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BISC330-MT2-Spring2008

# BISC330-MT2-Spring2008 - Page 1 Last name First name 1(42...

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Unformatted text preview: Page 1 Last name First name 1. (42 points) Complete the following models for Michaelis-Menten enzyme kinetics shown in (a) and complete the statements and equations shown in 0)) by ﬁlling in the blanks. (2 pts/blank) k1 k2 (a) Simple model (3 pts): E + S :‘A ES —> E + P k-1 k1 k2 k3 Model with EP (5 pts): E + S EES : EPHE + P k_1 k_2 (b) Both models ignore the backward reaction of E + P , by assuming that [P] is measured at a short enough time t such that [P] is much SMALLE K than [Q], so that v = [P]/_iL measures the initial velocity of P formation (V 0 in text) . The simple model predicts the following equations for v under steady-state conditions: (Eq. 2) Fae/Eu) [E] [i] (Eq. 3) v=k2[ET]é]/( Km +[ 5 ]) withKM=( k-, +i_<.&)/( i9 )andeax=(£gb)[[ET] In the model with EP, k2 = kcat (catalytic rate constants). By relating [EP] to I ES | at steady-state, one then ﬁnds that in the simple model k2 is equivalent to kcatk3/( k3 + k-2). Thus the simple model only has k2 = kcat when product release is much FASTER than catalysis. Page 2 Last name First name 2. (34 points) For simple Michaelis-Menten enzyme kinetics, illustrate a general model for the effects of a reversible inhibitor, including deﬁnition of equilibrium constants involved. (10 pts) El (\NAQ'HVE) ES]: (INAC’TIVE) + ,_ ~ WHQE 141i; KH=_._L§3_L§.L: STE/\B ~3TATE hissocmﬁom CsNSTAUT FOR 65 5 ‘ . kt [.5 i (N “axe—me c? 1:) MD 141 2, KI) Hue EqUthmoH NgmmATIcn consumgrbepwa) AS: A \ .‘c ' KT 2 ﬂ : Eqmo‘msoc. KT 1%.: EquiL, USSoc ‘ ,, LEI] cow: Fox E1 “ LEI] QQUS'T. FDR ESI Complete the following by ﬁlling in the blanks (1 pt/blank). If [I] is the inhibitor concentration, the general results are: me1=vmx/(1+g3_/ lg; ) KM1=KM<1+EI_1/V:. )/(1+_E11/ Q) Show the results for (a) competitive I (b) noncompetitive I and (c) uncompetitive I. (6 pts each) A ’ ‘ \ Q) Come—trove ,L MHbS To elm on To (:5) so KT : |UF|M1T€ ic. ’/ =0 ' ' I I ‘ 7 ki VQAX : VHAX (upCHAUCﬁB) KM = KH (1 ii 1%- (\mojc’kga)>‘ L , ‘ 13> NQNCLWETITWE: gums To E L, 55 ﬂed/04.") WM) 30 KT = K; vi _ VHAX HAA' ‘ , (1+ if) , h ( :) whom/Wynne? A, ENDS To ES M MDT To E) so Kl=wFMT€ ‘1.e.“‘":0 (/bfcpema) KEN : KM (warmest) l k: ,1 WM 1 K . i MW ’0? KM = :13 ( VMX 51, KM \$0M bang/Ba) "’T) (1*?) ‘ n We m) L ’, I , ‘ / VHszjfhi ”‘7‘“ {“075 OF V o: [53 CMAVL: ommﬁgaﬁicuc memoru t K“ WW‘G‘T‘. AT LOW [SLM‘EN ”make *8 (AWL? ES Maw? Folk T TO own. > Pug: 3 Lust name First mum: 3. (24 points) For each of the following, write the fonnula and brieﬂy explain the signiﬁcance nf each term. (a) AGinnlaﬁonhoKeqandQ.(8pts) K9, Ac’a mu(&),.2..s mmﬂﬁ) ;’a'n&6L-mukq) Me; A \= °4 Mme," “\K: Q‘s55 M M“ CMAIJgE u.) CHé‘l‘uCM. NEACTmu uYP‘a QAs Gavin/«rt MW iwwa'g Aug; 0\_ l- TM. |L) ‘GCA 1‘55 K“ Q,‘ “no of mamas To \EMT’WTS IN “sane R)" H #5 Kala «u ';;'5"MA “Nab 1x “\Kﬁf VMUC 0F q A1 EQUKMMNH Q. , {Kc} By G, 9%“ LA...) ‘0‘: VI.“ I A (b) The standard value ofAG when 0 =1, called AG". [8 pls) 3555231 N smote A9 \meub'» w macadammu of 14:11cher kuh lmM‘rgbCf n We 6%qu VALxE or- A9 mad EACH mud/:01 km: Egg newt? IS 4n 1J4 Comm-(toro- owl “mall Wm. Q=L In: M Tasman»; was, 43g”; rt QM” ) = ~l~T QM “1(6) k'x TRUE, {Fave W M'WLRWe' <91 He-sziur; Keg emmmanALL qua W WﬁL-UA'YE' J‘ Ag? (.1 a JSuIFICMJCE d" A§°mum1lou) (c) The biochemical standard M3“ (when Q= [H 1 =10' M) (8 pts) mum 05mg Emma-«Ems 1M1 mum-e N “:0 UJE usumﬂ “swag kg; N PH; EMS; ‘61}1 kin wag-3023: Arr 31G 18 C-XLLeb K9415: "L—iLTQ,‘A &(QA) 8°14JMM’S) mua' or» Ag Wn'ev‘ 9% met/km o~. mama-r 86131 I'M n; A“ 5—3 1M wcem’non wrm wa‘l bx“ Mug q; 10' u" H‘ 1: A news“: I T- l ‘06, 1"» 4.4K ox, - \o If ll 155 \e-ZACWJK. m U m R'Cr’gav '0' ( 1 AG .- emf ﬁlial-“ . Tint A—H’w H 96D (1,) =Ag‘+u1n :d‘kmz.m‘ (1w: W m mugs-now) ,5; 2'5 06 I ~ .1 2+ _r 1% EQUHW)’ Ag”- -.3 cg. kC—NLNJL a“? gig) Page 4 Last name First name 4. (25 points) Draw the structure of each of the followinilb (a) dCTP (7 pts) /: (D Tmmw HATE-5 (gm) H \@:\ *Nuen‘, ® “031 “RWQQW /\\0 «(341369 N WED vax -OH O o O 0L 01— wm -0" H M It “©-’P—o-'\>-o-’P—o~ @bebw? 809% CH“) \ l '\ O' 0‘ O“ @C @ 55W? (3 m) CAO (bmosine (I=rH) (6pts) 0 ”‘ owﬁi ‘mﬁgﬁwga k) \ . 0% WWW we HO )/0\N\ \ 3 @"Nmse sogkk (3 7“) (c)A:Ubase pair (6 pts) HO OH Hfmiwq a G @W (my N\/A\§‘~-H’N/U\ H @©'MS€ (oLWTS) “’07 N:{ >f’N @ convert H—immwc, (rm) (P 0 "Nb HOST HAVE MKLOAE ATTAcH% (d) G:U “wobble” base pair (6 pts) 8 TaKeIJCHEWSTR7 HOST 6“ ACCUQA’IC H é -—-H Mr} H (D @(SASE‘ (5013) 1 \$6? 0)“ \ @@ 6A8? (HTS) £2” -~ w» @W WW3 m2 n03? HAM: ”venom? ATTACHGB HH mamaxsw H031 w ACCUMTE TN; Page 5 last name 5. (25 points) Draw the structures of each of the following: (a) Dihyclroxyacclonc (2 pts) HO \ /CH"~ /(.’,H,OH (c) DRibosc (linear, 6 pts) 0 “ct/H CHO ‘ I H— o-cH 41— c; ~0H {vii/0H ”7U H-C: ‘0” H1}, —c-H Hg “OH 0an _ (thH e) [LO-Namely] gluumaminc (chair fotm, 8 pts) First name 0)) D-Glyccmldchyde (3 p”) O H \\o/ (t, HO K Wu K— c -OH H' IC‘O” \ CHLOH CPLOH (d) B-D-glucommose (Haworth, 6 pts) OH ‘74 HS .0“ HO\w/IH HOH H CW?“ 9 HC w . HO w ‘ 0” H l H C\ 0” 05 Page 6 Iasl mane K% First mm: 6. (25 points) Briefly mm the structure and function 0th of Lhe following: a e 059 l \ ”C uuél(sﬁ)4\/5-D(D-3§1:;;:MSQ Syn-Fhes'meaﬂ 14:1 Pia—n45 (”km“- 0'31 Whﬂ CO'HorL, £0004“) .4, w 5 Ea'mj £30916 ‘ P‘ M05? abunda’d: bIOPO mg” fdeammhﬁm ‘Q'Wahﬂzbm ‘10 7L”. . [giandl'w Wag!” H bMW-H ‘3‘ . F\$om _ QM (”L7 0 gate/k?) Wmh‘e’MH' “‘~ 7w +£3.1- owmsm' qgeutaﬂ H'bﬂJ ,puf- 4M (b) Hyulumniu acid (10 pls) (”Pinwholipidcglghlaﬁ ”mgr? 9’45””? [Lid/“’M'd r duh/rm itzémag amnobmln-Q K, «E-o—ETHZ /x_-,w-»%;;\ (Lip; 41% ﬂirﬁpé ‘ “ 10 o—ACHf-Cuz’N—ngf) [1:025 M HLC~0‘: "h to \cu KM MW ‘7 ’ P %w 3 . 10%;? . , men 5’ . 554% :15 gram C3 bphsm ' E1; “fad, ,‘Qs/uiur 0’1 cﬂbfhs'ﬂl-C - (It!) & A Mﬂ- L M Mpﬁoﬁ ...
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