Chapter 25: DNA Metabolism Flashcards

Question

Direct Repair

Answer 1

* damage is repaired without removing a base or nucleotide * Pyrimidine dimers result from a UV-induced reaction, and photolyases use energy derived from absorbed light to reverse the damage * one by DNA photolyases * contain 2 cofactors that serve as light-absorbing agents, or chromophores * One of the chromophores is always FADH2 * In E. coli and yeast, the other chromophore is a folate. * not present in the cells of placental mammals * reaction mechanism entails generation of free radicals * PDF pg. 1336

Answer 2

* in certain types of lesions, such as double-strand breaks, double-strand cross-links, or lesions in a single-stranded DNA, the complementary strand is itself damaged or is absent * Double-strand breaks and lesions in ssDNA arise when a replication fork encounters an unrepaired DNA lesion or from ionizing radiation and oxidative reactions * At a stalled bacterial replication fork, there are two avenues for repair In the absence of a second strand * genetic info must come from a separate, homologous chromosome, thus involving homologous genetic recombination * error-prone translesion DNA **error-prone translesion DNA (TLS) / SOS response** * desperation strategy * DNA repair becomes significantly less accurate * part of a cellular stress response to extensive DNA damage known as the SOS response * In bacteria * UvrA and UvrB are present at higher levels * UmuD protein is cleaved to a shorter form (UmuD') * UmuD' complexes with UmuC and RecA to create a specialized DNA polymerase, DNA polymerase V (UmuD'₂UmuC RecA), that can replicate past DNA lesions * umuC & umuD genes * fully induced only late in the SOS response * not activated for translesion synthesis initiated by UmuD cleavage unless levels of DNA damage are high and all replication forks are blocked * mutations resulting from DNA polymerase V kill some cells and create deleterious mutations in others * DNA polymerase IV, is also induced by SOS response * Replication is also highly error-prone * product of the dinB gene * In eukaryotes * Mammals have many low-fidelity DNA polymerases of the TLS polymerase family * most have specialized functions in DNA repair * DNA polymerase η (eta) promotes translesion synthesis primarily across cyclobutane T–T dimers * Few mutations result because it preferentially inserts two A residues across from the linked T residues * frequency of mutation is minimized by the very short lengths (often one nucleotide) of DNA synthesized * DNA polymerases β, ι (iota), and λ, have specialized roles in eukaryotic base-excision repair * all have 5'-deoxyribose phosphate lyase activity in addition to polymerase activity * After base removal by a glycosylase and backbone cleavage by an AP endonuclease, these polymerases remove the abasic site (a 5'-deoxyribose phosphate) and fill in the very short gap

Answer 3

* Homologous genetic recombination * involves genetic exchanges between any two DNA molecules that share an extended region of nearly identical sequence. * actual sequence of bases is irrelevant, as long as it is similar in the two DNAs * site-specific recombination * exchanges occur only at a particular DNA sequence * DNA transposition * involves a short segment of DNA with capacity to move from one location in a chromosome to another

Answer 4

* In bacteria, homologous genetic recombination is primarily a DNA repair process * directed at the reconstruction of replication forks that have stalled or collapsed at the site of DNA damage or during conjugation **Recombination Process** * replication fork encounters a damaged site under repair * one arm of the replication fork becomes disconnected by a double-strand break and the fork collapses * end of that break is processed by degrading the 5' ending strand * PDF pg. 1071 * RecBCD nuclease/helicase binds to linear DNA at a free (broken) end * RecB and RecD * helicases * molecular motors that propel the complex along the DNA * requires ATP * RecB moves 3'→5' along one strand * degrades both strands * cleaves 3'-ending strand more often than 5'-ending strand. * RecD moves 5'→3' along the other * RecC * binds to chi site ((5')GCTGGTGG) on the 3'-ending strand, preventing its further degradation and is looped out creating an extension * enzyme continues to unwind and degrade the 5'-ending strand * RecA * its active form is an ordered, helical filament of up to several thousand subunits that assemble * forms on single-stranded DNA, such as that produced RecBCD enzyme * SSB are normally present and impedes RecA from binding on first few subunits to DNA * RecBCD loads RecA loader, facilitating the nucleation of RecA filament on ssDNA coated with SSB * filaments assemble/disassemble in 5'→3' * promotes central steps of homologous recombination, including DNA strand invasion * enhancement declines as the distance from * the 3' end invades intact duplex DNA connected to the other arm of the fork and pairs with its complementary sequence * This creates a branched DNA structure (a point where three DNA segments come together) * branch can be moved (branch migration) promoted by RuvAB to create an X-like crossover structure known as a Holliday intermediate * Once a Holliday intermediate has been created, it can be cleaved by a specialized nuclease called RuvC * nicks are sealed with DNA ligase * PDF pg. 1070 **Replication Fork Reassembles** * After recombination steps are completed the replication fork reassembles in a process called origin independent restart of replication * Four proteins (PriA, PriB, PriC, and DnaT) act with DnaC to load the DnaB helicase onto the reconstructed replication fork * DnaG primase synthesizes an RNA primer * DNA polymerase reassembles on DnaB to restart DNA synthesis * requires DNA polymerase II (role unknown) this polymerase II activity gives way to DNA polymerase III

Answer 5

* In eukaryotes * has several roles in replication and cell division, including the repair of stalled replication forks * occurs with the highest frequency during meiosis * contributes to the repair of several types of DNA damage * provides a transient physical link between chromatids promoting orderly segregation of chromosomes at the first meiotic cell division * enhances genetic diversity in a population **Process** * PDF pg. 1074 * Step 1: begins with replication of DNA in diploid germ-line cell * textbook example has 4 homologous chromosomes * Step 2: Chromosomes duplicate * creates sister chromatids * held together by cohesins at their centromeres * Step 3: homologous chromosomes pair up * synapsis begins (very close pairing association) at the centromeres * three-part synaptonemal complex develops between homologous chromosomes holding them 2gether forming tetrads * each paired but not fused homologous pair has 4 chromatids * recombination, crossover occurs * not a random process; “hot spots” have been identified * frequency * roughly proportional to the distance between two points * allows determination of the relative positions of and distances between different genes * double-stranded DNA in each inner chromatid unwinds, is cut & repaired → segments of nonsister chromatids are exchanged  * two models: Holliday and double-strand-break model * double strand break model * synapsis breaks down * DNA brakes closed up by joining each broken end to segments of non-sister chromatid (paternal chromatid is joined to a piece of maternal chromatid) * These points of crossing over become visible as chiasmata after synaptonemal complex disassembles & homologs move slightly apart from each other * centromeres of paired chromosomes move apart * the two homologs remain attached at each chiasma * Step 4: First meiotic division * homologous pairs of chromosomes align along the metaphase plate * orientation of each pair w/respect to others is random → Mendel's Law of Independent Assortment * homologous pairs disjoin & move to opposite poles * If improper spindle fiber attachment occurs, a cellular kinase senses lack of tension, spindle attachments are removed and cell tries again * sister chromatids remain attached and travel together * Step 5: nuclear envelope reforms and 2 haploid cells * Step 6: Second meiotic division * chromosomes line up on metaphase plate * sister chromatids facing opposite poles * sister chromatids separate & pulled to opposite poles * each now a chromosome * Step 7: Meisosis II results in 4 unique haploid gametes * chromosome # has doubled **meiosis I - Prophase I** * has 5 stages: leptotene, zygotene, pachytene, dyplotene, diakinesis * after 5 stages, nuclear membrane breaks down & spindle forms * Leptotene * "thin thread" * chromosomes condense & become visible * homologous chromosomes come closer to each other * Double-strand breaks are introduced and processed * zygotene * “paired thread” * chromosomes continue to condense * homologous chromosomes pair up and begin synapsis (very close pairing association) * bivalent or tetrad: each paired but not fused homologous pair has 4 chromatids * pachytene * “thick thread” * pairing is complete * condensation continues * recombination, crossover * three-part synaptonemal complex develops between homologous chromosomes holding them 2gether * DNA synthesis happens during homologous recombination/cross-over * requires double-stranded DNA in each inner chromatid to unwind, be cut & repaired → segments of nonsister chromatids are exchanged * two models: Holliday and double-strand-break model * diplotene * broken end to segments of non-sister chromatid (paternal chromatid is joined to a piece of maternal chromatid) * These points of crossing over become visible as chiasmata after synaptonemal complex disassembles & homologs move slightly apart from each other * diakenisis * "moving apart" * centromeres of paired chromosomes move apart * the two homologs remain attached at each chiasma

Answer 6

* PDF pg. 1075 * step 1 * homologous chromosomes are aligned * a double-strand break in a DNA molecule is created * exposed ends are processed by an exonuclease * Strands with 3' ends are degraded less than those with 5/ ends, producing 3' single-strand extensions with a free 3'-hydroxyl group at the broken end * similar to "sticky ends" * Each of the 39 ends can then act as a primer for DNA replication * step 2 * and exposed 3' ends invade the intact duplex DNA of the homolog * it pairs with its complement in the intact homolog * The other strand of the duplex is displace * step 3 * invading 3' end is extended by DNA polymerase & branch migration * this generates a DNA molecule with two crossovers in the form of branched structures called Holliday intermediates * step 4 * *exposed non-invading* 3' end aligns w/displaced strand from intact duplex DNA homolog * *exposed non-invading* 3' is extended by DNA polymerase (using displaced strand as a template) * this replaces DNA missing from the original double strand break * step 5 * Holliday intermediate resolvases cleave the Holliday intermediate * RuvC-like nuclease * This generates two recombination products * If cleaved one way, the DNA flanking the region containing the hybrid DNA is not recombined * Product set 1 (Figure 25-35a, pg. 1075) * Picture cleavage of Holliday horizontally down the middle between the two homologs * if cleaved the other way, the flanking DNA is recombined * Product set 2 (Figure 25-35a, pg. 1075) * Picture the recombination of all homologs staying the same as the Holliday * Only the ends are recombined * The right ends of the top two chromatids combine with the left ends of the bottom two chromatids * The left ends of the top two chromatids combine with the right ends of the bottom two chromatids * middle portions stay the same

Answer 7

* Results in Precise DNA Rearrangements * limited to specific sequences * occur in virtually every cell * recombinase * acts on the recombination site * short (20 to 200 bp), unique DNA sequence * partially asymmetric (nonpalindromic) * align in the same orientation during the recombinase reaction * If the two sites are on the same DNA molecule * reaction either inverts or deletes the intervening DNA * determined by if recombination sites have the opposite or same orientation * PDF pg. 1079, Figure 25-38.a * If the sites are on different DNAs * intermolecular recombination * if one or both DNAs are circular, the result is an insertion * PDF pg. 1079, Figure 25-38.b * acts as a site-specific endonuclease and ligase * One or more auxiliary proteins regulate the timing or outcome of the reaction * two general classes: Rely on either Tyr or Ser residues in the active site * In both types of system, the exchange is always reciprocal and precise, regenerating the recombination sites **Tyr Reaction** * PDF pg. 1078, Figure 25-37 * step 1 * Tyr residues act as nucleophiles at the active site * reaction is carried out within a tetramer of identical subunits * Recombinase subunits bind to a specific sequence on recombination site * step 2 * One strand in each DNA is cleaved at particular points in the sequence * nucleophile is the —OH group of an active-site Tyr residue * the product is a covalent phosphotyrosine link between protein and DNA * step 3 * After isomerization , the cleaved strands join to new partners, producing a Holliday intermediate. * step 4 * a second reaction similar to the first occurs * The original sequence of the recombination site is regenerated after recombining the DNA flanking the site **Ser Reaction** * PDF pg. 1078 * both strands of each recombination site are cut concurrently and rejoined to new partners without the Holliday intermediate **On a circular bacterial chromosome** * PDF pg 1079, Fig. 25–39 * sometimes generates deleterious byproducts * resolution of a Holliday intermediate at a replication fork by a nuclease (RuvC) followed by completion of replication can produce either * two monomeric chromosomes * or a contiguous dimeric chromosome * covalently linked chromosomes cannot be segregated to daughter cells at cell division * XerCD system, converts the dimeric chromosomes to monomeric chromosomes so that cell division can proceed

Answer 8

* segments of DNA * found in virtually all cells * transposition * move, or “jump,” from one place on a chromosome to another on the same or a different chromosome * target is random * DNA sequence homology is not required * tightly regulated cuz insertion in essential gene can kill cell * molecular parasites, adapted to replicate passively within the chromosomes * bacteria have 2 classes * Insertion sequences / simple transposons * contain only the sequences required for transposition and the genes for the proteins (transposases) that promote the process * Complex transposons * contain one or more genes in addition to those needed for transposition * extra genes might confer additional capabilities likes resistance to antibodies * vary in structure * have short repeated sequences at each end that serve as binding sites for the transposase * In Eukaryotes * structurally similar to bacterial transposons * some use similar transposition mechanisms * others seem to involve an RNA intermediate **Transposition Pathway** * Direct or simple tranposition * PDF pg. 1080, Fig. 25–40 * cuts on each side of the transposon excise it * transposon moves to a new location * the double-strand break in the donor DNA that must be repaired * At the target site, a staggered cut is made * transposon is inserted * a short sequence at the target site (5 to 10 bp) is duplicated to form an additional short repeated sequence that flanks each end of the inserted transposon * These duplicated segments result from the cutting mechanism used to insert a transposon into the DNA at a new location * sequences are generally only a few base pairs long * replicative transposition * the entire transposon is replicated, leaving a copy behind at the donor location * cointegrate is an intermediate in this process, consisting of the donor region covalently linked to DNA at the target site * Two complete copies of the transposon are present in the cointegrate, both having the same relative orientation in the DNA * In some well-characterized transposons, the cointegrate intermediate is converted to products by site-specific recombination, in which specialized recombinases promote the required deletion

Answer 9

* PDF pg. 1336, Figure 25-42 * A human (like other mammals) is capable of producing millions of different immunoglobulins (antibodies) with distinct binding specificities, even though the human genome contains only ,29,000 genes * Recombination allows an organism to produce an extraordinary diversity of antibodies * the generation of complete immunoglobulin genes from separate gene segments is one example * The genes for these polypeptides are divided into segments, and the genome contains clusters with multiple versions of each segment * The joining of one version of each gene segment creates a complete gene

Chapter 25: DNA Metabolism Flashcards

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