Chem271_S07_x2_key

Chem271_S07_x2_key - Chemistry 271, Section 21xx Y ur ame:...

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Unformatted text preview: Chemistry 271, Section 21xx Y ur ame: '7 ’- University of Maryland, College Park Your ID #: General Chemistry and Energetics Prof. Jason Kahn Exam II (100 points total) May 2, 2007 You have 53 minutes for this exam. Exams written in pencil or erasable ink will not be re—graded under any circumstances. Explanations should be concise and clear. I have given you more space than you should need. There is extra space on the last page if you need it. You will need a calculator for this exam. No other study aids or materials are permitted. Generous partial credit will be given, i.e., if you don’t know, guess. Many useful equations are given on page 9. Honor Pledoe: At the end of the examination time lease write out the followino sentence and sion it or talk to me about it: “I pledge on my honor that I have not given or received any unauthorized assistance on this examination.” 1. DNA Thermodynamics (25 pts) The formation of the double—stranded DNA 20-mer oligonucleotide below from the separated single strands has been measured to proceed with AH° = —128 kcal/mole and AS° = —346 cal/moleK. Assume a total strand concentration of 3 yM (1.5 yM for each strand). 5 ’ —TATTAAGCGACCACACATAA VI ATAATTCGCTGGTGTGTATT—S ’ (a; 7 pts) What is the Tm for this oligonucleotide? What is the value of AG° at the Tm? W? gfitnar—m g‘HOHL w” fl '5 m7 : W Chemi tr'27l e tion21xx x mll /2/ 7 2/ (b; 8 pts) Briefly explain why AH° and AS° are each negative for the hybridization reaction. Mac/«W ' . [£56, AWOCO m i? an {WWE‘M‘Q ulW . “ H ar@ @ WA,”ng W arm-NM a»? Sink; Il/Ié/aafgw am (7.1m ZYOH/QJIM‘Q‘ -—) («\yl/dcummk‘i; AJ' «wt ’3“ w WELZWWW ~ "‘ +36“ 5. <0 W M ~3 we __12 Ww~ {2‘ (m Malawi My, “My. m UM SW 3 W agwgo 4% at PM“ ‘ % WW My “ W w. (c; 6 pts) Briefly and qualitatively explain how the nearest—neighbor rules for prediction of DNA melting thermodynamics are analogous to the bond—energy approximation used to estimate AH°r for previously unknown organic compounds. Tm 44% CM «ac oeou'l» Wadolzl’Cvk W ,VL\ “A fled” cW/{a/hw 5 {1A Wurlhnw 6/ Mfi'at 0W5. U0. 419qu w (Lu W [Bl/(0 “WAGON A-SOCl—vx (7C 06w Adv a *7. SW44 0V?” Mévfw W/(kkfmz/IM, ng‘ W M IMJJV (OM 4‘4 JV?“ 7 1 W M‘MMCIC’e/hfl A Score for the page h mi tr 271 e in21xxEx ml 2/ 7 3/9 (d; 4 pts) The DNA microarrays sold by Affymetrix typically use 25—mer DNA probes. One application is to detect the amount of a known mRNA sequence in a clinical sample. Each target mRNA (or fluorescently labeled copy of same) typically hybridizes to 20 or so different probe oligonucleotides on the array. Each of these perfect match probe oligos is accompanied by a “mismatch probe” on a nearby spot, with the idea being that the target mRNA will hybridize preferentially to the target probe over the mismatch probe. One doesn’t want to operate these sensitive and expensive machines at very high temperatures. Why don’t they use sequences much longer than 25 nucleotides long on the arrays? +1 Lava Mawwclulfiuok W J“ we? L‘gh Wt‘fi‘m H, So L w; wloL mi Ac 519% 7470 — Lek, Specta'Y/L'? W644» q 1%? f5 be 5M 2. Anaerobic Metabolism (25 pts) The “Anammox” reaction (anaerobic ammonia oxidation) is carried out by bacteria in the Black Sea. The Anammox bacteria are chemolithoautotrophs, which is considered an advantage for their possible use in wastewater treatment because they don’t have to be fed reduced carbon. (a; 8 pts) The Anammox reaction is as follows: NH4+(aq) + N02‘(aq) —> N2(g) + ZHZOU) Write down separate oxidative and reductive reactions for the individual conversions of ammonium and nitrite to elemental nitrogen. The beauty part is that the electrons and protons needed for Anammox are balanced. g 013 L_ V JEfiQNOJ +éefi+flht , MLkL-{Hw mg :3 Ml (*3 (a) Mama 6 (MM mm; ———> 20; + H Hue) / “me A (b;4pts) AG°’=—357 kJ/moleforthe Anammox reactiorpWhat is raw?“ E elm Mw‘rfik 6») M‘W— dug/Q. «(7% +ll 560‘: “A950, Mia/c \ 0 4— ,. :flm .— (-u .— 60 a n P“ (3) fiCSoodma / \fo +2. + \ NR“ _, .am Score for the page cin2lxx Exmll /2/7 _._ 4/9 (c; 2 pts) Is the Anammox reaction assimilative 0 Circle one.) +2_ (d; 8 pts) The source of nitrite (NOZ’) for this reaction is other bugs who reduce nitrate to nitrite: NO; + 2 H + 2 e‘ —> NO; + H20 This is an example of syntrophy. Briefly define syntrophy. It is obvious what the recipient in a syntrophic relationship gains. What is the advantage to the donor? @ my 0” W” Wfl {it} (,(Lolwv W CCMM arr/(1w M4442. Waff- 127 r M will (MO/12:44, & {1‘1 0/ / HM @ flvof 9m. {’«r awlw «11:7, 7,9 flaw Mug— 01/“; 0A 0W Wt>i(. (e; 3 pts) We have discussed the metabolism of methanogens, which transfer electrons from hydrogen to carbon dioxide to produce methane. They live on the edge of starvation, because the free energy available from this reaction is low. Why don’t the methanogens just burn the methane that they make and live high on the hog? MOW (Nu 04? tutu. Hex (W W (Le/vac; Score for the page ,hemi tr 271 ecti n21xx ml] 2 7 5/9 3. Redox Reactions to Live By (12 pts) The Breathalyzer works on the principle of oxidizing ethanol to acetic acid, with concurrent reduction of orange dichromate Cr2072't0 green C1”. mtandard red t1 n patqefitials. (a?) (I) gay-’9‘?" CH3COLO)H 41';+ + 4 e‘ —> Cl'lggHZOI-I(+ H20 E° = 0.058 V 5 "= v‘ ‘ . ' V 7' ‘ (kl 02072-314 Hal 6 e' —> 2 03+ +7 H20 ) 13° = 1.33 V * . a a;4pts What is tion? @ I @éKM-u. 3 (Jacky/(m + o - ~ 1’ {(5 2412/ ——~>3C(13C60H +W+ m (gyms/W $2020? +267?” +lle”——> “(03* afimo ('2; 7' (.33 \/ 5 Cuba/(M + ZC/207/ ’r (6 H+———)3CK3(OOV( +L(C.«~?” 4—H mo @ H H (b;2 pts) What is 13° for the overall reaction? E0 2 i3} ~OiO€ V 2 I (23V Q (c; 6 pts) What is the actual cell voltage E if all reagents are in their chemical standard states (1 M everything) except that the pH iS 4? 114 Vanna/41H. b E ; (2‘0 _ 7‘30; fi‘ @ n p @ _ 3 ‘(1\/\ +5 l-M :M'L?‘ “ 0050' lo D] D] (87 6» u. 6” ‘7 013mm”) My? we + 51 ill?- —- 0m VHF“ = (06“ Julia ‘1' Sim-4m rig; : quSV/ 4 60 ‘/ W Cr Chemi r 271 ti n21xx Ex ml] /2/ 7 6/ 4. Electrochemistry (14 pts) H1“ 6mm Mg, Mm“; ) The Leclanché cell (misnamed the dry cell) shown at the right has a g Pmitivc Inwlaring zinc anode (the negative terminal). The cathode is MnO2 dispersed “'“Cirmlc /Wh::_d MT in graphite throughout the body of the cell, with the central graphite /‘ ‘ electrode acting to return the electrons from the external circuit. The chloride in the picture is a spectator ion. (a; 10 pts) The anodic half—reaction for this galvanic cell is Zn(s) —> Zn2+ (aq) + 2 e‘. The cathodic half-reaction is 2 Mn02(s) + 2 NH4+ (aq) + n e‘ —> Mn203(s) + 2 NH3 (aq) + H20. Who is being reduced here? Calculate n for the reduction half—reaction. Write down the overall net galvanic reaction. MM 3W,“ “(5+3 “in .wul $.1er crisliion / (impliirc Cdlhluik‘ .\'ll4(',l '/,n(:l:, .\ln()_~ pns'u.’ Pom us SCPH “I for Zinc nnmic “rapper Negative ® n : 'L 45 Lorin ~ 2 MA x [2’ egg +5 for “Mt wier i“ . UP i .g 4 H20 ‘MY‘ILO'f, (2) , m; (4) (b; 4 pts) The Leclanché cell is headed for obsolescence, in part because its voltage drops off with time. It is being replaced by the alkaline cell with the overall reaction below, which uses the same redox couple. Why does the voltage of the Leclanché cell decline with use, whereas the voltage of the alkaline cell is constant until it is nearly dead (you need not calculate E°s)? Zn(s) + 2 Mn02(s) + H200) —> Zn(OH)2(s) + Mn203(s) fl 4, All a: M WM flrMQI/C gig/2,9 Maw cod, we?) M we?) saw .77 +1 ’ M Cell dvii‘ofic Score for the page r Lhemistry 27], section lex Exam ll, 5121127 7/9 5. Kinetics (24 pts) Kinetic partitioning describes the distribution of products for reactions under kinetic control, such as correc e incorporation by DNA polymerases. Forth elementary reaction A -> B, e differential rate law is —d[A]/dt = k[A]. (a; 3 pts) a [S ' g e law for a the first—order reaction A —> B assuming that we start with 100 % A at concentration [A]0 (not a trick question)? ” *3 [A] 3 M10 6 (b', 3 pts) Given that all of the starting material can always be found as either A or B during the reaction, derive an expression for the amount of B as a function of time. t61=wiloefifil= [aa]~[AJoe”“‘ +3 ~< L’TZ—FA gr (“I 7 ZAdJU’5/jw> H (,___,____,\ (c; 6 pts) Now consider the scheme at the right, where A can be converted to either B or C. Wm A L t What are the differential rate laws for 103 of [A], gain [B], and gain of [C]? k2 “ (m3 _ by; , y, m war/g =Qaww 1% J, M 3 H m M A La] \ ——-—.__ ’- obl‘ AW H fiLUfl H Score for the page , mi tr 271 t‘ n21xx x mII 2 7 8/9 (d; 3 pts) B and C are always produced at a constant ratio (in other words, for every mole of B we get x moles of C. From your answer to (c), what is [C]/[B] = x? VAN.» to] _ It,“ A?“ 0”} (e; 3 pts) By analogy with your answer in (a) for A —> B, what is the integrated rate law for the concentration of A as a function of time? TKA __ 6" [jut/$235 fit f/éb 0 5‘4qu +\ .cw-IM W #3 (f; 6 pts) The total concentration of products is given by [B] + [C] = [A]0 — [A] as in (b) above. Use your answers to (d) and (e) to derive the integrated rate law for the total amount of [B] as a function 0 ime- 2 VS [15]» ; [Mo ,. Mkthaauk . {far {c3 Score for the page hemi tr 71 i n21xx x m I /7 9/9 Useful Equations: AS — q/Tz 0 pH = — log([H+]) E 2 me2 S=kan AG=AH—TAS PV=nRT "D‘OH Ka = [H*][A‘]/[HA] AG° = — RTaneq 1e"! + 1 = 0 W = mm m!) nilneiexpL-(Ettfiol/fll Tm = AH°/[Cl/4>] °C = °K — 273.15 R = 1.987 cal/mole K' E = E° — 2.3 - T/n?)log,0Q AG = —n?7E a2 + b2 — 2abcosC = 62 , 1 atm gases 2.303RT/2’7 = 0.0592 Volts at 25 ° ? = 96500 C(oulomb)/mole 1 V = 1 J/C Chemical standard state: 1M solutes, pure liqu' Biochemical standard state: pH 7, all species in e ionic form found at pH 7 aneq = (—AH°/RT + AS°/R) R [A]=[A]O—kt m 13- 1Q rwguyow ln[A] = ln[A]0 — kt WM 1/[A] = 1/[A10 + 2k: Score for the page ...
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This note was uploaded on 09/07/2011 for the course CHEM 271 taught by Professor Staff during the Spring '08 term at Maryland.

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Chem271_S07_x2_key - Chemistry 271, Section 21xx Y ur ame:...

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