Midterm2Key (1) - Name AMWW W 1 Chemistry& Biochemistry...

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Unformatted text preview: Name AMWW W_______________ 1 Chemistry & Biochemistry 153C 2nd MIDTERM EXAMINATION 9 November 2007 (80 points, 3 pages, 50 minutes) PLEASE READ EACH QUESTION VERY CAREFULLY! POTENTIALL! van-UL consrms: R = 0.0083 kJ - mol‘1 . °K'1; Ambient temp = 298°K; F = 96.5 kJ ' mol‘1 ° voltd; 1 kcal = 4.2 kJ; h = 6.6x10-“ J ° 5; c= 3x10B m ° s'1 POTENTIALL! nan-m. REDUCTION POTENTIALS: E°' (Volts) 02 + 4e' + 4H” -) H20 0.815 s + 2e' + 2H* —) st -o.243 NADE+ + 2e' + 23" —) NADPH + H+ -0.320 1 (24 points) You have discovered a new species of bacteria that carry out E;*&‘vphotosynthesis in the presence of Hg) and “co,, but only if st is added and O2 is absent. During the course of photosynthesis, measured by formation of [“C]— carbohydrate, H28 is converted to elemental sulfur, but no 02 is evolved. (a) (12 points) This organism oxidizes H28 and passes electrons to NADP*. What wavelength of light would provide enough energy for'ihs to reduce NADP+ under standard conditions? H15 1‘3 dom’nné e’ 750 WWW A6“; (’0‘3ZOV3—G 0.2‘1’3‘0 féOIOHV AG“: ’2' 61196 U mX‘W/“I («OM40 ~ . W 7,. d, ‘i —l a oloé H dwmmch 0 A6 / ‘+/4.Ci kde‘e ‘6‘ F2: onS Witficm 6?) “Ropf ‘97 Hal/m _ . 23. V a: he \ +3 A a (wewo‘sbgeyol‘m‘sliw79%?) Z /\ #4360 J (mm 1 8 XIO_ m 7/ W 8550 hm (b) ( points) How does your answer compare with the wavelength requirement of plants t at use H20 to reduce NADP+ under standard conditions? “x INA/vets (WI M02. 11% 55 WW/L “£1qu (WW 2%» 69 beam/)2, Ywd’tm 6}; NADP“ W/lehwlvm 5mm mode/r (5 mm mm unM/MJCQ (enwgowc, ), (C)[(]X points) 02 must be absent for the bacteria to carry out photosynthesis. Briefly describe how the presence of 02 might act to impair or inactivate photosynthesis. 07, w‘“ bfl adVZVVJ“ WM ‘0‘} P‘Q6YV‘6/Vd’s (W6?) 0W cmfimfi 6k «\7\N>\'D§‘1n ‘ , -Flaw\’nsl scmtcfiw mmm e k) /\\N; 02" M w\\\ mm FL” ,EWM FMS cLuofm. elm/«1%. @1 6905‘ w&\\ a\8b dXsmWVL h) KM“ \XLD , me (“N/Wm“ 6mm «mm “107»? Ft” —; -0H + FR“= ”max (at/U cal w\\\ 615*”) “(rag kt WNW “km 135‘ ()hoh)$‘1l\Wl S _, NameMLw__ _ 2 2. (20 points) During periods of fasting skeletal muscle protein is degraded for the energy value of the amino acid carbons. Alanine is a major amino acid in proteins, and is transported to the liver where it provides a source of carbons for gluconeogenesis. Show a pathway for the conversion of Alanine to PEP. Include the complete structures of substrates and products, name the enzymes and cofactors and byproducts, and identify the cellular compartment in which they function . No electron pushing necessary here! \A WV; @ // P‘lwfmit @vbofiwm . 0* - CH3 am‘xofiwsfimfi / CA3 HLO3D‘O n LOO MAO“ NA $50 W a be 7r? I» W 2* GD ’ V665 c) v C20 NH; U50 s ta) M0 A00 \ . H “ OH f / M“ - Am w M?“ as @P‘ CO” 63‘ ED? 7 N 'L/Gn (EAL 41:1» \ _‘_—-_——-— ;;(3 ‘\ V . r); - may chou Cao-DO‘ \gx {65: (gm \ 0- CC? ‘ Earp ‘ ® ® ® 3. (10 points) Unlike Alanine, Lysine and Leucine cannot be converted to glucose in our cells because their degradation yields only AcetylSCoA. However, plants and many microorganisms can readily convert AcetleoA to glucose because they have two enzymes we lack. Show the reaction catalyzed by theseZenzymes.\Explain how the presence of these enzymes enable plants and many microorganisms to convert AcetylSCoA to glucose. {) . (99 r / u ~ by . A-p‘ \ A .u \ 5 + . WW umpu e SNUC o? g «(1/ O ,, .. {/044 {LC WWAL. ‘c ‘dx'dS' glfi€’/;q ’_C:~/ C,Cfl3 C’ */ Cli ~C,’S(D (7X0 J H ‘79:? g ’M‘" W (Alum _ (10L 050 \A , H b\ - i (8% (“0&033. +1 lefifiwk g—r M 260 ‘ @ vii,“ CW— ®fiM7pr 3L1 Mat lay—gm WVL MUM: LE \ » e) > COO MILD 5 Ma C6D \’ H 1 M M wco . It) 1 _\ X1) ~ \ 5‘” WW 0 wow _ Q?” WW 3 ($on Kim-“5;..— Mop’lfi’ 1:: “JV Name—'fl’B—W‘ZV K056 __ __ 3 4. (20 points) Our cells contain enzymes very similar to transketola e enzymes in the Calvin Cycle. However, the reactions catalyzed by the transketolases in our cells are exactly opposite as compared to the Calvin Cycle. Show a mechanism for the conversion of two 5—C sugar phosphates to one 7—C and one 3-C sugar phosphate in the reaction depicted below. Indicate movements of electron with arrows and draw just the ”business end” of any relevant cofactors.///_fifl (fl/ 1,/ J. i H\ K /) {by / {.\ ‘N‘ CH CH C=01 + / CH OH \ H \‘ofilm um»: /. '6 ,v ~ >C=—‘O 14+ H—C—OH \ C=O I 5 f? ,r 1 H— —OH «()9 @ HO—C—H + H—C—OH ——> HO—c—H + . M ? é / | l , | cwm mxw H— C—OH H—C—OH / H—C—OH I l / I glygeralfiiehyde- ' — t CH20P032’ CH20P032' H —c—0H P OSP “P @/ Xylulose—S-phosphate Ribose—S-phosphate I QR fin’Mfir’o" 5. 5 points) In plants the above reaction operates in the opposite direction (the 7—C and the 3-C sugar phosphates are used to produce 2 S—C sugar phosphates). What function is served by thi} transketolase enzyme? in the CALVIN CYCLE? Th6 59““ kW nmtox® mama“) @ 'fimm M 5 _ {4?le ? WUWCX b7 m MAW/k.) ' ' c - 1‘ - +. M it; v“ m Wm <29 _ _ ._,2 E /;£m' W. ...
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