121 MCB Spring 2011: Problem Set 5 Solutions 1. Over the years, a comprehensive set of temperature sensitive mutants involved in DNA replication in E. coli has been isolated and studied, defining the dna genes. These mutants can be divided into two classes, based on the rapidity with which their defects are observed following a shift to the non-permissive temperature of 42C. "Quick-stop" mutants cease DNA replication immediately upon a shift to 42C. These mutations generally occur in genes that are involved in the elongation phase of DNA replication and encode enzymes of the replication apparatus (e.g. Pol III ). In contrast, a second class of mutants, the "Slow-stop" mutants, can complete a current round of replication at the non-permissive temperature but cannot start another. These mutations generally reside in genes that are involved in initiating a cycle of replication at the origin (e.g. DnaA). Interesting, mutations have been isolated in DnaC that show either a "quickstop" or a "slow-stop" phenotype. Mutations in DnaC that give rise to a slow-stop phenotype are recessive whereas DnaC fast stop mutants are dominant. Both kinds of mutants affect the ability of DnaC to interact with DnaB, but in fundamentally different ways. What is the explanation for the existence of both fast stop as well as slow stop mutants in DnaC? [Hint: DnaC should interact with DnaB only at OriC, otherwise interactions between these two proteins is deleterious to the process of replication]. Answer: Slow stop DnaC mutants are impaired in their ability to bind to DnaB and recruit the helicase to OriC. Thus, initiation of replication is affected but not elongation. In contrast, the fast stop mutants bind constitutively to DnaB at the non-permissive temperature, and thus prevent the helicase from functioning properly during the elongation phase of replication. This "gain of function" mutation in DnaC is dominant as inappropriate binding to DnaB can occur even in the presence of wild type DnaC protein. In contrast, the slow stop mutants are recessive as they can be rescued if wild type DnaC is provided. 2. True or False: In eukaryotic cells, if an origin of replication is deleted from a chromosome, the DNA on either side will ultimately be lost, as well, because it cannot be replicated. Explain your answer. False. If one origin is deleted, the adjacent DNA, which would normally be replicated from that origin, will be replicated instead from a neighboring origin. Thus, replication of the DNA adjacent to the deleted origin may take somewhat longer than normal, but it will occur. 3. If replication of the human genome had to be accomplished within an 8 hour S phase and the replication fork moved at a rate of 50 bp/second, what would be the minimum number of origins required to replicate the human genome? (Recall that the human genome contains 6.4 x 109 nucleotides). For this answer, we will ignore the differences in the lengths of individual chromosomes and assume only that there are equally spaced origins covering the entire length of genetic material (i.e., let's assume there is only one giant chromosome). Therefore, in 8 hours (28,800 seconds) the two replication forks from one origin could move 2.88 x 106 nucleotides [(2 forks) x (50
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nucleotides/second) x 28,800 seconds]. Thus, to replicate the entire genome, it would take about 2222 evenly spaced origins [(6.4 x 109 nucleotides)/(2.88 x 106 nucleotides/origin)]. It is estimated that there are about 10,000 origins of replication in the human genome, more than enough to finish replication within the allotted time during the cell cycle. 4. True or False: According to the "moving train" model of DNA replication, when bidirectional replication forks from adjacent origins meet, a leading strand always runs into a lagging strand. Explain your answer. True. Consider a single template strand, with its 5' end on the left and its 3' end on the right. No matter where the origin is, synthesis to the left on this strand will be continuous (leading strand), and synthesis to the right will be discontinuous (lagging strand). Thus, when replication forks from adjacent origins collide, a rightward-moving (lagging) strand will always meet a leftwardmoving (leading) strand. 5. Another "quick stop" versus "slow stop" conditional lethal mutant question. 5a) Would a ts mutation in DnaB be expected to result in a quick-stop or slow-stop phenotype? How about a DnaA ts mutant? Answer: DnaB: because it is required for continued movement of the replication ts fork, mutants in this gene result in "quick stop" phenotypes. DnaA: because it is required for initiation of replication only, ts mutants in this gene result in "slow stop" phenotypes. 5b) Cell-free extracts of DNA replication mutants show essentially the same patterns of replication as in intact cells. Thus, extracts from quick-stop ts mutants halt DNA synthesis immediately at 42C, whereas extracts from slow-stop mutants do not stop DNA synthesis for several minutes after a shift to 42C. Suppose extracts from a ts DnaA mutant and a dnaB mutant were mixed together at 42C. Would sort of phenotype would you expect to observe regarding DNA replication with these mixed extracts? Answer: Hopefully by now you can recognize this as a trick question: the phenotype should be WT, since you are mixing extracts that each contain a wild type protein for one the two mutants in question. Of course this answer assumes that the mutations are recessive, and that as long as you have a wild type protein the system could work. However, what if either of the mutant proteins was derived from a dominant mutation? Then what would you observe in vitro at the non-permissive temperature?
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6. True or False. Each time the genome is replicated, half of the newly synthesized DNA is stitched together from Okazaki fragments. Explain your reasoning. True. At each replication fork the leading strand is synthesized continuously and the lagging strand is synthesized as Okazaki fragments. Since half the DNA at each replication fork is stitched together from Okazaki fragments, half the genome must be made this way. 7. How would you expect the loss of the 3'-5' proofreading activity of DNA polymerase in E. coli to affect the fidelity of DNA synthesis? How would its loss affect the rate of DNA synthesis? If the proofreading exonuclease activity of DNA polymerase were lost, you would expect the fidelity of DNA synthesis to be compromised. The proofreading exonuclease accounts for about a factor of 100 in the overall fidelity of DNA synthesis in E. coli, and its loss might be expected to lower overall fidelity by this amount. Loss of proofreading activity would also be expected to affect the rate of DNA synthesis. When a nucleotide is misincorporated, a normal DNA polymerase can quickly remove it with its proofreading activity and then continue on. By contrast, a misincorporated nucleotide might affect a DNA polymerase lacking a proofreading exonuclease more dramatically, since DNA polymerase requires a base-paired primer. Thus, a deficient DNA polymerase might be expected to pause or stall at each misincorporated nucleotide. Because the normal frequency of misincorporation is so low, about 1 in 105, it might be difficult to demonstrate such a rate change in practice. 8. It turns out that the Primase DnaG is a relatively "sloppy" enzyme that makes many mistakes. Eventually the RNA primers made by Primase are replaced by a polymerase (DNA Pol I) that has higher fidelity. Consider the factors that might explain this situation? While the process may seem wasteful, it provides an elegant solution to the difficulty of proofreading during primer formation. To start a new primer on a piece of single-stranded DNA, one nucleotide needs to be put in place and then linked to a second and then to a third and so on. Even if these first nucleotides were perfectly matched to the template strand, such short oligonucleotides bind with very low affinity and it would consequently be difficult to distinguish the correct from incorrect bases by proofreading. The task of the primase is to `just get anything down that binds reasonably well and not worry about accuracy.' Later, these sequences are removed and replaced by DNA polymerase, which uses the accurately synthesized DNA of the adjacent Okazaki fragment as its primer. DNA polymerase has the advantage--which primase does not have--of putting the new nucleotides onto the end of an already existing strand. The newly added nucleotide is held firmly in place, and the accuracy of its base pairing to the next nucleotide on the template strand can be checked. Therefore, as DNA polymerase fills the gap, it can proofread the new DNA strand that it makes. What appears at first glance as energetically wasteful is really just a necessary price to be paid for accuracy. 9. (True or False) In a replication bubble, a single parental DNA strand serves as both the template for leading strand synthesis for one replication fork and the template for lagging strand synthesis in the other replication fork. TRUE
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