Answer Key: Replication

BIOL230W Week 10 – Replication This week we will focus on the process of replicating DNA that occurs during the S-phase of the cell cycle. We will use this discussion to better understand the important molecular tool: the polymerase chain reaction (PCR). Day 1 Learning Objectives 1. Describe the DNA A,T,G,C content at origins of replication. 2. Explain how helicase is positioned at origins of replication. 3. Explain the timing of the helicase activation. 4. Define the components of the pre-replication complex. 5. Describe the timing of geminin function. 6. Explain the role of geminin in preventing re-replication during S-phase. 7. Identify a replication fork and describe how it is formed. 8. Interpret data relating to the timing and function of replication components. Mini-Lecture: ORC-Initiation 1. Will DNA at the origin be heterochromatic or euchromatic? Euchromatic 2. What protein complex binds to the origin before helicase is loaded? Origin recognition complex 3. During which cell cycle stage is the full pre-replication complex bound to DNA?? G1 4. What type of DNA modifying enzyme is required to separate the DNA strands at the origin? Helicase 5. What three components make up the Pre-Replication Complex (PRC)? Origin recognition complex, helicase loader, and helicase Small Group Activity: Cdt1 Rescues Geminin Mediated Inhibition of Replication After fertilization by sperm the DNA in the newly formed zygote will undergo replication. In this experiment researchers have added different concentrations of geminin protein to the egg. This figure compares the % of cells that have successfully completed replication of their genome in S-Phase with increasing concentration of supplied geminin. Percentage of replicated chromatin was recorded after 100. min BIOL230W Week 10 – Replication 6. How does the concentration of geminin relate to the ability of a cell to replicate its genome in this experiment? The more geminin is present, the lower the cell’s ability to replicate its genome. Mini-Lecture: ORC-Insuring single replication 7. When is the MCM double helicase activated? S phase 8. What is the role of geminin? Blocks addition of helicase (prevents loading of helicase to pre-replication complex) 9. When is geminin active? S-phase 10. Why does geminin not block replication during S-phase? It doesn’t prevent helicase from loading the first time, only the second time to prevent too much replication. Prevents reloading of MCM helicases until next S phase. 11. What would be the result if a cell lacked functioning geminin? MCM-helicase would rebind to DNA and cause too much replication. Day 2 Learning Objectives 1. Describe the different proteins involved in replication and how they function. 2. Label ends of DNA in a replication fork and on primers. 3. Describe the structure and functions of primers in replication and PCR. 4. Identify leading vs. lagging strands. 5. Explain where and why Okazaki fragments are required for replication in eukaryotes. 6. Compare and contrast the mechanisms for replication that occur at the leading versus the lagging strands. 7. Predict changes in function due to mutations in proteins of the replisome. Small Group Activity: Compare and Contrast Transcription and Replication 5’  3’ 3’  5’ RNA Polymerase RNA DNA 5’  3’ 3’  5’ DNA Polymerase DNA DNA Mini-Lecture: The Replisome BIOL230W Week 10 – Replication 12. Which component of the replisome keeps separated strands of DNA from reannealing (coming back together)? Single stranding DNA binding protein 13. How many replication forks are there in 10 replication bubbles? 20 forks 14. What type of macromolecule is the primer for DNA polymerase? RNA 15. How many primers are present on the leading strand compared to the lagging strand? Only one in leading, many in lagging 16. Okazaki fragments are found on which strand (leading or lagging)? Lagging strand 17. In your own words describe processivity in relation to the action of DNA polymerase. The number of times an enzyme adds nucleotides before falling off 18. If a polymerase enzyme has increased processivity, how would you expect this to change the amount of time for replication to complete? It would decrease the amount of time replication takes to be complete. 19. What is the final step of replication after the RNA primer has been removed and replaced with DNA? DNA ligase forms a phosphodiester bond to fix nick. Small Group Activity: Replication Fork Origins of replication are recognized and bound by ORCs. ORCs help load helicases at the origins. DNA replication requires that the hydrogen bonds holding the strands of DNA together are broken (by helicases) so that the replication machinery can access each of the template strands. The helical nature of DNA leads to a build-up of rotational tension as these bonds break and topoisomerase activity relieves this tension. Single-strand binding proteins help keep the template strands separated as DNA polymerase replicates the DNA. As with all polymerases (be it RNA or DNA) nucleotides are added to the 3’ end of an existing strand. Primases add a short stretch of RNA, which serves as the initial strand upon which DNA polymerases will elongate (this short, “starter stretch” of RNA will later be removed by a polymerase and replaced by DNA). Because DNA polymerases only add to the 3’ end, the synthesis of new strands from two template strands occurs bidirectionally with one strand BIOL230W Week 10 – Replication continuous (leading strand) and other being discontinuous (lagging strand). Ligases serve to connect the discontinuous lagging strand together. Day 3 Learning Objectives: 1. From Sci-Lit PCR paper:  Compare and contrast Taq polymerase and Klenow fragment.  Describe the three steps of a PCR cycle.  Explain what effects primer melting temperature.  Describe the relationship between melting temperature and specificity and analyze figures from paper using figure legends. Mini-Lecture: PCR 20. What are the three steps of PCR? BIOL230W Week 10 – Replication Denature, Anneal, and Extend Small Group Activity: PCR Remember our previous discussions about the relationship between absorbance of UV at 260 nm, temperature, and the single-stranded vs doublestranded nature of DNA. Use the figure below to recall how these three factors are correlated and then answer the following questions. 21. What happens to the DS vs SS nature of the DNA as you add heat (increase temperature)? As temperature increases, Strand absorption increases (denaturation) 22. What would happen if you were to decrease the heat? As temperature decreases, Strand absorption decreases (reannealing) 23. How is DNA denatured in PCR? It is exposed to heat to break hydrogen bonds and make it single stranded. 24. What happens to the temperature in the PCR as it moves from the denature step to the annealing step? Temperatures decrease so that primers can anneal. The figure below is a comparison of Klenow and Taq DNA polymerase-catalyzed PCR amplification products of the human B-globin gene. Electrophoretic analysis of the PCR products obtained with Klenow polymerase (lanes 1 to 6) and Taq polymerase (lanes 7 to 12) after: 0 cycles (lanes1and7) 20 cycles (lanes 2 and 8) 25 cycles (lanes 3 and 9) 30 cycles (lanes 4 and 10) 35 cycles (lanes 5, 6, 11, and 12) of amplification. The DNA samples that were amplified using primers specific to the B-globin gene and were prepared from the human cell lines MOLT4 (lanes 1 to 5 and BIOL230W Week 10 – Replication 7 to 11) and GM2064 (lanes 6 and 12). MOLT4 is homozygous for the wild-type 3-globin gene. GM2064 possesses a homozygous deletion of entire the B-globin gene. The reactions using Klenow as the polymerase were annealed and extended at 37˚C while the reactions using Taq polymerase were annealed and extended at 70˚C. 25. What two conditions are different between lanes 1-5 compared to lanes 7-12? In lanes 1-5, E. coli DNA polymerase is used. In lanes 7-12, Taq DNA polymerase is used. 26. Mark where the specific band should run on the gel. 27. What is causing multiple bands to be amplified in lanes 1-6? The temperature is lower, and the primer can anneal at multiple different spots in the genome. 28. What is the purpose of lane 6 and lane 12? Lanes 6 and 12 remove the regions that are complimentary to the primers (region supposed to be amplified is removed) 29. What are the two purposes of a primer? Defines the region to be amplified and provide a free 3’ OH.