MCB 250 discussion WS Week 3 SP24-Jackie Jelderks

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Feb 20, 2024

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MCB 250 Spring 2024 Discussion Worksheet Week 3 – Feb. 8-12 1. a. Chromosomal DNA of both bacteria and eukaryotes is negatively supercoiled.   Why does negative supercoiling facilitate both replication and transcription of DNA?   Replication: reduces energy needed to separate the DNA strands for replication due to the DNA not wanting to be supercoiled as it introduces stress. Transcription: Negative supercoiling counteracts the downstream tension that RNA polymerase introduces when needing to read the DNA strand. b.   In bacteria, Gyrase adds negative supercoils to DNA while other topoisomerases relax DNA.   These two reactions oppose each other.   Why do you think the cell is both adding and removing negative supercoils all the time? Simultaneously adding and removing the supercoils ensures that the structure of the DNA remains stable, while also making it accessible for proteins relating to replication and transcription 2. a. A closed circular DNA in its relaxed form has a linking number (Lk) of 500. A supercoiled form of this DNA has Lk = 495. This supercoiled DNA is treated with E. coli topoisomerase I and samples are withdrawn from the reaction at various times and electrophoresed in an agarose gel. Sketch how these gels would look. Include a lane with the relaxed form, a lane with the supercoiled DNA before the enzyme is added, and lanes at various times after the enzyme is added. The final lane should show what the gel looks like when the reaction has reached completion. b. Repeat the experiment using E. coli topoisomerase IV. What do you need in the reaction mix besides the enzyme to make it work? When the topo IV reaction is complete will the product be the same as for topo I? 1
c. Discuss the mechanism Topo I. Why is ATP not required to reform a phosphodiester bond after strand passage? Top I causes a single strand break in DNA, forming a covalent intermediate with the 3’ end of the broken strand. This makes it so the DNA can now rotate around the break and reseal, but ATP is not needed for the resealing process. 3. In order to fit DNA into the nucleus of a eukaryotic cell it must be greatly compacted. To accomplish this, several levels of compaction are required. Discuss each of these levels beginning with the nucleosome. What proteins are required at each level and what structures are formed? The details of the lower level structures are much better understood than those of the higher level structures. Discuss. At what point in the cell division cycle is DNA most highly compacted and when is it least highly compacted? Nucleosomes: Beads on a string structure as DNA is wrapped around histones, linking the DNA strand with multiple histones. Each nucleosome contains 8 histones: two H2A, two H2B, two H3 and two H4. ~146 bp are wrapped around each histone unit, with H1 binds to 20-60 bp linker DNA between each bead. 2
30 nm fiber: compacts the DNA structures 100-fold by creating a “chromosome scaffold.” This level of folding is not completely understood, but it is believed that many H1 histones, Topo II, and possibly other proteins are used to maintain this structure. In cell division, DNA is most highly compacted during metaphase/anaphase and least highly compacted during S-phase. 4. The necessity to compact DNA presents a problem for the cell since DNA has to be accessible for replication (all of the DNA), transcription (some of the DNA), and repair (only damaged regions). Define heterochromatin and euchromatin in terms of accessibility and the level of compaction. How does the cell make DNA in highly compacted states available for these processes to occur? Heterochromatin: tightly compacted form of chromatin, lowering accessibility for replication, transcription, and repair. Heterochromatin contains nucleosomes that are closely packed together, and DNA is tightly wrapped around the histones. Euchromatin: a less compacted form of chromatin, meaning an increased accessibility for cellular function. Nucleosomes are farther apart and the DNA is less tightly wrapped around the histones. During replication, helicase unwinds the double helix, dismantling the high-order structure and allowing for DNA polymerases to go through replication. Afterwards, the hetero/euchromatin structure is restored. During transcription, the chromatin structure is able to lower is condensed state in certain areas, as proteins modify the histones to open up the structure, allowing RNA polymerase to bind. When a protein specialized in repair binds to a segment of DNA, the chromatin structure is decondensed to allow for fixing. Once repair is completed, the structure is restored. 3
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