First Midterm Exam Key

First Midterm Exam Key - Biochemistry 10013 FIRST MIDTERM...

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Unformatted text preview: Biochemistry 10013 FIRST MIDTERM EXAM Winter 2007 Bogomolni / Page 1 Name: ' Student Numberflast four digits}: Score: 1- 40 pts 2- 20pts 3- 30pts 4- 30pts 5- 15pts 6- 15pts Total 150pts This is a closed book examination. K you are not sure about the ineaning of a question please do not hesitate to ask for clarification. Show all your work and include all appropriate units in your answers. You may use a calculator. fit more space is needed for your answers use the reverse side of the page, but please indicate that the question continues on the reverse! *Mnrk all pages with your name. 1- Now that ligand binding and thermodynamics has been covered in detail you should be able to define the following in 20 words or less:(5 points each). a- AllféfiricLTW hiWLQ“? fl’r 041M Cm log, M ' p . . ‘ (astigmatism). it b- Intrinsic in apparent bin 32 ing( nd di sociation) e ilibrium constants fl .5 mike ' mam“ a, t “hm of “magma A Wefi’ Cons'l' " £2 ALL Arm-rasng 'flltfiocimw‘ ween b Srl’bfla (5- Transition state The. Liz/{gsi' Sistine. aloft? Wk 4;} 0L met-rem- a d- Free energy of activa GHQ.» r: We ml El; 60"; in M15 hé‘ll'wéiiw W3 ' S-i'HQ, 15“ 6% EC! &,rmr:ci e- Homotropic allosteric effect 1' L? A 40¢ a. Nan PNIYQJH can, rug lob, pLC‘l'VL‘h' ' "+5 0 i ' ' ‘ ’br f— Scatchar . ‘ i own fig" JL'W“ CL} '3 Mt?ng ) I CM. 3-7.5 a. “‘W teal ~ iys'l'tm 'i'L floh \tQ u.- I Q . . 3- Hill plot r x . . f; a, (o Pl+ h" 5 Shift ‘ x ' , I . , n 3 ' f " .' “wahfle New 9’63“: ]« «Isl Effifimag' (3‘2" a h- Cooperative binding fPILIl-Mmgfldfin (:91 (155mg Amrflnj 77$ dlfflVVflJ “JIM lamggm (TC I i A MW‘HS‘fi’ hm iri—ué ailfiSQC wt New!) Milena; . Biochemistry 10013 FIRST MIDTERM EXAM Winter 2007 Page 2 Bogomolni K I . Name: U Z Student Numberflast four digits): 2- 20pts a- A certain reaction has the following stoichiometry: A + B -—> 2C The standard free energy change (at 298K) for the reaction is observed to be -l_Q_.§,_k._I[mol . The reaction is set up with 0.50M of reactant A, 0.10M reactant B and 2.0M of product C. /0f£( What is the free energy change under these starting conditions, and will the reaction proceed spontaneously as written? R = 8.315 J/molK .v _ eo , __ «o tcj‘ Ag .419 +t'LTan .— A0 + KTln mm _ 1 AG’: “forgifl‘mVl 'i’ AYiSIS'XIUl \nr ( g ' ° _l %K > [05709.0 AG 2 3‘9; X‘0 Ida-r SWUTANEOWS .’ b- A certain process is observed to be spontaneous. The AS for this process is observed to be a negative value. What general conclusions can you draw regarding AS, AH and AG for this (0 re process? What might happen if we raise the temperature? Provide the relevant equati0n(s) to 3 support your conclusions _ AS 2? (-) AG = AH -TAS SOB ~ Ag '5:- nafimlvwo. mime M91798 om incfwlnaw- Class dfissornfir") AG 3' WKM SfOnWS Vail/Ll is [archw- 2m, - .. ._. ELO :1: MMT T50. 6L Maw Mam/mic Valid. AH ovwcnmz ’pasicizm. Q7355 {Sn/amt A61 ’AH'TASi T incrqasqr TAS b‘LCOmLS .wnn larder Qt") Vault/UL... 74am, *IMPES Pfiarfihom bJ‘iIHflJUJ smn‘i'mow’ M A” “TAS afme are er (4—) Viva/9‘1" 109-65 fits l3 Biochemistry 100B FIRST MIDTERM EXAM Winter 2007 Bogomolni ’ Page 3 Name: z Student Number(last four digits): 3- 30 pts. Studies of oxygen transport in pregnant females have shown that the oxygen saturation curves of fetal and maternal blood are markedly different when measured under the same conditions. This phenomenon arises from the fact that fetal erythrocytes (red blood cells) contain a structural variant of hemoglobin (hemoglobin F), while maternal erythrocytes contain normal hemoglobin. a)- Which hemoglobin must have a higher affinig for oxygen to allow transfer of oxygen from mother to baby? Explain your answer. Hi5 F mus r HAVS (:aem-aa 4FFIA/L‘TY {10¢- 0; into-«92¢ 'l'a: oui'compt'l’c MA {3:}: O @6641. ?aflb‘m9~ Wu.st <99 (9; 066 Céluwflwfl- The graph shows the 02 saturation curve for human hemoglobin (Hb A). Hemoglobin U; Binding 1.0 - r Fraction Saturation 140 100 6D 30 120 pl]; (turrs) b) Draw a curve on the graph for the 02 saturation expected for hemoglobin F. SQ Gr SW” '- Q Md HLF,‘ an 4 730;, um 0) Draw accurately a second curve on the graph for a hypothetical hemoglobin mutant ("Hb hm") with the same Ed gor P50 as Hb A, but with a Hill coefficient, nH = 1.0. To draw this curve more accurate y, 1rst ca en a e its fractional saturation at the p02 values in the lungs and tissues. . f)“ :1 Tmpiits IMD COOPW‘LNFI'V Pg“) HLA; P50 = 120; Hum = 25' em r ,1 _ 20 fear 1 J r . am: Pg: +31; 20a.» + New O'LH 3 A??? 4“ 50¢ W “n Pl“L {ooeam W WILL __.__._::.- — games 2 P Io 014w raflwr Biochemistry 1003 FIRST MIDTERM EXAM Winter 2007 Page 4 Bogomolni - Name: K f z Student Numberflast four digits): 4- 30pts A dimeric allosteric receptor that follows well the MCW concerted model has an equilibrium constant for the conversion of the RD into the T0 states, L= To/‘Rfl = 104 . The intrinsic ligand dissociation constants for single and double occupied states of R states and T states can be assumed to be the same for each step of substrate binding. When the enzyme is saturated with substrate (such that both T and R states are fully occupied with substrate) the ratio of T to R state concentrations (T21R1) is 102. a- Qualitativelx does this observation indicate that the ligand binds more loosely or more tightly to the R state than to the T state? (in other words is the intrinsic dissociation constant TKi larger or smaller that RKi? Explain b- Quantitatively find the factor by which the ligand binds more strongly to the R state than to the T state (or viceversa, is such is the case). Show all your work. Lime—:00“ q) L211; when? Tzilslal: Tic-ll L, ll “K: L14 L0 is? a slat. km N l :CQOH L7. ctmmii .) f; RK; (Lama/a We. ‘i—t M L1 ‘9 Ring-'8 ver‘CL “leak-811th RE _-_-v 1:3... 1:. —= [0’6 Lo (Onl Biochemistry 1003 FIRST MIDTERM EXAM Winter 2007 Bogomolni 4 f Page 5 Name: l C Student Numbeflast four digits): 5- 15 pts.- The following reaction proceeds to the right spontaneously in normal metabolism; Explain how can a reaction with such unfavorable AG“ be spontaneous in a‘living cell.. Citrate —+ isocitrate AG“ == +133 kJ/mol A 6’0 z 491+ "RT In..__ml:l‘““l'“3l‘g_ ” Mass AJeq” in liar/5’0; 24inch CC} >>> ‘I “‘3' 3:” M lid-mm)me (idle.ch (Macm‘irfliiéno bi: “isoc'L‘lraJi'Q 61/qu CW am he. Wn‘i' gm“ "15* IMA'U ACTIBJW :3 wafiunhmny mam +0 ‘9me w AG“ hammer. 6- 15pts The KMfOI an enzyme is 1.0 x 10'5 mol L" . At a substrate concentration of 0.01mol L" the rate of reaction is 37 umol min—I. The same result is obtained at a substrate concentration of 0.001 mol L'] What's going on? VW 7" $51 mi [52] um walks )mjni/‘owii lay“ éi‘afi Km, Evan; I‘S [aka/X offar‘c‘vi‘mrj @ fivepwa, emfl madden (Wars ‘50 lJQ, Ntlepj, (Sika (51C J :1pr rough Ecnl c9pr on Edr- ...
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This note was uploaded on 04/07/2010 for the course BIOC 100B taught by Professor Sethrubin during the Winter '10 term at UCSC.

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First Midterm Exam Key - Biochemistry 10013 FIRST MIDTERM...

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