Answer7 - MCB 161 Lecture 8 DNA Recombination and...

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Unformatted text preview: MCB 161 Lecture 8: DNA Recombination and positional cloning MBOC Problems Book: Study Problems 5-75 Repair of a double ­strand break by homologous recombination requires an intact homologous chromosome as a template for repair. In a haploid cell in G1, each chromosome is present in only one copy. Thus, when a break occurs in G1, there is no intact homologous template to use for repair. In haploid cells in G2, there are two copies of each chromosome (sister chromatids), so that a broken copy can be repaired from the intact sister chromatid. 5-76 The variable in these experiments is light: the brighter the light, the less the observed killing. Thus, visible light can reverse the effects of UV irradiation. Direct reversal of UV damage is common in microorganisms and is called enzymatic photoreactivation. The enzyme from E. coli has two chromophores that cooperate in capturing photons from sunlight and using their energy to unlink pyrimidine dimers. The account here is not much different from the original discovery of photoreactivation by Albert Kelner in the 1940s. While investigating the effects of postirradiation temperature on UV survival, Kelner was plagued by another variable. In his own words: “Careful consideration was made of variable factors which might have accounted for such tremendous variation. We were using a glass ­fronted water bath placed on a table near a window, in which were suspended transparent bottles containing the irradiated spores. The fact that some of the bottles were more directly exposed to light than others suggested that light might be a factor… Experiments showed that exposure of UV ­irradiated suspensions to light resulted in an increase in survival rate or a recovery of 100,000 ­ to 400,000 ­ fold. Controls kept in the dark… showed no recovery at all.” Reference: Friedberg EC, Walker GC & Siede W (1995) DNA Repair and Mutagenesis, pp 92–103. New York: WH Freeman. 5-96 A. The first labeled restriction fragment to appear after starting the reaction is fragment c, and label appears in the other fragments progressively, with fragment a the last to become double stranded. This order of appearance matches the order of the fragments in the 5 ‘ ­to ­3 ‘ direction on the + single ­ stranded circle shown in Figure 5–38B, starting from the top. Since pairing between DNA strands is antiparallel, invasion must start at the 3 ‘ end of the minus strand of the linear DNA, and branch migration must proceed in the 3 ‘ ­to ­ 5 ‘ direction along the minus strand (Figure 5–65). B. It takes about 20 minutes for the last fragment to become well labeled. This indicates that the rate of movement of the branch point is about 350 nucleotides/minute (7000 nucleotides/20 minutes), or about 6 nucleotides/ second. Compared with the rate of replication, which is about 500 nucleotides/second, the rate of branch migration catalyzed by RecA is very slow. C. The presence of a 500 ­nucleotide ­pair patch of nonhomologous DNA would inhibit branch migration severely. Nevertheless, RecA can catalyze branch migration through such a nonhomology at a low frequency and produce a double ­stranded circle with the nonhomologous DNA looped out as a single strand (Figure 5–65). Reference: Cox M & Lehman IR (1981) The polarity of the recA protein ­mediated branch migration. Proc. Natl Acad. Sci. U.S.A. 78, 6023–6027. 5-98 The double Holliday junction that would result from strand invasion is shown in Figure 5–66. Two versions are shown, both equally correct. The upper one looks simpler because the invading duplex has been rotated so that the marked 5 ‘ end is on the bottom. This arrangement minimizes the number of lines that must cross; which is why most recombination diagrams are shown in this way. The lower representation is perfectly correct, but it looks more complicated. DNA synthesis uses the 3 ‘ end of the invading duplex as a primer and fills the single ­strand gap by 5 ‘ ­to ­3 ‘ synthesis, as indicated. See the link to the on-line problem, found on the “Lecture Info” web page. Answer the below question: 1) Why is Holliday junction formation during DSB repair a dangerous thing for higher eukaryotes? Although your textbook indicates that DSB are most frequently repaired via homologous recombination and the two Holliday junction mechanism, this is only true for organisms with relatively simple genomes, like E. coli and budding yeast. Large genomes (such as those found in mammals and angiosperms) contain large amounts of repetitive ‘junk’ DNA. RecA/Rad51mediated HR relies upon finding relatively small regions of homology between two chromosomes (roughly 100 bp). Although in a compact genome 100 bp of homology would be a good indication that the appropriate homologous chromosome had been found, this is not true in complex genomes. In these types of genomes, this degree of homology is likely to be found by chance in a non-homologous chromosome (see Figure 11-17 to see how repetitive sequences may be distributed across a complex genome; compare the high number of repeated sequences in maize and humans compared to the low number in budding yeast and E. coli). The resolution of a Holliday junction between two non-homologous chromosomes could result in the translocation of two different chromosome arms, potentially resulting in cells with the wrong number of copies of each chromosome arm after the next mitosis. ...
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