA double helix and a homologous region of single-stranded DNA.
    • 1.       In the first step, the RecA protein intertwines the DNA single strand and the DNA duplex in a sequence-independent manner.
    • 2.       Next, the DNA single strand “searches” the duplex for homologous sequences.
    • 3.       Once located, a strand invasion occurs—the single strand displaces one strand of the duplex as it forms conventional base pairs with the other strandà heteroduplex
  31. 1)      Explain homologous repair.
    • a.       It can repair double-stranded breaks without any loss or alteration of nucleotides at the site of repair. There is a double stranded break. An exonclease degrades the 5’ ends. A strand invasion occurs and branching occurs. DNA synthesis and migration occurs at the branch pint, followed by pairing of newly synthesized DNA with top strnd and top strand DNA synthesis. DNA ligation is the last step.
    • b.      Loss of heterozygosity is a problem tat can occur.
  32. 1)      What is a control over recombination repair? 
    a.       These enzymes are present in high levels and are dispersed throughout the nucleus. In response to DNA damage, they rapidly converge on the sites of damage and form “repair factories” where many DNA lesions are brought together and repaired. This rapid mobilization is controlled, two of which are Brca1 and Brca2. Too much or too little recombination can lead to cancer in humans. 
  33. 1)      How can homologous recombination be used to generate DNA olecules of novel sequence? 
    a.       During this process, a special DNA intermediate forms that contains four DNA strands shared between two DNA helices. In this key intermediate, called Holliday junction (Cross strand exchange), two DNA strands switch partners between two double helices. The Holliday junction can adopt multiple conformations, and a special set of recombination proteins binds to, and stabliziles, the open, symmetric isomer. Via ATP hydrolysis for branch migration, the protiens can move the point at which the two DNA helices are joined rapidly along the two helices. To regenerate two separate DNA helices and end the recombination process, the strands connecting the two helieces in a Holliday junction must be cut, called resolution. 
  34. 1)      Explain meiotic recombination. 
    a.       A specialized protein (Spo11) breaks both strands of the DNA double helix and covalently binds to the broken DNA. A specialized nuclease then rapidly processes the ends bound by Spo11, removing the protein and leaving protruding 3’-single-strand ends. Strand invasions and branch migrations take place that produce an intermediate with two closely spaced Holliday junctions, called a double Holliday junction. These functions are similar. 
  35. 1)      What are the two different ways to resolve the double Holliday junction?
    a.       The original pairs of crossing strands are cut at both Holliday junctions in the same way, which causes the two original helices to separate from one another in a form unaltered except for the region between the two junctions. If, on the other hand, the two Holliday junctions are resolved oppositely, the outcome is more profound and the portions upstream and downstream from the two Holliday junctions are swapped, creating two chromosomes that have crossed over. Few are crossovers. A majority are noncrossovers. 
  36. 1)      What is the reason for the few crossovers? 
    a.       The few crossovers that form are distributed along chromosomes such that the presence of a crossover in one position inhibits crossing over in neighboring regions. This is crossover control. 
  37. 1)      What can lead to gene conversion? 
    a.       First, the DNA synthesis that accompanies the early steps of homologous recombination will produce regions of the double Holliday junction where three copies of the sequence on one homolog are present; these will produce sites of gene conversion once the Holliday junction is resolved. In addition, if the two strands that make up a heteroduplex region don’t have identical nucleotide sequences, mismatched airs will result and can be repaired by the cell’s mismatch repair system with no distinction between paternal and maternal strand, causing one to be lost and another duplicated and leading to the conversion of one allele to another.