CH339L.09.Test2.Key-m - CH339LT€st209 Name XE Z(1 10...

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Unformatted text preview: CH339LT€st209 Name XE Z (1; 10 points) Adjacent you see Figure 25-43 from your book. All recombination junctions can be resolved in two different ways. If the recombination junction shownhad been resolved in an opposite manner (that is, not according to the red arrows), then: (a) A dimeric genome would still have been produced, and would have been resolved in the same way. (b) A dimeric genome would have been produced, but since it would be in a configuration different than that shown it would have required a different enzyme system for resolution. © Two monomer genomes would have been produced by normal replication and termination. (d) This alternative recombination would have led to rolling circle replication. (e) This alternative recombination would have led to the polymerases colliding, and only one of the two strands being replicated. Fork undergoing recombinational DNA repair termination of replication Dimeric genome resolution to monomers by XerCD system (2; 10 points) If present at high concentrations, which of the following might be expected to competitively inhibit the proofreading activity of DN_Apolyme_r_ases? (a)dADP (b)dTTP (jdGMP OflrADP (3; 15 points) In the replication complex shown below, indicate the different features. Many of the symbols (hexagons, rings) have meanings similar to those you have seen before. Black is RNA. ii), at -..,.-,..= -. Draw a solid arrow at or near the replication fork P '“"-_..--‘ -- indicating which direction overall the polymerase dj :5; .3" complex is moving in. Lending J g: ‘_§J.1‘5H“Troymbone “m" P 3* _ ' "‘6" / / Draw a dashed arrow at or near the polymerase . K) .' -,;s B. on the lagging strand showing the direction {1}"? .3, of synthesis. (”J”) “is: Put an ‘X’ where the next ligation between -' 1. Okazaki fragments will occur. r“ X 1 fl“ ‘1 {'1 “i 5- 3‘ 5 3. Lesion (4; 20 points) The diagram above shows recombinational bypass of an unrepairable lesion (blue dot). During bypass, the polymerase has made an additional mistake (inserted a mismatch). Assume that mismatches are repaired at the conclusion of recombinational repair by the mismatch repair machinery. Further assume that all repair is in favor of pyrimidines (that is, pyrimidines serve as a template for mismatch repair). What will be the final sequence on both sides of the replication fork? Draw it in (four letters total). {5: 15 paints} Transpdeens ear: {1| digest gene-nit.- tans be Efltflilf'r'IEl'l'f useful teels fer Tmnspflgnn M 1 Meiesular hieiegy. Adjacent is m as :- a scheme for identifying genes. , Trflnfimsflnfi. [fir-“15ml? Gemmiereetdstren site- Mdei restriction sites insert inte e gene-me. disrupting . ' - - genes and yielding phenetrpes. 121 “93'“ @ Armin-s en _ The site at insertien can he Frm1 pfimers indicate : determined by elentng eut the eh 5' '3' 3’ transpesen junctieri, as shown. 5. PrunerE Cine intermediate in the process is emitted- Based on feur knewledge et FER, drew the missing preduet- Fiepreduse the different thicknesses. shading sr and symbeis es app repriete. [at inverse PEI-i What gees here? |5| Get perineum-I- Te-slenIn-g et nested PCFI pureed-dis. FIE-H en single eei'enies {i {5: 1!] pdinlsi In melting up “3" WWW ““11 WWI-5'3 the sequencing reactinn mix. the dideesyETP sempeund was inadvertently iefi nut. What effect wduid this have? Fltedraliliur the indicated pertidn at the diagram. FiE-bmw ”4.1—er E} 'Humrs‘ i7: 30 paints} Eielaw WU will find a diagram indicating a itinetir: scheme for DNA flut‘j’l‘l‘lEflEfltiU-I‘l. Adjaeent te it re a diagram shearing a kinetic plat for the first insertien at dATF b1; the pditrrnerase. “L “New tarts—Te] _ infants :5 "a LEELE rmFaETArJT H _ {a} if! paints} Assuming that the balymerase "‘ F'F’l primarily discriminates against mismatches during the eatelytie steel [as apprised tn the binding step} a. draw a dashed line an the diagram indleatlng what happens at the nerd step. as the polymerase attempts te insert a aardss tram C. [t]; 1t) paints} Eeiaw lll'fl-Li twill find a raaet'ian eaerdinate diagram, with 1.rarieLra energyr leuelsalready drawn in. Draw the apbmbriate lines sen nesting these energyI levels fer bath a insertien aeress from T {sdlid line} and a insertian across from It: {dashed IIne}. is: ti] paints] New assume that dissrimtnatiea was based primarity en dATP binding rather than eatahsis. Draw a third set at energy states and lines far A insertibn aetass trdrn C: use a detted rather than a dashed line ta shew this- Keep yeur diagram neat eneugh ta grade; yau may want te draw a practice drmWer. I' I 'r, " “—i—u- Heaetien eaerdinate 1511-, amino- M (8; 10 points) Above you see an adenosine in the syn conformation, mis—pairing with a tautomer of adenosine (not shown). Draw in the tautomer of adenosine. Keep in mind that this mis-pair will be most readily incorporated if you keep the configuration roughly isosteric with a normal base—pair. ...
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