Chem+153B+-+Homework+3

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153B

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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? 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. 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. 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). c. Is the formation of this 21 bp bubble favored, disfavored, or neither? Mia Kim 005896550. -> r T = 58(34) = n48 -I W = - 3 - L = 48 - ( 3) = 45 & & 445 T = (23)(34)+138 = 44 36 45.81 -1.9 = T W Not 44 45 = change W = 1 A -1 A - =1035) = 163 W = L = 103-3 = 10 & T = 1059 21335 + 0) = 10 360 L = 100 = 10 1 + W W = - 1 formation of bubble was favored because II was decreased from 1-31-1-1)
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? -lest * right & Es T = -2 T = 0 W = 0 W = 0 W = - 2 - -> & S T 08 W 0 = 8 4 - origin of replication has several sequences with affinity to the replication initiation factor DnaA. It binds to these sequences, creating supercoiling that open up a replication bubble. Draft also recruits draB helicases which binds to SSDNA in the replication bubble and begins unwinding the double-stranded DNA at each replication fork.
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. b. Which nucleotide is about to be added to the newly synthesized 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. 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? The orange strand A guanosine phosphate is in position to be added to the new strand. This is the guanine-cytosine right below the about to be added guanosine triose phosphate. Arg573 and Gin (34 are interacting with the minor groove of this base pair. There's H-bond between quanidium group of Arg373 side chain and the or atom of cytosine and an H-bond between amide group of the GIn734 side chain and the N3 atom of guanine. All watson-Crick base pairs have 2 H-bond acceptor in the same part of the minor groove edge. Due to their symmetry around the dyad axis, the exchange positions whenever a purine: pyrimidine pair is swapped for. a pyrimidine: purine pair. Most non-Watson-Crick base pairs lack H-bond acceptors in one or both of these positions.
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Name _______________ ID _____________ 7. How does the baseline intracellular pyrophosphate concentration promote DNA synthesis? What would happen if this concentration was increased and why? 8. List (and explain) the 5 paradigms of DNA replication. 9. Explain the two roles of topoisomerases in DNA replication. 10. How do the metal ions in the active sites of DNA polymerases reduce the activation barrier to phosphoester bond formation? DNA Synthesis produces pyrophosphate as a product. Give this, the relative concentration of pyrophosphate can impact the favorability of DNA synthesis. Conveniently the baseline concentration of pyrophosphate is low enough (around 0.5MM) so the DNA synthesis proceeds forwards by Le Chatelier's principle. If the pyrophosphate concentration were to be increased (to 200 times the dNTP concentration) the rate of DNA degradation would match the rate of synthesis, as we have reached the equilibrium concentration. as helicase unwinds the DNA duplex, it decreases the natural positive twist of DNA (reduce it to of Due to topological constraints, this increase writhes eventually increasing the free energy needed to continue unwinding the strand. &one role of topoisomerases to move a heat of the replication fork, uncoiling the DNAs as to keep the free eners be them, we are stranded break is required. The metal ions stablize the substrates (dNTP and DNA strand) and hold them (as well as the nascent covalent bond between them ? In optimal geometry, which stablizes the reaction, as well as the eventual transition state. They also stablize the oxyanion form of the nucleophile (3 hydroxyl) that initiate the reaction.
Name _______________ ID _____________ 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. b. Which strand is the lagging strand for this replication fork? c. Indicate the direction of movement for the replication fork on the diagram 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. 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. 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? * top Strand bottom strand. 5 CCA AAA 3 c. replication fork template (and leading strand) i & ACC lagging strand. wherever a mismatched base pair is formed the DNA duplex is destablized. 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.
Name _______________ ID _____________ 13. What are Okazaki fragments and why are they necessary? Draw a diagram incorporating the lagging strand to assist your explanation. 14. Why is the β 2 sliding clamp a necessary feature of DNA Pol III? What role does it serve? one strand (the 3' tos' leading strand) can have DNA added in the 5' to 'direction (which is necessary for polymerizations but the lagging strand (5 to 3') is in the wrong orientation to add DNA towards the replication fork. Okazaki fragments any short strands of replicated DNA formed on the lagging strand by discontinuous replication events, each primed by distinct RNA primer. They solve the issue of the lagging strand as each is polymerized in a 5'ts 3' direction (away from the replication fork) but they're created sequentially moving in a 3 tos' direction. DNAtaki fragment · mer 5/ 5 I DNA helicase 3 DNA is made continuosly
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