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CH07 - DNA Structure and Replication BASIC PROBLEMS 1 The...

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Unformatted text preview: DNA: Structure and Replication BASIC PROBLEMS 1. The DNA double helix is held together by two types of bonds, covalent and hydrogen. Covalent bonds occur within each linear strand and strongly bond the bases, sugars, and phosphate groups (both within each component and between components). Hydrogen bonds occur between the two strands and involve a base from one strand with a base from the second in complementary pairing. These hydrogen bonds are individually weak but collectively quite strong. Conservative replication is a hypothetical form of DNA synthesis in which the two template strands remain together but dictate the synthesis of two new DNA strands, which then form a second DNA helix. The end point is two double helices, one containing only old DNA and one containing only new DNA. This hypothesis was found to be not correct. Semiconservative replication is a form of DNA synthesis in which the two template strands separate and each dictates the synthesis of a new strand. The end point is two double helices, both containing one new and one old strand of DNA. This hypothesis was found to be correct. A primer is a short segment of RNA that is synthesized by primase using DNA as a template during DNA replication. Once the primer is synthesized, DNA polymerase then adds DNA to the 3’ end of the RNA. Primers are required because the major DNA polymerase involved with DNA replication is unable to initiate DNA synthesis and, rather, requires a 3’ end. (It is the 3’ -OH group that is required to create the next phosphodiester bond.) The RNA is subsequently removed and replaced with DNA so that no gaps exist in the final product. Helicases are enzymes that disrupt the hydrogen bonds that hold the two DNA strands together in a double helix. This breakage is required for both RNA and DNA synthesis. Topoisomerases are enzymes that create and relax supercoiling in the DNA double helix. The supercoiling itself is a result of the twisting of the DNA helix that occurs when the two strands separate. 170 Chapter Seven 10. 11. 12. Because the DNA polymerase is capable of adding new nucleotides only at the 3’ end of a DNA strand, and because the two strands are antiparallel, at least two molecules of DNA polymerase must be involved in the replication of any specific region of DNA. When a region becomes single-stranded, the two strands have an opposite orientation. Imagine a single-stranded region that runs from right to left. The 5’ end is at the right, with the 3’ end pointing to the left; synthesis can initiate and continue uninterrupted toward the right end of this strand. Remember: new nucleotides are added in a 5’—> 3’ direction, so the template must be copied from its 3’ end. The other strand has a 5’ end at the left with the 3’ end pointing right. Thus, the two strands are oriented in opposite directions (antiparallel), and synthesis (which is 5’—> 3’) must proceed in opposite directions. For the leading strand (say, the top strand) replication is to the right, following the replication fork. It is continuous and may be thought of as moving “downstream.” Replication on the bottom strand cannot move in the direction of the fork (to the right) because, for this strand, that would mean adding nucleotides to its 5’ end. Therefore, this strand must replicate discontinuously: as the fork creates a new single-stranded stretch of DNA, this is replicated to the left (away from the direction of fork movement). For this lagging strand, the replication fork is always opening new single-stranded DNA for replication upstream of the previously replicated stretch, and a new fragment of DNA is replicated back to the previously created fragment. Thus, one (Okazaki) fragment follows the other in the direction of the replication fork, but each fragment is created in the opposite direction. No. The information of DNA is dependent on a faithful copying mechanism. The strict rules of complementarity ensure that replication and transcription are reproducible. Helicases are enzymes that disrupt the hydrogen bonds that hold the two DNA strands together in a double helix. This breakage exposes lengths of single- stranded DNA that will act as the template and are required for DNA replication. Therefore, the absence of helicases would prevent the replication process. Theoretically, DNA could be replicated this way but not with the replisome, which is organized to replicate both stands simultaneously. Further, this would leave one strand of the DNA single-stranded where mutagenic events would be more likely, and it would certainly take longer. The chromosome would become hopelessly fragmented. c. DNA synthesis might take longer. (1. It does not have to adapt to each of the body’s tissues. b. The RNA would be more likely to contain errors. Chapter Seven 171 13. Part of the replisome is the sliding clamp, which encircles the DNA and keeps pol III attached to the DNA molecule. Thus pol III is transformed into a processive enzyme capable of adding tens of thousands of nucleotides. 14. d. Replication would take twice as long. 15. 3. Prior to the S phase, each chromosome has two telomeres, so in the case of 2n = 14, there are 14 chromosomes and 28 telomeres. b. After S, each chromosome consists of two chromatids, each with two telomeres, for a total of four telomeres per chromosome. So, for 14 chromosomes, there would be 14 x 4 = 56 telomeres. c. At prophase, the chromosomes still consist of two chromatids each, so there would be 14 x 4 = 56 telomeres. d. At telophase, there would be 28 telomeres in each of the soon—to—be daughter cells. 16. If the DNA is double stranded, A = T and G = C and A + T + C + G = 100%. If T = 15%, then C = [100 — 15(2)]/2 = 35%. 17. If the DNA is double stranded, G = C = 24% and A = T = 26% . 18. Six. The first replication start would have two replication forks proceeding to completion, and the now replicated origins would each start replication again. Each would have two more replication forks, for a total of six. 19. Only the DNA molecule that used the poly-T strand as a template would be radioactive. The other daughter molecule would not be radioactive, because it would not have required any dATP for its replication. Because each strand of the second molecule contains T, both daughter molecules would require dATP for replication, so each would be radioactive. 20. Yes. DNA replication is also semi-conservative in diploid eukaryotes. 21. a. The bottom strand. 172 Chapter Seven Ee3aggéfiea38a“ ooaoocd\\ \ RNA primer/ KCTGACAGTC—3C... FGACTGTCAG-SU... g/saseesscesecescs gzsveecvseesvecsea ..y—ATTCGTACGATCGACTGACTGACAGTC—3C... .y-TAAGCATGCTAGCTGACTGACTGTCAG—5K. ..W—ATTCGTACGATCGACTGACTGACAGTC—3K.. .Y—TAAGCATGCTAGCTGACTGACTGTCAG—3.... (1. Yes, but the other replication fork would be moving in the opposite direction and the top strand, as drawn, would now be the leading strand and the bottom strand would now be the lagging strand. 22. The bottom strand will serve as the template for the Okazaki fragment so its sequence will be: 5’uuCCTTAAGACTAACTACTTACTGGGATCNN3’ 23. a. 24. 25. Chapter Seven 173 e. Model (b) is ruled out by the experiment. The results are compatible with semiconservative replication, but the exact structure could not be predicted. The other models would all give one band of intermediate density. Sample A must be live S cells because it kills mice when injected and S cells are recovered from the mice. Sample C must be live R cells because it has no effect on mice when injected but live R cells are recovered. Sample B must be DNA from S cells as it transforms sample C when co- injected. If all samples remain the same, only the line will change when samples B + C are co-injected. Because B contains DNA, it will not transform the R cells, so injected mice will have no response and live R cells will be recovered. CHALLENGING PROBLEMS 26. 27. 28. Without functional telomerase, the telomeres would shorten at each replication cycle, leading to eventual loss of essential coding information and death. In fact, there are some current observations that decline or loss of telomerase activity plays a role in the mechanism of aging in humans. a. A very plausible model is of a triple helix, which would look like a braid, with each strand interacting by hydrogen bonding to the other two. b. Replication would have to be terti—conservative. The three strands would separate, and each strand would dictate the synthesis of the other two strands. c. The reductional division would have to result in three daughter cells, and the equational would have to result in two daughter cells, in either order. Thus, meiosis would yield six gametes. Chargaff’s rules are that A = T and G = C. Because this is not observed, the most likely interpretation is that the DNA is single—stranded. The phage would first have to synthesize a complementary strand before it could begin to make multiple copies of itself. ...
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