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Unformatted text preview: UNIVERSITY OF CALIFORNIA "this exam should have 8 pages
CHEM C130 I MCB C100A" ' ' including the cover MIDTERM EXAMINATION #3 November 12, .2009 INSTRUCTORS: John Kuriyan and David Wemmer TIME LIMIT FOR THIS EXAM: 1 HOUR 20 MINUTES SIGNATURE: é _
Please SIGN you ame above in INDELIBLE INK YOUR NAME: Please PRINT your name clearly in INDELIBLE INK PLEASE CIRCLE the name of your GSI:
Anna Asmundson Geoffrey Feld _ Zhijuan Gao Patrick Visperase PLEASE: Write answers as CLEARLY as possible, ILLEGIBLE answers will not be given
points. SCORING  the number here are the number of points possible for'each part of each problem page 1 Problem 1. Movement of biological molecules can be roughly approximated as jumps between points
on a grid. Although real molecules move in three dimensions we can learn about aspects of movement using simpler models, even steps on a one dimensional line starting at the origin. Suppose that a
molecule takes a ‘step' 1000 times per second, and each step can randomly be to the left or to the right with equal probability and has unit size, is. one step goes from 0 to 1 or 1, etc. To think about the
average movement of molecules during such a process we can look at the distribution of probabilities of different outcomes, for example ending up a speciﬁc number of steps from the origin. a) if a molecule moves for 1 second (1000 steps):
What is the most probable location of the molecule: 0} ‘H/kﬁ 95%)“:
C1 (lug{GUM it Mgrtwill“ {ti fix 2' :3 n; “Misfit” {pdt? b) if the distribution is examined after 1 second and again after 100 seconds, does the probability of
being Exactly 10 steps away from the origin increase or decrease at 100 sec. relative to 1 see? (explain briefly) bottle. Ci is. let; titan lot“ :2 ’30 step? mixers)
bi ill (EQQQQEQ (2w M St, vwﬁte {3 isi‘t'it (gift disbllwiém {in i £13.}. miti E‘j’i item “is. c) What is the relative width of the distribution of ﬁnal locations at 100 sec. vs. 1 sec. times?
(be quantitative) I _ ........ mt (algalit, m. Mﬁiﬁi a?“ at ling{spit so in lK. ago wt» was it) “trim3" :1) Suppose that the cell taking the steps is sensing food in its environment, and biases its steps to
move toward the food. if there is a food source to the right (positive numbers), and the probability of
stepping toward food is 2'3 and the probability of stepping away from food is 113, what is the most
probable location for a cell after 1 second? (think carefully about the distribution and its relation to ﬁnal location in this case).
(ME. mil: its. 7 re...‘ K M3 “3%”? ﬂ «size it.
is? with... W; “"4 ‘1 "M page 2 Problem 2. Thenucleobase adenosine is a weak acid, when protonated on the N1 atom of the Watson
Crick edge with a pKa = 3.5 observed in an unstructured RNA molecule.    a) a particular adenosine was thought to play an important role in the activity of a catalytically active RNA molecute. if its pKa were normal as in the unstructured RNA, what fraction of this adenosine would
be protonated in a solution with a pH of 7.0 ? E3 5” RN ass} a e W amid K6. la : “ﬁlms”
at gift ‘7 almﬁt suit wilt be: A 59 Lﬁi'i’ttit‘xj’k UM H ‘ ; an“;
i. Mi _,, tom“
"ﬁg11" “ is?“ “my;th b) in the actual structured RNA the pKa of the adenosine of interest was followed by NMR, and its pKa
value was determined to be 6.5. Describe two interactions which could contribute substantially to this
shift in pKa value (be brief but speciﬁc in your description) i’VxOﬁi' ﬁlmth it 5R, ﬁeﬁﬁ' it: {: Oc‘itﬁ PQSEQWQ A‘i’vﬁvxﬁ Hoboné “it; Cilia gm UREE} QM N rm a limit” {of unisex—1‘) A, t—iwlrcmé i. +0 Ce... wet set. for ts? tilt3th.. sit tittiestitted c) Calculate the value of AG for the protonxatipn of a normai adenosine in an unstructured RNA if the
RNA is present at 1 uM concentration and 'the'pr is 7.0. 25 c“G. AC7: eat t m as? er is. ml 2 PT In Kt AG : “lOLJ/ + H" + A reﬁt» HA is prettyvast"; wt
I W e x [M mi: .... mvatae. at acid suede9m
g‘mklxgﬁo \016 ‘16 K91: VKQ tardy“; g; itk.€..._%gkﬁ : — 1‘ LFOILMSXWQ 1" ‘k‘rlbt‘gt’mii baa C “(ZO kT/MOKQ aw t
it" ' d
d) Will the AG value for protonating the adenosine in the folded ribozyme structure be: a 1 more positive the same (more negative} (circie one)?
pKQ 1g“ {in A63 if Wt"? Mattiet2;
more ﬁwﬂwﬂé’ “is; ﬁrme Watt? Beam .5153 f p “l
sew wags at» it [i
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f
l t' .«I page 3 Problem 3. DNA sequences have been studied in which a ‘core' duplex region is ﬁxed, but bases are
added to make either one or two loops which are unstructured, giving molecules such as:    (a) . (b) ' to) Consider the thermodynamic parameters below all referring to the ‘melting’ transition (dupiex to random
coil transition) _ mm... Answer the following questions by circling a letter or letters (if more than one is correct) or no letter
(if no answer is correct) corresponding to the appropriate structure(s): part A) Structure (a) ; (b) ; (c) ' will have the highest Tm value
part B) Structure ; (b) ; (c) will have the most negative value of AG" at 25°C
part C) Structure ; (b) ; (c) will have the largest AS" associated with melting part D) Structure ; (b) ; (c) will have a lower Tm if the sample is diluted 2 fold part E) Structure ; ; will have a AH°vaue that is independent of concentration A) {megatic». ﬁatM. teem}? new??? I B) Unite L31; alth am [met 1L AC1 “J t .
it? aaﬁtegt“ +0 mﬁob}; —_ (a)
6) his at state chase tide tug—tie estt mtgﬁt:de ' “it Git
it.) 33) out tj Le) titer “ha—it? gtetutttlt r: tgwi'tt ‘t t .
) Wkai Put dearu git... AGES net i“ «siteQR. behest at“ “’t‘krﬁ" tﬁﬁ {.ti‘t‘ttﬁ‘t' 2% at?" (JR, pagezt E Note. at: if kjW Emit» aged. KeﬁfC'er Problem 4) The melting of the selfcomplementary DNA duplex hexamer CGATCG to two identical
single strands was studied by calorimetry at a concentration of 0.10 mM of DNA strands, and it was found that AH“ = 177 lemol and the Tm was 38.3 c‘C. (reminder. for nucleic acids ACp=0) For this
problem consider the reaction direction to be duplex going to single strands (i.e. the unfolding direction) part A) What is the value of the equilibrium constant for unfolding at Tm ? duvlgx #9 2 SW 3.3 out"T... ﬂama3=zfévpieﬂ
iijq ’3’— K f. Lil a,
a1 {lbrz‘k thmg'ﬁr lLdthlex ] ":3. M part B) Calculate the A8" for the melting (unfolding) of this duplex. AGO: AW«TAP : .. RT [MKQR a 33.1 r; = 3M K
sl‘iiiia‘ywvjll A5" 3 U3 Vim/wt A39 2 éLtEjymut/l< part C) In analogy to the calculation done for protein folding, assume that each unconstrained single
bond along the backbone has three equally iow energy conformers, and that bases have two allowed
conformations relative to sugars both when in a single strand state, while there is one commtiqu for
bonds between nucleotides or connecting to the base in the duplexgtate. Estimate the conformational entropy change (on a per mole of duplex DNA melted basis) for the melting precess.
0 We: “2. ‘Dﬂ‘f b» duptﬁx S [email protected]: betwam ngﬂxéq’ )3}
Eye—tat Cmshaieeﬁ , easy with “Wolff?
S (matte. (met, + " as Z SW35
_ t : pigwﬁ _ W» ” WWW“!
:1. 50 W3 a :. O acid, set {if Q tater ye! 5pm.},
[ﬁshnets x Z rower stater N"R NBSO 2”) part D) Liﬁttiﬁher contributions to the AS” value that have been ignored in the calcuiation
in‘part B), explain these very brieﬂy. «w CKMEC/S 1:“ {JQQéW fe'lusxxf {pow} ‘ I f a t k“ mfztﬁiih t
4mm M... 6913’ ex. is? thQ guaranties 923?. .c... 9K M My»: éuﬂ... is W6“ pew tightened
balmtee uS‘WS page 5 Problem 5) A DNA otigomer was observed to be a duplex with all 12 bases in Watson—Crick base
pairs. However in low salt concentrations when temperature was increased each duplex convertedto two hairpins, and then at yet higher temperature to random coil single strands, as shown in the diagram
below. F is the fraction of the DNA strands in the higher temperature form for each of the transitions, i.e. 
for the left curve it is the fraction hairpin. The total strand concentration was 3.8 lel. part A) Estimate the free energy. AG“. for conversion of duplex to hairpin at 35 °C. . ._. l as;
lead Same as 4A a h? “I
act: “laws 16% Sines Q 2&3” «this "it 0K WM {45$ Mtg” we RC3 inS‘il’ﬁ‘ﬂé. HP part B) Use these data to estimate the value of AH” for the transition from the dupiex to single strands. at, =AR»”T"M f r s " i = "
Abba; w (A R m K Aﬁif‘t‘ﬁ Mt
wa WW» 5 .4 T Mi“ \ am 1 “it; {A 3
a "f" 1' ‘ "‘m‘” “t”
1 . l / ‘3. ‘ Kt... _, “is. b w 3:339 { ’5 KESO "32% LN} cit”: Myth 3mm; 9 a),an MM. :51“ [DH 3: l ‘f 1‘“? éf‘iiuﬁt a) a} 9 ,9 ﬁx a; Wig", :, Eligible part C) The raterof conversion of the duplex to hairpin was measured as a function of temperature, and
the rate constants were used to generate the plot at the right. From these data determine the activation energy for this reaction. A ~ & 9w numb L" {,9 "" «#3:? A >4 k {’4.
_ a ﬂ, 5M.
was.“ t ht “ ca
53):; 5‘3 ‘2 m {:65 ’7." ’3 I','M.H~NQ I. '
‘fr {:03 part D) The estimated value of AH" between the duplex and random coil single strands was estimated
to be 377 lemol. The most obvious mechanism for the transition from the DNA duplex to individual
single strands is for the strands to dissociate to make two random coils, which then each folds up to make the hairpin. Do the available data support this mechanism? (explain brieﬂy)
$3 SCN dupe 1x as a: “a 9"? ‘1 ga tﬁﬁmé W ‘2 apr What? The
EU f ﬁbng M1332 < gm?" if? Ufgfﬁllia‘; 5%}
a; t Most ml: aﬁo 'W gum? “MS g k ' " w“  1 ','.L"':.’CL. .. r" thmhjiaﬁ: Emirate Qatar as“? m S a? use? VWM its “the; “{“\mé'i+:z a“ page 7 Problem 6. Some bacterial proteins are phosphorylated by a protein called ahistidine kinase, HK. The
kinase reacts with ATP to make a phosphointermediate HKP. The HKP reacts with a receiver
domain, RD, transferring phosphate to give RDP, which can hydrolyze to give RD plus phosphate. k1 '
HK + ATP > HKP + ADP k2
HKP + RD —» HK + RDP k3
RD—P + H20 * RD + Pi Suppose the ATP concentration in the cell is held constant by other processes. Part A) Since RD is used in one step but then recovered in the next step, the concentrations of RD and
RDﬁashould reachageadystate: Write a diﬁemmt‘di equation describing this steady state in terms of
the reactions given, and give an equation for the ratio [RDPIIIRD]. lib stata. re Ag? :1 cs 1. ltkteﬂUlK—ﬂ + kﬁRDﬂtwﬂ Part B) Suppose that the reaction is done in vitro, starting with known concentrations of pure HK and _RD. and then adding ATP. Write the differential equation describing the change in concentration of HK
as a function of time. '   . w 53"“; f, «Blotter? mot—tithe} ‘ Part C) Suppose the rate of the reaction RDP + H20 also increases with increased [M92+]. Under what conditions would you expect the addition of Mg2+ to the solution to substantially affect the rate of
disappearance of ATP ? (explain brieﬂy) ' Hun61.x. “to See. i.) Mi await; J.) “title. [4,3 feta3:; must he: at} least thidu‘m mi? 1M r
m M. , m . ping. f‘t «a " _ E ‘ ﬁg 3 Kb”? j @133} g kiwi1K; i: REE} sate; kﬂdiﬁiﬁ’ﬂ'i j 25.: at? kt wed in “$9 Li; LatL oslu “$1:
' _ it is”...
See cm, ed: page 8 ...
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