exam3key - [AZ 141“ 200K [8]. = _'__.'2'.0;<ID‘3...

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Unformatted text preview: [AZ 141“ 200K [8]. = _'__.'2'.0;<ID‘3 =' (9.772(10'3 x10 . [A] 2~é no“ .5 A [—83 = 6 65 [A3 AG" =—-8-3z4 x 200 )6. 6:65 :mef . = —N058 3m" =-//-/ RI m»! "r _.Nwa m m . 3 almucAM A 5’0 ILL W Jam: "1;" Amara» 42m.” [31 CA] . == - 8-31” 3’00 £7 (“0 Xian?) _ _ _ U /-0)&/0_‘3 __ = 0 @7651?” W W“ ‘62 1(8) " ' Afl/zwz)“: AH" —-2002<AS° I (I) ' _ 46-”[20010446'068’0020 = —//-/ @7qu ' " " " _ '_={—200 +300) A50 =' +/0043 " "__ % '4503 -//-1 29204123 mt"! ' é”) . 33 A' :ifir 1’, 1?; 57'szfo )6 m4 £29 I. w—C mm M 52/8 29 mm,” yaw/3:21".- " 5W7%3= 40”“? 2. W 6 .023 xzaz’x4-o mo” —! J»? = 924.1 x/oz J me" = 241 AJ‘ MHZ" Atfgruflafirim) N2_ = 6‘02: UD/RT if}: 2005, 27‘: Egifix3ao= 2.9 A? m—e" " . __N_z_ =—. g 2" = 0.331 (1) N: N,+N..,_ = 2 (a) (Ab/U2. mw in AM 1. 912) a.) Na: 2~N, (3) I fFram 1: 331;; = 2m. = 9.331 N! NI =1} 2." N: = N1 Q 2 = KN; => N: F" '2..- ='- )‘45 (‘39! . 6328.. 1. (HM AUO " U03): 0 :4 3:. RT 00%) = a 37 xRT =9 4w" = (aw—pager: —¢7-23;<£7‘ -— xv/ea M; 7’ N? am; a: k3_.=__.@m5t . __ Nag?! Wfl 1 M NXAB .' 1;; _' .-.i;__._s_(A)_ = —~RZA'&~[>L = — R 0.63&&463 4- 9.37%“ 0- s7) __ = -€.(—0-2‘7 - 0-37); _ ' ' ' —R(o-/4I&0-/%+4261&9-30 - R(——_ 0.27 45.1.?) u.— -—- e +12 (0-?) _ AS" = 5(8) 44 554) = 0.42 —- $666 . _ = —J-26R '.. '75:" L:- ——o-2é 27- = —0-26xz-J‘=-—a.65 . fAflé-aAJ— 7215'" = ——_fl-5‘8 +0.65“ "if." ' ‘ M. 2:;ij “ZOEZAWMJ- ; A M M m w M M :3 . . . .. MWfl ). . _ii__ __sM-s¢uja€%wfm_7‘ _ " 8:. 4-5a W __ /Ma@a~lx. 49f 1;: 93 W . ' ' W: W3 =2 Z5141: /529XZ«3 _'—-. MS" SD = NAX¢£3 : Nfixlég I = ./é$‘_‘xR. _ _ xx? 7 «lg/WM; as (5020; jj A50 = —-/é-S‘R_ mic-- IK" r:- "'/372— _ I I M) fold" .1625. man/doc Wz. my“ :AJH fit/M m4; Wages! : 42m”. imp/qu ab W, 7, 'ZéwA' M MM 4;, 2. =9 “39%! :2/144 AB NH A 2 .. ,_ __ -_-_.-_ n2. = QXRXJ'é?X/U .. fi’wafi" M -——-- //-47x/Ug __ .. _ JWIt‘, j' yérmfiw = (RR/UH X/8 =2679x 41,, .. W1 dek’ A6;:______S‘.0_7 X300=-/-52 A; “‘4‘ Ag = WWW/“1:2? We“ W242. Q = (m — 4031mm“) 20"‘= Riva/:51:- 25‘ —-/-6/ = —4-02 x/ ) IQJ' Ins-6" AG" = #RTflm K = ~3/4Jm-efg/MQ '. AG = -—-3/ —-4‘~02. 2—35-01 .‘ Mam/W fl/Wrém/ufi _ a 5-3; “f K #[Lj 5%? D /&"‘ =..» gum (Md—é 9/, (9-02. '1 W—J :02) mmfl yA6r°7£afll WW fim :; {£4 47: j: A602 +£T/64 k1, = gmfinmzxza") f .. ' _ :M44-3AJ‘M16’T" ' 13-4360 mev‘h/a/zfid . f a4:——44A3 1/;0 = —/4c-3érmaé“’ wife;— _ 1% (my: Wife) = Q 61mm) Maud. . {[wm-wa%t) = /0Fé /.M33 + /0"6 N --6 ~ /0 = Sas‘xzo o 3 fl” 3 = 5?.fl0030r5" UC Berkeley. Chem 0130ch 100A, Fall 2006. Exam #3. Your Name Question 5, continued. 053. (7 points) Y 4 The graph above represents the binding isotherm for a non-allosteric interaction between a ligand and a protein. (SB.i) What parameter is represented by the vertical axis (labeled Y)? (3 points) _ 70 3' l éouuj, 74 {jg} ‘/ L game” mead/J (SB.ii) Based on this graph, what is the value of the dissociation constant for the interaction? Clearly justify your answer. (4 points) lyi/if) wink/236m. :Pwéw4-f m, = Z9;ho => K10 was“: 050. (3 points) We can, in principle, increase the specificity of a drug for its target by increasing its affinity. Which of the following modifications to a drug is most likely to increase the affinity? Circle your answer. (a) increase the size of the drug molecule (b) increase the number of hydrogen bonding groups (c decreasgfitflenurmeflf h dr en bonding groups in the drug d) increase the hydrophobici of the drug I (e Increase the rigidity of the drug molecule Pagellofll ...
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exam3key - [AZ 141“ 200K [8]. = _'__.'2'.0;&amp;lt;ID‘3...

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