Exam 1- 2009

Exam 1- 2009 - NAME_KEY MIDTERM EXAM I BIOLOGY 231 17...

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Unformatted text preview: NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 17 September 2009 ________________________________________________________________________________ GENERAL INSTRUCTIONS: 1. Answer all questions IN INK. 2. NO CALCULATORS. 3. Confine each answer to the BOX or SPACE provided. You are free to use the backs of the pages for scratch work, but ONLY what is written in the box or space will be graded! 4. SHOW YOUR WORK in all calculations. 5. Make sure that your name is PRINTED LEGIBLY on EACH PAGE. 6. This exam consists of 7 questions on 16 pages and it is worth a total of 100 points. Information that may be useful: ￿ For the reaction A + B <==> C + D, ÄG = ÄGo + RTln([C][D]/[A][B]) and RTln(X) = 5.7 kJ/mol￿log(X) ￿ At equilibrium, ÄG = 0 and ÄGo = ￿5.7kJ/mol ￿ log(Keq) ￿ log (10X) = X; 10log(X) = X ￿ surface of sphere = 4ðr2; volume of sphere = 4/3ðr3; area of circle = ðr2; surface of cube = 6e2 ￿ 1M = 103 mM = 106 ìM; 1 mole = 103 mmoles = 106 ìmoles ￿ For an enzyme that obeys Michaelis-Menten kinetics, the relationship between V and [S] is: V = VMAX × 1 V = [S] ([S] + KM) 1 + VMAX OR, KM 1 × VMAX [S] ￿ An amino acid has the following structure: ￿ On this date in 1787, the US Constitution was signed in Philadelphia by 38 of the 41 delegates present. It is now the longest-standing, longest-observed written constitution in the world. -1- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 17 September 2009 ________________________________________________________________________________ R groups of some amino acids, shown in their charged form, if relevant: ala (A ): g ly (G ): s er (S ): g lu ( E ): ly s (K ): a rg (R ): v a l (V ): ty r ( Y ): h is (H ): trp (W ): le u (L ): a s p (D ): -2- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 17 September 2009 ________________________________________________________________________________ p K v a lue s o f s om e a m ino a c ids a m in o a c id pK o f s ide chain o r term inal g roup uncharged 1 2 .5 uncharged 3 .9 8 .3 4 .3 uncharged 6 .0 uncharged 1 0 .8 uncharged uncharged uncharged 1 0 .9 uncharged 7 .8 3 .6 alanine (ala or A ) a rg in in e (a rg o r R ) asparagine (asn or N ) a s p a rtic a c id (a s p o r D ) C y s te in e (c y s o r C ) g lu ta m ic a c id (g lu o r E ) glycine (gly or G ) h is tid in e (h is o r H ) leucine (leu or L ) ly s in e (ly s o r K ) Phenylalanine (phe or F ) s erine (s er o r S ) tryptophan (trp or W ) ty ro s in e (ty r o r Y ) V aline (val o r V ) pe ptide N te r m inus pe ptide C te r m inus -3- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 17 September 2009 ________________________________________________________________________________ (1) T h e d ra w in g s b e lo w s h o w a c e ll in w h ic h a lip id b ila y e r is re p re s e n te d a s a s in g le lin e . T he cell in (1) contains an en do som e w ith a trans-m em brane p rotein represented as a “lollipop” and a free soluble mol ecule “X ,” both orientated as s how n. Im agine that a fis s ion or b udding event s plits the endosom e into tw o s eparate vesicles: a large one, “ A ” that retains both the mem brane protein and the soluble mol ecule, a nd a sm aller v esicle, “ B ,” that c o n tain s n e ith e r. T h is is s h o w n in d ra w in g 2 . N e x t, th e v e sic le “ A ” u n d e rg o e s e x o c y to sis : it fu s e s w ith th e p la s m a m e m b ra n e , w h ile v e s ic le B re m a in s in th e c y to p la s m , a s s h o w n in (3 ). (a) p ts . In th e d ra w in g 2 , D R A W in B O T H th e tra n sm e m b ra n e p ro tein a n d th e so lu b le m o le c u le X a s th e y w o u ld b e o rie n ta te d in o rg a n e lle A . 2 ( b) 2 pts. In the draw ing 3, D R A W in B O T H the transm embra ne protein and the s o lu b le m o le c u le X in th e ir p ro p e r lo c a tio n a n d o rie n ta tio n . ( c) 2 p ts. A s s um e that the organelle in draw ing 3 is s pherical. If the volum e of the cell is 1 0,000 ì m 3 a n d th e ra d iu s o f th e o rg a n e lle is 1 /1 0 th e ra d iu s o f th e c e ll, w h a t is the volum e of the organelle? A n s we r in th e b o x b e lo w. (1/10)3 x 1 0,000 ì m 3 = 1 0 ì m 3 -4- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 17 September 2009 ________________________________________________________________________________ (2 ) 9 p ts . Y ou have learned about b oth covalent b onds a nd non-covalent o r “ w eak” interactions . In the 3 boxes b elow , N A M E A N Y T H R E E o f the 4 non-covalent interactions that w e have d is cuss ed. In a phrase or tw o, D E SC R IB E T H E B A SI S of e ach interaction. (all 4 h ere) Inte ra c tion 1: h y d rog en b o n d s B a s is : A h y drogen (or p roton ) is shared betw een tw o electron egative atom s such as O an d N . In te ra c tio n 2 : Ionic b onds B a s is : T h e attractio n o f gro u p s w ith o p p o site electrical c h arges In te ra c tio n 3 : v a n d e r W a a ls B a s is : Ind uced dipo les (snapshot a sym m etries in charge distribu tion ) b etw een ne a rby a tom s , a t s ha rp optim a l d is ta nc e [im porta nt w he n m ole c ule s h a ve c o m p le m e n ta ry s h a p e /s tru c tu re ]. In te ra c tio n 4 : h y d rop h o b ic in teractio n s B asis: H y d ro p h o b ic g ro u p s a s s o c ia te w ith e a c h o th e r b e c a u s e e n tro p y fa v o rs th is o v er th e o rderin g o f w ater (3) 5 pts. A “sizes” question. Number each of the following molecules so that 1 is the LARGEST and 5 is the SMALLEST. _ __ 4 __ glycine ___ 2 __ H IV p rotease, a n enzym e _ __ 1 __ a cell n ucleus _ __ 3 __ histidine ___ 5 __ H 2O -5- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 17 September 2009 ________________________________________________________________________________ (4) 8 p ts. Y ou have studied the properties o f the four m ajor c lass es o f m acrom olecules found in the cell. B elow a re lis ted the four k inds o f m acrom olecules a nd a lis t o f p roperties. In the bo x n ext to each m acrom olecule, w rite the letters o f the T W O p roperties that best c o rre sp o n d to it: M ACRO M OLECULE PROPE RTIES carbohy drate E, F protein B, F lipid A, D nucleic a cid C, F PO SSIBLE PR O PERTI ES: (A ) r e s p o n s ib le fo r f o rm in g m o s t o f th e b o u n d a rie s in a n a n im a l c e ll (B ) p e rf o rm s m o s t o f th e e n z y m a tic a c tiv itie s o f a c e ll (C ) p olym e r is h e ld toge the r b y c ova le nt p hos phodie s te r b onds (D ) h ig h ly h y d ro p h o b ic (E ) can fo rm a b ranch ed p o ly m er structure (F ) a p olym er b u ilt u p o f identical o r sim ilar b u ilding b lock m olecules -6- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 17 September 2009 ________________________________________________________________________________ (5) C on sider the sho rt p o lyp eptide, a tripep tide, a bb reviated as: N -term – lys – g ly – arg – C -term (a) 7 p ts . D R A W T H E S T R U C T U R E o f this tripeptide in the box below . F or e ach fu n c tio n a l g ro u p th a t c a n b e c h a rg e d a t a n y p H, S HO W IT IN TH E CH AR GED FO R M in your d raw ing: (b ) 2 p ts . In y o u r d ra w in g , C IR C L E th e p e p tid e b o n d s (o n ly ). -7- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 17 September 2009 ________________________________________________________________________________ 5(c) 6 p ts. T ake into account a ll o f the functional g roups o n this tripeptide that c an b e c h a rg e d , a n d d e te rmin e w h a t th e T OT AL C H AR GE O N T HE T RIP E P T IDE W IL L B E A T E A C H O F T H E F O L L O W IN G p H v alues. B e sure to include the C O R R E C T S IG N (po sitive o r ne g ativ e). N -term ly s (pK 7 .8) (10.8) +1 +1 +1 +1 +1 ￿1 0 W ha t is the tota l c ha rge a t p H = 4 ? arg C -term (12.5) (3.6) +1 ￿1 0 +1 ￿1 +2 W ha t is the tota l c ha rge a t p H = 7 ? +2 W ha t is the tota l c ha rge a t p H = 1 1? 0 -8- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 17 September 2009 ________________________________________________________________________________ (6) Y ou are searching through a m olecular library, trying to find for a p o tent inh ibito r of the H IV a s p a rty l p ro te a s e . Y o u h a v e two prom is ing candidates, c ompound A a nd compound B . Y ou ass ay their p rotease inhibition by m easuring the activity (velocity) of the H IV p rotease over a range o f su b stra te c o n c e n tra tio n s, w ith A , B , o r n o inhibitor. Y ou get the d a ta a t rig h t: (a) 1 8 p ts. C a lcu late the V m ax a nd K m o f the p rotease. S ho w y o ur w ork and c ircle yo ur final answ er. B e careful w ith u nits. V max w ith n o in h ib ito r: K m w ith n o in h ib ito r: V max = 1 /y-intcp = (1/10 (um ol/sec)￿1) K m = ￿ (1/x-intc p) = ￿ (1/￿ 1 0 0 m M ￿1) = 0 .1 ì m ol/sec = 0 .0 1 m M V max w ith in h ib ito r A : K m w ith in h ib ito r A : V max = 1 /y-intcp = (1/10 (um ol/sec)￿1) K m = ￿ (1/x-intc p) = ￿ (1/￿ 5 0 m M ￿1) = 0 .1 ì m ol/sec = 0 .0 2 m M V max w ith in h ib ito r B : K m w ith in h ib ito r B : V max = 1 /y-intcp = (1/20 (um ol/sec)￿1) K m = ￿ (1/x-intc p) = ￿ (1/￿ 1 0 0 m M ￿1) = 0 .05 ì m ol/sec = 0 .0 1 m M -9- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 17 September 2009 ________________________________________________________________________________ (b) 3 pts. C irc le w h ic h o n e o f th e s e in h ib ito rs (A o r B ) is a N ON-CO M P E T IT IVE inh ibito r: A B B riefly, e xp lain h ow y o u d edu ced this: In the presence of inhibitor B, the protease has a lower Vm ax but the same Km , indicating that B is non-competitive. ( c ) 2 p ts. F or inh ibitor A , c ou ld y ou o v ercom e the inhibitory effect o n the p rotease by adding high concentrations o f s ubs trate? A ns w er Y E S or N O a nd explain in a s entence o r tw o: YES. Because A is a competitive inhibitor, it competes with S for access to the active site. As [S]/[A] increases, the probability that A can get into an unoccupied active site diminishes and V approaches Vm ax. -10- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 17 September 2009 ________________________________________________________________________________ (7) Y ou have is olated a new m etabolic enzym e from a slim e mold found in your s how er a t h o m e . Y o u d isc o v e r th a t it c a taly z e s th e re a c tio n : f r u c to s e 6 - p h o s p h a te + A T P < = > f r u c to s e 1 ,6 - b is p h o s p h a te + A D P In th e lab o ratory , y o u a llo w the re ac tio n t o p roc ee d T O E Q U IL IB R IU M a n d t hen m ea su re th e c o n c e n tra tio n s o f th e re a c tan ts a n d p ro d u c ts. T h e y a re : [fructos e 6-phos phate] = 1 0 ￿ 5 M [A T P] = 1 0 ￿ 4 M [fructos e 1,6-bis phos phate] = 1 0 ￿ 3 M [A D P] = 1 0 ￿ 3 M (a) 4 p ts. C A L C U L A T E the K eq for this reaction. S how y our w ork: K eq = [f 1 ,6-bisP] [A D P] / [f 6 -P][A TP] a t equilibrium = (1 0 ￿3M)(1 0 ￿3M)/(1 0 ￿5M)(1 0 ￿4M) = 10 ￿6/10 ￿9 = 103 (b) 4 pts. CALCULATE the ÄG￿ for this reaction. Show your work: at eq u ilib riu m , Ä G o = ￿ 5.7 kJ/m ole ￿ log (K eq) so Ä G o = = = = ￿ 5.7 kJ/m ole ￿ log (K eq) ￿ 5.7 kJ/m ole ￿ log (10 3) ￿ 5.7 kJ/m ole ￿ 3 ￿ 1 7 .1 k J/m o le -11- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 17 September 2009 ________________________________________________________________________________ (c) 3 p ts. Y ou w ant to unders tand how far from e quilibrium this reaction is in the intact c ell, s o you develop an ass ay for e ach of the reactants a nd products a nd find that in the intact c ell their c oncentrations a re: [fructos e 6-phos phate] = 1 0 ì M [ATP ] = 1 .0 m M [f r u c to s e 1 ,6 - b is p h o s p h a te ] = 1 0 m M [ADP] = 0 .1 m M C ALCUL ATE q , th e ra tio o f p ro d u c ts to re a c ta n ts fo r th e s e c o n d itio n s . B E C A R E FU L O F Y O U R U N IT S. S how y our w ork: q = [P]/[R] = [f 1,6-bisP] [A D P ] / [f 6-P ][A T P ] = (1 0 mM)(0 .1 mM)/(0 .0 1 mM)(1 mM) = (1 / 0 .01) = 1 0 2 (d) 3 pts. G iven y o u r va lues f or K eq a n d q , w o u ld this reac tio n p roc ee d L -to -R s p o n ta n e o u s ly u n d e r th e c o n d itio n s e x is tin g in th e c e ll E VE N W IT HOUT T HE ENZY M E? A nswer Y ES o r N O , a n d in o n e s e n te n c e , e x p la in W HY o r W HY N O T: Y E S or N O : _ __ YES ____ W h y o r w h y n o t: B ecause q < K eq, there is an excess o f reactants relative to the eq u ilib riu m co n d itio n , so th e reactio n w ill p ro ceed sp o n tan eo u sly L -to -R. (e) 3 p ts. If m any reactions in the cell a re favorable, w hy are enzym es n eces s ary at a ll? E X PLA IN in 1 o r 2 S EN TEN CE S. A lthoug h enzym es d o no t c hange the Ä G ￿ o f a reaction, they low er the activation energy and vastly increase the rate o f the reaction. -12- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 17 September 2009 ________________________________________________________________________________ (8) 8 p ts. O n the follow ing page, the protein ribonuclease (PD B id#1a2w ) is s how n as y ou saw i t i n lecture , u sing the P rotein D ata B an k W eb M o l view er. T h e d isplay m o d es a re show n n ext to each im age. R eview these im ages an d an sw er the follow ing q u estion s: W hich secondary s tructure element(s ) d oes this p rotein contain? Ribonuclease has both alpha-helices and beta strands / beta sheets D o e s rib o n u c lea se c o n tain a sig n ifica n t am o u n t o f u n stru c tu re d , “ra n d o m ” c o il? Yes [apparent in the 20 structure image] H ow m any dis ulfide bonds d oes ribonuclease have, P E R S U B U N IT ? 4 per subunit [8 total in the complex, apparent in 40 structure image] H ow m any polypeptide chains m ake up each molecule of ribonuclease? TWO -13- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 17 September 2009 ________________________________________________________________________________ R IB O N U C L E A S E : “ A ll A tom s ” display mode, “C olor/A ll A tom s ” mode “B ack Bone” dis play mode, “C olor/Sec” mode, for d is play of sec o n d ary s tru cture “B ack Bone” dis play mode, “C olor/C hain” mode, for d is play of q u atern ary stru cture -14- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 17 September 2009 ________________________________________________________________________________ (9) T h e c a rto o n a b o v e sh o w s s c h e m a tica lly th e c A M P b in d in g site th a t w e sc ru tin ize d in lecture. O ne protein that c ontains this s ite is the regulatory s ubunit o f P r otein K inase A , a n im portant s ignaling enzyme that is n orm ally s w itched on w hen it b inds c A M P and regulates the activity of m any other e nzymes in the cell. It is n ow k now n that a t least 2 d ifferent ly m p h o m a s (im m u n e -c e ll-d e riv e d c a n c e rs) re su lt fro m m u tatio n s in th e c A M P b in d in g site o f p ro te in k in a s e A . ( a) 3 p ts. T he firs t m utation replaces the arginine (arg) in the diagram w ith a lys ine (lys ). T his results in a mut ant p rotein kinase A that s till b inds c A M P, b ut it requires 4-5x higher [cA M P] to s tim ulate the same respons e as in the norm al p rotein. In a s entence o r tw o, e xplain this o bs ervation. L y sin e h as th e sam e p o sitiv e ch arg e as arg in in e at n eu tra l p H , b u t it is smaller an d sh ap ed d ifferen tly . T h u s it p ro b ab ly p articip ates in th e same ionic bond w ith negatively-charged O o f the phosphate group as arg does in the n orm al p rotein, b ut w ith a reduced affinity. -15- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 17 September 2009 ________________________________________________________________________________ (b) 3 pts. T he second mutation replaces a g lycine (gly), found in the active site at the pos ition m arked X in the diagram, w ith a glutami c acid (glu). T his results in tum or cells in w hich protein kinase A d oes n ot b ind cA M P at a ll. In a sentence o r tw o, explain this o bs ervation. G lycine is a s m all, u ncharged R -group, w hile glutam ic acid has a n eg ativ e ch arg e at n eu tra l p H . In th e m u tan t p ro tein , th e g lu p ro b ab ly repels the n earby negative charge o f the p hosphate in cA M P, p reventing its b inding. ( c) 3 p ts S om etim es p oint m utations in residues that a re N O T a p art o f the binding site w ill no n ethe less d isru p t cA M P b ind ing . E x p lain h o w t his is p o ssible in 2 -3 sentences: The p ro tein m u st fo ld co rrectly (3 o stru ctu re) in o rd er fo r th e co rrect R groups to be b rought together in the right p ositions to form the b inding site. A m utation aw ay from the b inding site that g rossly affects folding could disrupt the site a nd thus d isrupt cA M P binding. -16- ...
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