2006 Exams and Keys - amen-acwm-Mmyuw» “u Juneau-M....

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Unformatted text preview: amen-acwm-Mmyuw» “u Juneau-M. emsue‘iimii’é?) “Ma—Wmmumzwuuwm.pauwmmhn1m Wm H. _ I _ "m CHEM C13D/MCB C100A. FINAL EXAM FALL 2006. UC Berkeley. YOUR NAME: v m. UNIVERSITY OF CALIFORNIA, BERKELEY CHEM C130/MCB C100A. FINAL EXAM. December 14, 2006 INSTRUCTOR: John Kuriyan THE TIME LIMIT FOR THIS EXAMINATION IS 2 HOURS and FIFTY MINUTES SIGNATURE: Please SIGN your name (in indelible ink) on the line above. YOUR NAME: ___________________________————————————————————- Please PRINT your name (in indelible ink) on the line above (& on the top right hand corner of every page). Also, please write all of your answers as leginy as possible. PLEASE CIRCLE THE NAME OF THE GSI FROM WHOM YOU WILL PICK UP YOUR GRADED MID—TERM EXAM: Natasha Keith Kyle Simonetta Jonas Lee James Fraser Allen Liu This exam consists of 10 questions. The points for each question are indicated below, for a total of 100 points. This exam counts for 400 points out of the final score of 1100 for the course, and so your score on this exam will be multiplied by 4.0 when the final score is calculated. Question 1” Part (A)_ ParflB Part 0 Part _(D) Total I 1 . (4) (6) xxxxx xxxxx ( 10) 2. (4) (2) 4) xxxxx 10 3. 3 3 4 10 4- t—w ——@——4 —<m> 5. 4 3 3 xxxxx 10 7. 2 4 4 xxxxx 10 8. 6) (4) xxxxx xxxxx 1 O 10. 6 10 Page I of 22 CHEM C130lMCB 6100A. FlNAL EXAM. FALL 2006. UC Berkeley. YOUR NAME: Question 1. (10 points total) 01A (4 points). I Shown below are two oxygen binding isotherms, labeled A and B. One Isotherm is for hemoglobin and one is for myoglobin. ‘ i i l l ‘ Ogrygen flange oi’A”(i’)‘Wfii¢h“6he) A or B, is the hemoglobin isotherm? Explain how you made your choice. (2 points) B is m 80+ ES giWa (a Moshe A is #12. wjiijfsgt savings, . 3% i3 he; (MOW - a. ll 0 swic Q1A (ii) A solution containing hemoglobin is treated so as to remove all the bisphosphoglycerate (BPG). On the diagram above, sketch the resulting binding isotherm for BPG—free hemoglobin. Explain briefly the mechanism by which BPG alters the binding isotherm of hemoglobin. (2 points) We. stale; is; (23m rm. amzrwmmrw (‘7‘ be , ,(Z.€Axm&l«u.§..: B We in ow.m¢,,g file. 02, w‘ (W 35gb? it» Page 2 of 22 CHEM ClSO/MCB C100A. FINAL EXAM FALL 2006. UC Berkeley, YOUR NAME: Question 1, continued. (B) (6 points) I _ _ . I Shown below is the binding isotherm fora ligand binding to a protein. Q'iB(i) Does the protein exhibit positive or negative cooperativity? Explain the 8 basis for your answer. (2 points) " 0' At r5040 L5». c.9wt..4va.,trla."m' Kb :3: ~02 a RD : IO "8 4f cammfaufiw D a; [£2 .33; Kb : lo 24ch Kb rnwmwfq w’fim ligand Linnea, 7%} l3 NEMWVE MPERAYMM umr-uWalP;n—u rrwm -»~ Q1B(ii) Estimate roughly the difference in standard binding free energy (AAG°), at 300K, in kJ mol'1 between the ligand and the protein at very low ligand concentration and at very high ligand concentration. That is, calculate the value of: AAG° = AGEM (high ligand concentration) — AGEW, (low ligand concentration) (4 points) AQOZEfifix Mumfrmn‘m) : ——RT,ZYL (W's) l 3 ’9 g: r\ m if“ C2 Aéo/Law mammflmfgtm) .3 m [Q 7" KM (IO—0.2) % 4’456/[13' Mi”, 4‘46 3 4’7“? king?" Page 3 0f 22 CHEM C130/MCB C1ODA. FlNAL EXAM FALL 2006. UC Berkeley. YOUR NAME: Question 2 (10 points total) Q2(A). (4 points) . Q2A(i) What is the major contribution to the energy of DNA double helix formation? Explain briefly the nature of this energy contribution. (2 pomts) I L 0 ‘ ‘ M ‘ i I". [aflfifivw +9 an ,égflflwézzfl Singer/Ac p 149%.; ab «,5 MW M Wm (it): Mantle Ongté‘ib’é’g— 12W cm W MAM-rem, ,5ij SW 02A(ii) Explain briefly the critical distinguishing feature between correct and incorrect basepairs that enables DNA polymerases to make fewer than about 1 error in 105 nucleotide incorporations. (2 points) We gwa g?" 65W r'cci" awe Page 4 of 22 CHEM C130/MCB C100A. FINAL EXAM, FALL 2006. UC Berkeley. YOUR NAME: Question 2, continued. Q28. (2 points) Shown below is a Lineweaver~Burke plot for an ezyme catalyzed reaction that obeys Michaelis-Menten kinetics, in the presence and absence of an inhibitor. Does the inhibitor exhibit competitive or non—competitive inhibition? Explain the basis for your answer. inhibited reaction normal reaction l/[S] WM“) @WLJAW L-M M‘VW 55616) Mammy?!» 41%» mMMc’A. @ L 113le“: 33> imkét‘wfi? )3 Afltmj Page 5 of 22 CHEM C13OIMCB C1OOA. FlNAL EXAM. FALL 2006. UC Berkeley. YOUR NAME: Question 2, continued 02C (4 points) The enzyme reaction is carried out in the presence of another small molecule, labeled B. A graph of reaciton velocity versus substrate concentration in shown below. The concentration of B is kept fixed at the same constant value during all the measurements. reaction in presence of B \ velocity (v) reaction in absence of B substrate concentration [S] 020 (i) ls B likely to be binding at the same site on the enzyme as the substrate? Explain your reasoning. (2 points) 25 is mafia ta AMJ mi" m SW .512. Keenan.“ ' VW is in cahed he ffiaawm BU WWW B «I'D «6%me a1” gamma... was 5156 9v!» sWAIE, , 5W m if mite-15’” m.w mwfllrt [We II icy 020 (ii) Explain the effect of B on the reaction rate. and describe how this could occur. (2 points) (B ’3‘ 0M fljm"&1¥€x aclitvdam ' Add A? £14. gaww, )n j My 61 ‘Mfi’ M 723. Mar at.“ 01>” Hun}? %9‘2 A}: m omit (we: Add; 1th if”: 0 m )‘S tun/WW1 / . Page 6 of22 afifiafigwwwaw 1W M‘ CHEM C‘lSOIMCB Cl DOA. FINAL EXAM, FALL 2006. UC Berkeley. YOUR NAME: Q3 (10 points total) 03A. (3 points)The graph below shows the velocity as a function of substrate concentration for an enzyme-catalyzed reaction that obeys Michaeiis-Menten kinetics. per minute) initial velocity (micromoie o 0 40 so so 100 120 substrate concentration (HM) What is the value of KM, the Michaeiis constant, based on the data shown above? Page 7 of 22 CHEM C130/MCB 0100A, FINAL EXAM FALL 2006. UC Berkeley. YOUR NAME: Question 3, continued. 038 (3 points)The value of k2 is 1000 min'l. What is the total amount of the enzyme in the solution? VW x £2“ [El-m7. 0w l/W, A... «w :2 0' gfiuims-Em x: to“ 3 :1 ’3' Emma: Witt/“£5 03C (4 points) The value of k.1 is ‘107 sec". What is the value of the dissociation constant (KD) for the enzyme-substrate complex? Page 8 of 22 CHEM C130lMCB C100A. FINAL EXAM. FALL 2006. UC Berkeley. YOUR NAME: Question 4 (10 points total) Q4 (A) (4 points) Two metal electrodes, labeled A and B, are inserted into a lemon (see figure). The positive pole of a voltmeter is attached to electrode A. The negative pole of the voltmeter is attached to electrode B. When A is copper, and B is zinc, the voltage recorded by the voltmeter is positive. For each pair of electrodes in the table below, write down whether the voltage measured by the voltmeter is expected to be positive or negative (4 points) ’ ? _ A table of standard redox potentials is given below. Oxidizing Agent Reducing Agent Reduction Potential(V) A1+3 + 3e’ = A1 —1.66 Zn+2 + Ze‘ = Zn —O.76 Ni+2 + Ze' = Ni —0.23 Sn+2 + Ze‘ = Sn —o.14 Pb” + 2e‘ = Pb —0.13 2H* + e' = H2 —0.00 c:u*2 + Ze' = Cu +0.34 Ag" + e' = Ag +0.80 Page 9 of 22 CHEM C1BO/MCB C100At FINAL EXAM. FALL 2006. UC Berkeley. YOUR NAME: Q4, continued. Q4(B). As a lemon ripens it becomes more acidic. A pair of metal electrodes are used to measure the voltage from an unripe lemon and from a ripe lemon. Which lemon will display the larger absolute value of voltage? Explain your reasoning. (2 points) W W\ 35 (“u/am.) flit] increased. 751 WJMI 665' We, Ag’acfirm/m )5 DNI+ g" .s ill, jawma it?» i?“ "'3 09mm mm ma an. M. lie MAM») so film W” VWm‘irfi- / will“) m‘l/ Wm 04(0) A coupled chemical reaction that is very important in biological energy production is the oxidation of NADH by pyruvate to yield NAD” and lactate. The relevant redox potentials (under biochemical standard state conditions) are as follows: Pyruvate + 2H” + 2e" —> lactate (E04 = —O. 185V) NADJr + 2H+ + 2e‘ ~—>NADH + HJ' (EOK = —0.315V) Consider the following condition: [lactateJ/[pyruvate] = ‘l; [NAD+]/[NADH] = 1 Calculate the free energy change for the reaction (in the direction pyruvate -> lactate) at pH 7 for this condition. Will pyruvate convert spontaneously to lactate? (4points) “ ‘ i .- NAtsH + W ———> Nl«bi‘"+ 2i‘4*+— 2e E.°’:+ 0 3H: l/ is“: ~04?“ Mt 2” l t 2 a ' “' ~M.W.g. up.“ tuneup. any... .. A “News.” '11:,» a, ' = i av Pgmmfir NA-verwK-s- Nisan imam. ,5 013 ’ 4G” —~ vaE" = —— 2 Memo x new 0 AG: 1360+”ng .2: »25-»i+ Pr/émé'iiditlsl [if :1" NW“ (W‘Qm: . :: ~ 25‘! w,flfiMtv-/wa,é”0«nl since“ f: ‘ LS Page 10 of22 flew ‘ am WM CHEM C130/MCB C100A. FINAL EXAM. FALL 2005. UC Berkeley. YOUR NAME: QUESTION 5 _ I ‘ I 05(A) (4 points) Neuronal cells transmit electrical lmpulses m the form of action potentials (voltage spikes) that are generated by voltage-gated ion channels. The membrane potential as a function of time at one particular position in the cell is shown below. A *50 > E I? E n 2 8 A B a: E 50 i n E cu E-too _____, time Two time points are denoted A and B in the diagram. For each time point, circle the best description of the Na+ and K+ currents below (2 points each): Time point A: , .. t.ela:.fl9ytl.§..i.ntgemecetlwa s into the cell. 2 Na” flowewinto the cell, K+ flux is zero. wa5 into t e ce , o owe the cell. (4) Na+ flows out of the cell, K' flows into the cell. (5) Na+ flows out of the cell, K” flux is zero. (6) Na+ flows out of the cell, K+ flows out of the cell. (7) Na+ flux is zero, K+ flows into the cell. (8) Na” flux is zero, K+ flux is zero. (9) Na+ flux is zero, K+ flows out of the cell. Time point B: (1) Na+ flows into the cell, K+ flows into the cell. (2) Na+ flows into the cell, K" flux is zero. (3) Na+ flows into the cell, K+ flows out of the cell. (4) Na+ flows out of the cell, K+ flows into the cell. (5) Na+ flows out of the cell, K+ flux is zero. (6 .maitflgmtegutgtthengellilf of the cell. 7) Na“ flux is zero, K+ flows into the cell. Autumn -rn'-mu=I-M 'hhn. a ux ls EEFETMFCTFJYIS zero. (9) Na+ flux is zero, K+ flows out of the cell. Page 11 of22 CHEM C130/MCB C‘lOOA‘ FINAL EXAM. FALL 2006. UC Berkeley. YOUR NAME: 05(8) (6 points) Consider a cell in which the membrane potential is —50 mV (negative inside, positive outside), and in which this potential is due solely to Na+ ions. QSB(i) lfthe concentration of Na+ outside the cell is 150 mM, what is the concentration of Na+ inside the cell? (3 points). ‘3 [Na *3 IN +£92.51?”— =7 {Milo-Jr -.-. QjAE/o'mm = 6 0.0237: 7'0 [Mil/.4 :2 N F. = = 2/, >[ elm “7,0 QSB(ii) What is the difference in chemical potential (molar free energy) for sodium ions inside and outside the cell (i.e., calculate the value of um - mm for sodium ions arising from the membrane potential and the concentration difference). (3 points). -= 4 awarw‘ (2) Wax/w / w < /M 0W.) xxo'3 A r 3 Voflflqém x cfiémgc = ’50X75 50%], fl 3'4«32k3m&€”‘ ,'. 15/1“ “9.7 LTVMK,“ Page 12 of 22 CHEM C130/MCB C1OOA. FlNAL EXAM. FALL 2006. Up Berkeley. YOUR NAME: Question 6 (10 points total) Q6(A). Consider a peptide with the sequence: Val—Leu-Leu-Val-Leu-Leu-Leu-Leu This peptide converts from a random coil conformation to a helical conformation when transferred from water to non-polar solvent (e.g. chloroform). Explain why. (2 points) elm We». my. [mum 0.0 W M 749444 W £49....pr F40 Warm, vat/I a may) fiat/2.4 $33.5»,th $2127 WE H~ 190ml lo 410 11%, mng H"’Aé5"?’ifi£& $55.2qu tin 0k [Na/64f» 06(8). Consider two peptides: ‘ .“,,...r«'m*"""’“”"""-<~n.....Mm....,.....mm«~~w““""“’“"M'" ' "‘"Wfi Peptide 1: Glu-Val-Leu-Leu-Arg-Val—Leu-Leu-Glu-Leu—Val-Leu-Arg \ \. fl... 4... m "M n p ...‘I w _ , .. ' "m rWA‘I-l». AIM Jig—M."mrrm KWWMQ‘w _, .. 1m, “71 n- Peptide 2: Glu-VaI-Leu—Leu-Glu-Val-Arg-Leu-Arg-Leu-Val-Leu-Arg One of these peptides is more stable as an a—helix in water than the other. Explain which one. (2 points) KLf/ifl'ofiq 4. 4M Gm. Wee.“ Av; MW g,” "£7"ng Wk [37, {if 4) :50 (3;... aw: Page 13 of 22 CHEM C130lMCB C1ODA. FINAL EXAM. FALL 2006‘ UC Berkeley. YOUR NAME: QUESTION 6 (continued) 06(0). Consider two peptide . 1...,«u-M.‘ .«Wmfim, if-” “,7. 1 V ' ' ""'"-~——-‘.__._ Peptide 1: ASp-Leu—Leu—Leu-Va|-Leu-Ala-Leu—Leu-Aia—Arg .,__, .,._."m-ntmm.m.—vwmmwwwwm m... hwy...” Peptide 2: Arg-Leu-Leu-Leu-Val-Leu-A|a-Lue-Leu-Aia-Asp One of these forms an a—helix in a nonpolar solvent while the other one does not. Which one is helical in the non-polar solvent? Explain your reasoning. (2 points) )we awe cm, was arm c—W3 M c—WM E m. aa/mxz,9a [’1‘ / Mag 6L0 A I443)»: 06(0) (4 points) Shown below is the energy of a hydrogen bond as a function of hydrogen-oxygen distance in vacuum. Energy ikJ/molel Distance (NH --- o=c> (A) What are reasonable values for the well depth (a, kJ moi") and the optimal distance (r, A)? r w, 1/50ng Page 14 of 22 MM CHEM C130/MCB C100A. FlNAL EXAM. FALL 2006‘ UC Berkeley. YOUR NAME: Question 7 (10 points total) 07(A). An aspartate residue and a lysine residue in a folded and water soluble protein form an ion pair. The energy, U(r), of the ion pairing interaction is given by: 91% U“) = 477:808r Where q1 and qz are the charges, r is the distance between the charges, a. is a parameter known as the permittivity, s is the dielectric constant (a dimensionless pure number). If ris expressed in meters, qi and q2 in Coulombs, and energy in Joules, what is the unit for the permittivity, so? (2 points) I ch => EO=CZJAW 50 x M (Wk) ’ ‘- QT(B). By substituting values for the charges and the permittivity, and expressing energy in kJ mol" and distance in A, we get the following expression for the energy: —1380 Br U0“): When the distance between the lysine and the aspartate is increased from 3 A to 4 A, the energy increases by1.4 kJ mol“. is this ion pair located near the surface of the protein or near the interior? Explain the reasoning behind your answer. (4 points). . :2 fi‘ 2 I 2) 2 A? A M 4,7 wt? ciflvi'c $41“ . ' Jaw“ » - c, a writ-wt" ‘" “a m , mm (40)» fit“ . 5;”... ca. A» My? ’9 91/? m Page 15 of 22 I CHEM C130lMCB C1ODA. FINAL EXAM. FALL 2006. UC Berkeley. YOUR NAME: Question 7 (continued) Q7(C). What is the value of the electrostatic force on the lysine when the ion- pairing distance is 3 A? Express the force in units of kJ mol'1 A'1. Indicate with an arrow on the diagram below the direction of the force (i.e, whether the force is away from or towards the aspartate. (4 points) in “1* 3A Page 16 of 22 CHEM C130IMCB C100A. FINAL EXAM. FALL 2006. UC Berkeley. YOUR NAME: Question 8 (10 points) 08(A). (6 points) A protein undergoes a folding transition as follows: U :F where U is the unfolded state and F is the folded state. The concentrations of U and F at two temperatures is as follows: — [u] (11M) [mum 300K __ 350K QSA(i) What are values of AH° and AS" for the folding reaction, assuming that these are temperature-independent? (4 points) 0 , )0 = —-s.75L3m92“ A6 :~,Q71énlglj =_qunT @130“ A60 = - ETA; - ~8-3MX3SOX(’ 2-3o)><io'3 '0 r. +6.59 lefywof'ai' 3523K. => AHO- Booxaso : —5-75@@D AHO— 3b—DXASO :+é‘6q I 0 -l - _ = — (D—é): 50XA5°= 42444me 9‘15 0 1’ r' AH”: —5-75 +300 xaso = —80' 4S éJ—MM-‘ Q8A (ii) What is the value of the melting temperature of the protein (i.e., the temperature at which half the protein is unfolded at equilibrium)? (2 points) x46 7,; AGO = Zora ) z) AHov-TMXAso =0 => T : AH“ -.-. ~8’D'4‘S‘ = $23.; l< Page 17 of 22 CHEM C130/MCB C100A_ FINAL EXAM. FALL 2005. UC Berkeley. YOUR NAME: 08(8). (4 points) The schematic diagram below shows the positions of 3 histidine sidechains in a protein, labeled A, B and C. The pKa values of the histidines are 5.0, 6.5 and 7.5. Identify which histidine sidechain has which pKa value. pKaoinsA= 7'5 pKaoinsB= 5-0 pKaoinsC= Page 18 of 22 CHEM C130/MCB C1ODA. FINAL EXAM. FALL 2006. UC Berkeley‘ YOUR NAME: Question 9 (10 points total) Consider a molecular system containing A type molecules and B type molecules. A and B can interconvert, and each has 2 energy levels, shown below. The number of moles. at equilibrium in each of the energy levels of states, A and B at 300K are shown. n1= 0.119 Energy = ? m: 0.100 Energy = ? no = 0.881 Energy = 0 n9 = 0.900 Energy = 0 A B 09(A). What is the molar change in energy in going from pure A to pure 8?. Show all the steps of your calculation. (5 points) gr A) 7’11 :_ JUl/ET _ U] = —£7‘/g¢ = Q‘OXZ'S 7/1.: _> Afr/1191'. :_ 5/13 M“. .. - 0'] (W 8) U1 _ Er’gy‘ W Page 19 of 22 CHEM C130/MCB C100A. FINAL EXAM. FALL 2006‘ UC Berkeley YOUR NAME: 09(3). What is the molar change in entropy in going from pure A to pure 8?: Show all the steps of your calculation. (5 points) r3? 4 ,5}. 2 mi A; = —[m% 0.06} +4. est/£01281] 3 Nk _ = 0,253 +ZD-l12 = “"5 (Li/Lome “A72 , 0325 , ,4: = “Ml Nk gt NA: Q; 4% 1m9 , -. / p0 Lune ‘fl’ESZJK . :r "‘ Page 20 of 22 Ci-iEM C130/MCB C100A. FINAL EXAM. FALL 2006. UC Berkeley. YOUR NAME: Question 10 010(A). (6 points) An inhibitor binds to a desired target with an association constant of 1010. It binds to an undesired target with an association constant of 107. Both targets are present in a solution. Q10A(i) What is the maximum fractional saturation of the desired target that yields no more than 0.5% saturation of the undesired target? (3 points) f 4405: [Ll/m (74w WWW {AW 1+ [LE/g1, M X : EL;ED a) 0.005 : l :) fl ,00§‘(/+X> = X M I+-x :3) 0,005 gquqs‘x wt». m = :> x ; a. a 0 50 L H, seems/wk .1— : i = 83% a», m = MDWZK /0_ H—5 9 5 502“? MO 7 z s Xio'wM Q10A (ii) What is the concentration of the drug that satisfies this condition? (3 points) 77.: CM mM/LWZEM is SEXIEHOM Page 21 of 22 CHEM C130/MCB 0100A, FINAL EXAM FALL 2006. UC Berkeley. YOUR NAME: QUESTION 10 (continued) 010(8). One mole of an ideal gas is contained in a 2L volume chamber. What is the probability that all the molecules are in the left hand quarter of the chamber relative to the probability that the molecules are distributed everywhere in the chamber? (4 points) ’ Total A volume = 2L Left hand quarter (volume = 0.5L) (95' Qatar”; (§;’ér9«MlLAu: lA/ZL' W05!- J = 54L _ 5L “gang‘s—L '2’ 2%” END OF EXAM Page 22 of 22 ...
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This note was uploaded on 01/12/2010 for the course MCB 100A taught by Professor Kuryian during the Fall '09 term at University of California, Berkeley.

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2006 Exams and Keys - amen-acwm-Mmyuw» “u Juneau-M....

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