Questions from chapters in the textbook that are relevant for the final exam

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1 Questions from chapters in the textbook that are relevant for the final exam Chapter 9 Replication of DNA Question 1. Name the two substrates for DNA synthesis. Explain why each is necessary for DNA synthesis. Answer: The two substrates are (1) the deoxynucleoside tri- phosphates (dgtp, datp, dttp, dctp) and (2) the primer:template junction. The primer contains the 30-OH where the new nucleotides are added. The single-stranded DNA that is part of the junction serves as the template to direct what nucleotide is incorporated into the growing strand. Question 3. Explain why DNA synthesis is coupled to the hydrolysis of pyrophosphate. Answer: By coupling DNA synthesis to the hydrolysis of pyrophosphate, the free energy change goes from 3.5 kcal/mol ( for the addition of 1 nucleotide to the growing strand of DNA and release of pyrophosphate) to 7 kcal/mol for the net reaction including the hydrolysis of pyrophosphate. The net equilibrium constant (Keq) is 105, which strongly favors products making the net reaction essentially irreversible. Question 11. A. Describe the role of a DNA helicase at a replication fork. B. Explain how topoisomerases help DNA helicases function more efficiently. C. During PCR you do not have to add DNA helicase to the reaction. Explain why not? Answers: A. DNA helicase separates the two strands of DNA at a replication fork in an ATP-dependent, processive manner by pulling single-stranded DNA through its central pore. DNA helicases either possess 50 to 30 or 30 to 50 directionality with respect to the strand that the helicase encircles. B. As a result of DNA unwinding by DNA helicase, positive supercoils accumulate in front of the replication fork. Topo- isomerases relieve positive supercoils by breaking and re- sealing one or both strands of the DNA, allowing DNA helicase to continue separating the DNA strands at the replication fork. C. During PCR, the reaction is heated and cooled to cause DNA strand separation, the annealing of primers, and extension by the DNA polymerase. Strand separation by heat eliminates the need for a DNA helicase.

2 Additional questions a. Describe two ways by which DNA polymerase ensures that the correct base is added to the growing polynucleotide chain during replication, addressing what these mechanisms have in common and how they differ. What is the contribution of these mechanisms toward the overall fidelity of DNA replication? The first mechanism used by DNA polymerase to ensure that the inserted base is correct, called kinetic proofreading, relies on the exclusive ability of correctly paired bases to sit properly within the active site of DNA Polymerase III. If the incoming nucleotide is not complementary to the template base at the active site, then the nucleotide cannot properly align with the template and the polymerization reaction becomes much slower. This gives the incorrectly matched nucleotide time to leave the active site and to be replaced by a correctly matched one. Kinetic proofreading reduces polymerase's error rate to about The second proofreading mechanism uses the 3ʹ to 5ʹ exonuclease activity of many DNA polymerases. This activity allows polymerase to backtrack and remove the nucleotide that was most recently incorporated into the primer strand. Although this 3ʹ to 5ʹ exonuclease activity works on both correctly and incorrectly-matched nucleotides, it is much more active in removing incorrectly matched ones, in part because incorrectly paired bases slow the polymerase down and give the exonuclease more time to work. This second level of proofreading increases the accuracy of replication to about 1 error per 10 7 bases. b. Briefly summarize the palm, thumb, and finger domains of DNA polymerase III, and explain how they contribute to replication. What is the role of the metal ions in the palm's activity? The "palm," "thumb," and "hand" refer to three different subdomains of the DNA polymerase III enzyme. The palm, made up of a β sheet that includes the key components of the polymerase's active site, contributes to replication in several critical ways. First, it binds to metal ions that help remove the hydrogen from the OH group at the 3ʹ end of the primer, producing a 3ʹ O that can attack the α phosphate of an incoming nucleoside triphosphate. The metal ions also help to counteract the negative charges of the β and γ phosphates of incoming nucleoside triphosphates as well as the pyrophosphate that is released when a nucleotide is added to a growing primer chain. Finally, the palm contributes to replication accuracy by stabilizing newly synthesized DNA that is correctly base paired; if an incorrect nucleotide is incorporated, the absence of palm-mediated stabilization slows down the polymerase allowing its 3ʹ to 5ʹ exonuclease activity to correct the error.

3 The fingers contribute to replication by binding to the incoming nucleoside triphosphates. If the incoming nucleoside triphosphate is complementary to the nucleotide present on the template strand, the fingers move by approximately 40 to "close" around the base pair. This closed state creates a tight pocket that holds the base pair in place and allows catalysis to occur. The fingers also bend the template strand in a way that exposes the nucleotide present at the catalytic site, helping to ensure that the correct base is used for base pairing with the incoming nucleoside triphosphate. The thumb does not contribute directly to catalysis but instead interacts with recentlysynthesized DNA to keep the active site in the correct position and to inhibit the dissociation of the polymerase from the DNA. c. What prevents ribonucleotides from being incorporated into the growing DNA strand during replication of the genome? In what aspect of replication are ribonucleotides used, and what ultimately happens to them? Polymerase can effectively avoid incorporating ribonucleotides into the growing DNA strand because of the precise way in which the enzyme binds deoxyribonucleotides. Specifically, polymerase has two amino acids whose side chains project into the nucleotide binding pocket of the enzyme and occupy the same space that the 2ʹ hydroxyl group of a ribonucleotide would if it were to enter the pocket. Accordingly, ribonucleotides are sterically excluded from the active site of the enzyme. Ribonucleotides are used in vivo to make up the primers that DNA polymerase requires to carry out DNA synthesis. These short RNA primers are synthesized by an enzyme called primase and RNA polymerase, and they are required to start both leading and lagging strand synthesis. The ribonucleotides that are incorporated into the strand as a result of primer synthesis (and the extension of the primers by DNA polymerase) are ultimately removed by the 5ʹ to 3ʹ exonuclease activity of the DNA polymerase I enzyme. DNA polymerase I binds just upstream of the RNA primer, uses its 5ʹ to 3ʹ exonuclease activity to remove the ribonucleotides, and then replaces them sequentially with deoxynucleotides. d. How is the DNA unwound at the replication fork? What effect does this have on the DNA upstream of the fork, and how does the cell deal with this effect? DNA upstream of the replication fork is unwound by a class of enzymes called DNA helicases. Helicases, which are typically hexameric protein complexes, encircle one of the two DNA strands upstream of the fork and use the energy provided by ATP hydrolysis to separate the strands. The single-stranded DNA produced by the helicase is rapidly coated by proteins called single-stranded binding protein (SSB), helping to prevent the separated strands from reannealing before the DNA polymerase arrives.

4 The helicase-mediated unwinding of the DNA introduces positive supercoils into the doublestranded DNA upstream of the fork. Enzymes called type I topoisomerases remove the supercoils by nicking one of the two strands of the DNA, allowing the strands to rotate until they regain a relaxed state, and then re-sealing the nick. This topoisomerase I activity is critical for replication because, in its absence, the positive supercoils would prevent the replication machinery from advancing. e. Describe the various steps involved in the initiation of replication in eukaryotes. What proteins are involved in each of the steps, and when in the cell cycle does each of them occur? How does their temporal segregation contribute to the limitation of origin firing to at most once per cell cycle? Replication initiation in eukaryotic cells is separated into two distinct steps. In the first step, occurring during the G1 phase of the cell cycle, multiple proteins bind to potential replicators to form a pre-replicative complex (pre-rc) that includes the ORC complex, Cdc6, Cdt1, and the MCM proteins. While pre-rc assembly marks the replicator as a potential site for replication initiation, it does not lead by itself to unwinding of the DNA, nor does it guarantee that the replicator will ultimately serve as a replication origin during S phase. The second step, marked by the appearance of the kinases Ddk and Cdk, occurs at the beginning of S phase. These kinases cause other replication factors to join the origin: first Sld3, Cdc45, and Mcm10, and subsequently the DNA polymerases, sliding clamp loader, primase, and the sliding clamps. Once all of these proteins have joined the origin, replication can begin. The restriction of these two steps to different phases of the cell cycle helps limit replication to once per cell cycle because of the dual role of Cdk in inhibiting pre-rc formation and in promoting origin firing. Specifically, pre-rcs can form only during G1 phase, when Cdk activity is absent, but at the same time the lack of Cdk activity during G1 means that the pre- RCs that form cannot complete origin assembly and initiate replication. Subsequently, when Cdk activity appears at the beginning of S phase, already assembled pre-rcs can proceed with origin assembly and begin replicating, but no new pre-rcs can form because of the continuous presence of Cdk (which persists until the following mitosis, disappearing as cells re-enter G1 phase). f. Explain why the replication machinery is incapable of completely replicating the ends of the chromosomes. What is the practical effect of this? How do eukaryotic cells get around this problem? While leading strand synthesis can proceed until the last nucleotide on the chromosome is copied, the replication fork is incapable of copying the very end of the lagging strand because primase cannot function at the end of a DNA molecule. This is a potentially major problem for the cell because it could result in the loss of several bases from the end of one of the chromosomes with every round of DNA replication.

5 Eukaryotic cells have several strategies for avoiding this potential loss of terminal nucleotides. For example, some bacterial cells and viruses use a specialized protein that can replace the primase at the end of the chromosome by attaching to the chromosome end and priming synthesis of the terminal nucleotides. Another example is seen in eukaryotic cells, where an enzyme called telomerase helps maintain chromosome ends by extending them through the addition of multiple copies of a specific repeated sequence (for example, TTAGGG in humans).