Chem+153B+-+Homework+3+solutions

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Chemistry

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Feb 20, 2024

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Name _______________ ID _____________ Chem 153A Homework 3 due 8/28 @ 11:59 pm 1. Envision a 508 bp double-stranded piece of closed, circular DNA with an average helical twist of 34° . The path of the backbone is shown in the figure on the right. a. What is your best estimate of the linking number of this loop? T = (508 bp)(34°)/360 ° ≈ 48 W = -3 L = T + W = 48 3 = 45 b. A 23 bp segment of alternating purine/pyrimidine residues in our loop spontaneously changes to the Z conformation (helical twist angle of -30° ). The DNA backbone remains intact throughout this transition. Estimate the new writhe of the loop. T = (485 bp)(34°)/360° + (23 bp)(-30°)/360° = 45.8 1.9 ≈ 44 L = 45 = T new + W new = 44 + W new W new = 1 2. Envision a 1059 bp -long piece of closed, circular DNA with an average helical twist angle of 35° . The path of the backbone is shown in the figure on the right. Contained within this sequence is a 21 bp long AT-rich region that has a tendency to melt, forming a denatured bubble. a. Determine the twist, writhe, and linking number for this DNA loop prior to bubble formation. T = (1059 bp)(35°)/360 ° ≈ 103 W = -3 L = T + W = 103 3 = 100 b. Determine the twist, writhe, and linking number for this DNA loop after bubble formation (assume the helical twist angle in the bubble is and that the DNA backbone remains intact). T = (1038 bp)(35°)/360° + (21 bp)(0°)/360° = 100.9 101 L = 100 = T new + W new = 101 + W new W new = -1 c. Is the formation of this 21 bp bubble favored, disfavored, or neither? Favored because the absolute value of W decreased!
Name _______________ ID _____________ 3. The following strand of circular dsDNA (twist = -1, writhe = 0) is acted upon by a type I topoisomerase. What are the two possible outcomes? Draw both and note the resulting twists and writhes (hint: Look closely at the diagram!). 4. The following strand of circular dsDNA (twist = 0) is acted upon by a type II topoisomerase. What are the two possible outcomes? Draw both and note the resulting twists and writhes. 5. In prokaryotes (like E. Coli) what determines where replication starts? How is replication initiated? A segment of DNA (called the origin of replication) has several sequences with affinity to the replication initiation factor dnaA. It binds to these sequences, creating supercoiling that opens up a replication bubble. It also recruits dnaB helicase, which binds to ssDNA in the replication bubble and begins unwinding the dsDNA at each replication fork. OR T = 0 W = 0 T = -2 W = 0 OR T = 0 W = 0 T = 0 W = -4
Name _______________ ID _____________ 6. PDB DNA/Protein Scavenger Hunt Search for 6FBI on rcsb.org and visualize with Pymol or rcsb’s 3D View. Use this structure of klentaq DNA polymerase bound to template DNA to answer the following: a. Which strand is being synthesized? Use a description or a screenshot for your answer. The “orange” strand on the rcsb GUI b. Which nucleotide is about to be added to the newly synthesized strand? A guanosine triphosphate is in position to be added to the new strand c. Find the two metal ions vital to the polymerization mechanism. Take a screenshot showing the groups they’re interacting with d. Isolate the duplex base pair most recently synthesized by klentaq. Two amino acid side chains make sequence-independent H-bond contacts with two atoms in the base pair. Draw the base pair shown in this scene along with the two amino acid side chains, representing proper geometry, accurate distances, and any intermolecular interactions. This is the Guanine-Cytosine right below the about-to-be-added guanosine triphosphate. Arg573 and Gln754 are interacting with the minor groove of this base pair. There should be an H-bond between the Arg573 side chain nitrogen and the N3 atom of adenine, and an H-bond between the Gln754 side chain nitrogen and the O2 atom of thymine.
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Name _______________ ID _____________ e. Explain how the H-bonds you added in part d allow the DNA polymerase to distinguish between WC and non-WC base pairs. Consider how the positions of the two H-bond acceptors in the bases makes these H-bonds sequence- independent. Why is access to the minor groove useful here? All Watson-Crick base pairs have two H-bond acceptors in the same parts of the minor groove edge. Due to their symmetry around the dyad axis, they exchange positions whenever a purine:pyrimidine pair is swapped for a pyrimidine:purine pair. Most non- Watson-Crick basepairs lack H-bond acceptors in one or both of these positions. 7. How does the baseline intracellular pyrophosphate concentration promote DNA synthesis? What would happen if this concentration was increased and why? DNA synthesis produces pyrophosphate as a product. Given this, the relative concentration of pyrophosphate can impact the favorability of DNA synthesis. Conveniently (or by cellular design) the baseline concentration of pyrophosphate is low enough (around 0.5 mM by one estimate) so that DNA synthesis proceeds forward, by Le Chatelier ’s principle . If the pyrophosphate concentration were to be increased (to 200X the dNTP concentration) the rate of DNA degradation would match the rate of synthesis, as we would have reached the equilibrium concentration. 8. List (and explain) the 5 paradigms of DNA replication. 9. Explain the two roles of topoisomerases in DNA replication. As helicase unwinds the DNA duplex, it decreases the natural positive twist of DNA (reduces it to 0°). Due to topological constraints, this increases writhe, eventually increasing the free energy needed to continue unwinding the strand. One role of topoisomerases is to move ahead of the replication fork, uncoiling the DNA so as to keep the free energy barrier to unwinding low. The second role is reserved for topoisomerase II. Once replication is complete, two full circular genomes are interlinked. To extricate them, a double-stranded break is required.
Name _______________ ID _____________ 10. How do the metal ions in the active sites of DNA polymerases reduce the activation barrier to phosphoester bond formation? They stabilize the substrates (dNTP and the DNA strand) and hold them (as well as the nascent covalent bond between them) in optimal geometry, which stabilizes the reaction, as well as the eventual transition state. They also stabilize the oxyanion form of the nucleophile (3 hydroxyl) that initiates the reaction. 11. Suppose that the double-stranded DNA shown below is part of the circular E. Coli chromosome. One DNA polymerase III is currently traversing this sequence left to right (and has been for more than 2 kilobases). a. Indicate which strand is the template being used for DNA synthesis. (indicated on figure below) b. Which strand is the lagging strand for this replication fork? Bottom strand (also indicated) c. Indicate the direction of movement for the replication fork on the diagram below. (indicated on figure below) d. The asterisk below indicates the 5’ end of a primer adjacent to an Okazaki fragment. Write the first 6 base pairs of its sequence. 5’ CCA AAA 3’ (remember that this is RNA! Uracil could also be present) e. A fraction of a second prior to the arrival of Pol III, the DnaB helicase moves through this area. At one moment it is bonded to the area boxed in below (I’ve boxed both strands but it may not be associated with both!). The “top” subunit of the helicase hydrolyzes ATP and moves to the “bottom” of the hexamer. Circle the base pairs it associates with at this point. * replication fork lagging strand template (and leading strand)
Name _______________ ID _____________ 12. Describe the proofreading mechanism of Pol I. Why is movement by the nascent (newly created) DNA critical for the editing process as carried out by this enzyme? Whenever a mismatched base pair is formed, the DNA duplex is destabilized. The movement of the DNA allows the duplex to slide into a region of the channel where the mismatched strand can dissociate and flip into the 3’ to 5’ exonuclease active site. This active site degrades the mismatched base pair, allowing for synthesis to replace it. 13. What are Okazaki fragments and why are they necessary? Draw a diagram incorporating the lagging strand to assist your explanation. One strand (the 3’ to 5’ leading strand) can have DNA added in the 5’ to 3’ direction (which is necessary for polymerization, as discussed in problem #2) but the “lagging” strand (5’ to 3’) is in the wrong ori entation to add DNA towards the replication fork. Okazaki fragments are short strands of replicated DNA formed on the lagging strand by discontinuous replication events, each primed by a distinct RNA primer. They solve the issue of the lagging strand as ea ch is polymerized in a 5’ to 3’ direction (away from the replication fork) but they’re created sequentially moving in a 3’ to 5’ direction. 14. Why is the β 2 sliding clamp a necessary feature of DNA Pol III? What role does it serve?
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