2010 Exam 2 - NAME: ________KEY_________________ MIDTERM...

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Unformatted text preview: NAME: ________KEY_________________ MIDTERM EXAM II BIOLOGY 231 18 October 2010 ________________________________________________________________________________ READ THESE 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 blank pages for scratch work, but ONLY what is written in the box or space will be graded! 4. Show your work in all calculations. Render numerical answers in decimals and check your units. 5. Make sure that your name is PRINTED LEGIBLY on EACH PAGE. 6. This exam consists of 8 questions on 13 pages (plus one blank page for scratch work) and it is worth a total of 100 points. Here’s some information that may be useful: ! 1M = 103mM = 106µM; 1V = 103mV ! log(10X) = X; log(10X/10Y) = log(10X￿Y); 10X/10Y = 10X￿Y ! for a transmembrane gradient of any solute X, the free energy of the gradient is found by: ∆GX = RT ln([X]in/[X]out) + zFVm = 5.7 kJ/mol ￿ log([X]in/[X]out) + zFVm “in” = cytoplasmic; “out” = extracellular or non-cytoplasmic space; z = valence of the ion; F = Faraday constant = 102 kJ/V￿mole ! at equilibrium, the Nernst equation gives us: VX = (59mV/z) ￿ log([X]out/[X]in) ! for a charged molecule X: ∆GX = zF(Vm ￿ VX) ! for a proton gradient: ∆G = F(∆p) ! the protonmotive force, ∆p = ψ ￿ (59mV ￿ ∆pH) where ψ = Vm of IMM; ∆pH = (pHIMS ￿ pHmatrix); IMS is the intermembrane space ! when a negative number increases, it becomes more positive ! when a positively charged ion leaves the cytoplasm, the Vm becomes more negative ! On this day, October 18th, 1867, the US took possession of Alaska, which it had purchased from Russia for 2 cents an acre. -1- NAME: ________KEY_________________ MIDTERM EXAM II BIOLOGY 231 18 October 2010 ________________________________________________________________________________ THE GENETIC CODE (Single letter abbreviations for amino acids) 2nd U C A G 1st U 3rd C U S Y C C S term. term. A L S term. W G L P H R U L P H R C L P Q R A L P Q R G I T N S U I T N S C I T K R A M (init.) G Y L A S F C F T K R G V A D G U V A D G C V A E G A V A E G G -2- NAME: ________KEY_________________ MIDTERM EXAM II BIOLOGY 231 18 October 2010 ________________________________________________________________________________ (1) 20 pts. The diagram below depicts three organelles in a plant cell: the ER, a mitochondrion (M), and a chloroplast (C) . It shows 12 different sites or paths of movement in the cell, numbered 1 - 12. These are: (1) = anywhere in thylakoid lumen; (2) = within thylakoid membrane; (3) = anywhere in stroma; (4) = movement from stroma across thylakoid membrane into lumen; (5) = opposite direction to (4); (6) = anywhere in matrix; (7) = within IMM; (8) = mitochondrial IMS; (9) = movement from matrix across IMM to IMS; (10) = opposite direction to (9); (11) = cytoplasm; (12) = lumen of the ER; 13 = OMM. In each box below, write the number(s) of the correct location(s) where the item takes place. Note that either one or two answers may apply to each; choose ONLY the appropriate answer(s) for each. FUNCTION LOCATION CO2 + ATP + NADPH => carbohydrate 3 H+ flow in this direction through ATP synthase 5,10 glucose => => => pyruvate 11 ATP accumulates due to action of ATP synthase 3,6 chlorophyll captures photons 2 ions cross freely between compartments via porins 13 electron transport chain undergoes redox reactions 2,7 ½ O2 + 2 H+ + 2e￿ => H2O 6 (or 7) isocitrate => alpha-ketoglutarate + CO2 6 NAD+ + H:￿ => NADH 6,11 ionic composition very similar to cytosol 8 pH increases due to electron transport 3, 6 favorable direction of H+ flow when metabolism is active 5, 10 H2O => ½ O2 + 2 H+ + 2e￿ 1 (or 2) -3- NAME: ________KEY_________________ MIDTERM EXAM II BIOLOGY 231 18 October 2010 ________________________________________________________________________________ (2) 8 pts. Below are 8 examples of a nucleic acid template (at left) and a possible result of synthesis by polymerase activity on that template (at right). For each, choose whether the enzyme most likely to have generated the product at right from the template at left is: ( A ) D N A - d e p e n d e n t D N A p o l ym e r a s e ( B ) D N A - d e p e n d e n t R N A p o l ym e r a s e (C ) R N A p rim a s e ( D ) a c o m b i n a t i o n o f m o r e t h a n o n e o f t h e s e 3 e n z ym e s ( E ) c a n ’ t b e d o n e w i t h t h e s e t h r e e e n z ym e s a l o n e WRITE the CORRECT LETTER (A-E) for each in the BOX at the far right. S T A R T IN G T E M P L A T E : R E S U L T O F S Y N T H E SIS : 5' CCAUGA 5'CCAUGACCAGAGCTATTA 3' 3' GGTACTGGTCTCGATAAT 5' 3'GGTACTGGTCTCGATAAT 5' _____________________________________________________ 5' CCAUGA 5'CCATGACCAGAGCTATTA 3' 3' GGTACTGGTCTCGATAAT 5' 3'GGTACTGGTCTCGATAAT 5' ______________________________________________________ A E 5' CCAUGA 5'CCAUGACCAGAGCUAUUA 3' 3' GGTACTGGTCTCGATAAT 5' 3'GGTACTGGTCTCGATAAT 5' ______________________________________________________ B /c 5'CCAUGACC3' 3' GGTACTGGTCTCGATAAT 5' 3'GGTACTGGTCTCGATAAT 5' _____________________________________________________ C/b 5'CCAUGACCAGAGCTATTA 3' 3' GGTACTGGTCTCGATAAT 5' 3'GGTACTGGTCTCGATAAT 5' ______________________________________________________ D 5'CCATGACCAGAGCTATTA 3' 3' GGTACTGGTCTCGATAAT 5' 3'GGTACTGGTCTCGATAAT 5' _________________________________________________________________ E 5' CCATGA 5'CCATGACCAGAGCTATTA 3' 3' GGTACTGGTCTCGATAAT 5' 3'GGTACTGGTCTCGATAAT 5' _____________________________________________________ A 5' CCAUGACCAGAGCTATTA 3' 3' GGTACTGGTCTCGATAAT 5' E 5'CCATGACCAGAGCTATTA 3' 3'GGTACTGGTCTCGATAAT 5' -4- NAME: ________KEY_________________ MIDTERM EXAM II BIOLOGY 231 18 October 2010 ________________________________________________________________________________ (3) The double stranded DNA below contains just ONE open reading frame, encoding a short protein of MW ￿ 700 D. The DNA is repeated on the next page so that you can work on it as needed. 3'CAATTACTAACTACTGAATCTTACAATTCGTAC 5' 5'GTTAATGATTGATGACTTAGAATGTTAAGCATG 3' (a) 5 pts. Identify which strand contains the open reading frame and WRITE in the box below the ENTIRE mRNA that will be transcribed from that strand, open reading frame plus any flanking sequences, from 5' to 3'. LABEL the 5’ and 3’ ends. UNDERLINE the initiation and termination codons. 5' GUUA AUG AUU GAU GAC UUA GAA UGU UAA GCAUG 3' ( T r a n s c r ib e th e to p s tr a n d , tr a n s c r ip t o f b o tto m s tr a n d h a s n o O R F .) (b ) 4 p ts . U s in g th e s in g le le tte r a b b re v ia tio n s f o r a m in o a c id s , W R IT E b e lo w th e d e d u c e d a m in o a c id s e q u e n c e o f th e s m a ll p ro te in e n c o d e d b y th is o p e n - r e a d in g f r a m e . B e s u r e to L A B E L th e N - a n d C - te r m in i. (N-term) - M - I - D - D - L - E - C - (C-term) -5- NAME: ________KEY_________________ MIDTERM EXAM II BIOLOGY 231 18 October 2010 ________________________________________________________________________________ 3'CAATTACTAACTACTGAATCTTACAATTCGTAC 5' 5'GTTAATGATTGATGACTTAGAATGTTAAGCATG 3' (c) 3 pts. You could make a single base pair change that would result in a protein that is 3 amino acid residues shorter than the normal one. What change you would make, and exactly HOW it would give this result: If the DNA triplet, 3'-AAT-5', that encodes the codon for L (UUA) were mutated to 3'-ATT-5', then the new codon would be UAA, a stop codon. As a result, translation would terminate 3 amino acid residues too soon (MIDDLEC becomes MIDD). [or, change A:T pair at position 18 L-to-R to T:A pair] (d) 2 pts. The first amino acid residue incorporated into a protein is always methionine (M). Are there any constraints on the last residue of a protein? Answer YES or NO and explain briefly. No. The last (most C-terminal) residue will be the one encoded by whatever codon precedes the stop codon – and this can be any of the 20 amino acids. -6- NAME: ________KEY_________________ MIDTERM EXAM II BIOLOGY 231 18 October 2010 ________________________________________________________________________________ (4) You are studying the membrane properties of an epithelial cell that has numerous transmembrane proteins. So far you have determined that it has a Vm = ￿70 mV, and the following solute concentrations: S o lu te [ I n tr a c e llu la r ] [ E x tr a c e llu la r ] C a+ + 10￿7 M 10￿3 M N a+ 45 m M 450 mM K+ 200 m M 2 mM g lu c o se 100 mM 1 0 0 µM (4a) 9 pts. Calculate the equilibrium potential (VX) for EACH of the IONS. Show your work and circle your final answers: a t eq u ilib riu m , V X = 5 9 m V /z × lo g ([ X ] out/[ X ] in) V C a = 5 9 m V / 2 × l o g (1 0 ￿ 3 M / 1 0 ￿ 7 M ) = 5 9 m V /2 × l o g (1 0 4 ) = +118 mV V N a = 5 9 m V × l o g (4 5 0 m M / 4 5 m M ) = 5 9 m V × l o g (1 0 ) = +59 mV V K = 5 9 m V × l o g (2 m M / 2 0 0 m M ) = 5 9 m V × l o g (1 0 ￿ 2 ) = ￿118 mV -7- NAME: ________KEY_________________ MIDTERM EXAM II BIOLOGY 231 18 October 2010 ________________________________________________________________________________ (4b) 6 pts. Calculate the ∆G for the gradient of EACH ION. Show your work and circle your final answers: f o r a c h a r g e d m o l e c u l e X : ∆ G X = z F ( V m ￿ V X) , w h e r e F = 1 0 2 k J / V ￿ m o l e ∆G C a = (2 ) (1 0 2 k J / V ￿ m o l e ) (￿ 0 .0 7 0 V ￿ 0 .1 1 8 V ) = 2 x 1 0 2 k J / V ￿ m o l e (￿ 0 .1 8 8 V ) = ￿ 3 7 .6 k J /m o le ∆G N a = 1 0 2 k J / V ￿ m o l e (￿ 0 .0 7 0 V ￿ 0 .0 5 9 V ) = 1 0 2 k J /V ￿ m o l e (￿ 0 .1 2 9 V ) = ￿ 1 2 .9 k J /m o le ∆G K = 1 0 2 k J /V ￿ m o l e (￿ 0 .0 7 0 V ￿ (￿ 0 .1 1 8 V ) ) = 1 0 2 k J / V ￿ m o l e (+ 0 .0 4 8 V ) = + 4 .8 k J /m o le (4c) 4 pts. Calculate the ∆G for the glucose gradient. Show your work and circle your final answer: F o r a n y s o lu te , c h a r g e d o r u n c h a r g e d : ∆G = 5 .7 kJ/m o le × lo g ([ X ] in/[ X ] out) + zF V m (F o r g l u c , z = 0 , s o z F V m = 0 ) so : ∆G = 5 .7 kJ/m o le × lo g ([ g lu c] in/[ g lu c] out) = 5 .7 kJ/m o le × lo g (1 0 0 m M /1 0 0 µM ) = 5 .7 k J / m o le × l o g (1 0 3 ) = + 1 7 .1 k J /m o le -8- NAME: ________KEY_________________ MIDTERM EXAM II BIOLOGY 231 18 October 2010 ________________________________________________________________________________ (4d) 3 pts. Look carefully at the values you have calculated for the gradients of Na+ and glucose. Using the conventional Na+/glucose symporter that we have discussed, is it possible to drive the formation of the glucose gradient using the energy of the Na+ gradient? Answer YES or NO and briefly EXPLAIN why or why not. YES. T h e s y m p o r te r u s e s th e m o v e m e n t o f tw o m o le s o f N a + p e r m o le o f g lu c o s e . T h e fa v o r a b le m o v e m e n t o f 2 m o le s o f N a + a t ￿ 1 2 .9 k J /m o le , o r ￿ 2 5 .8 k J t o t a l , i s m o r e t h a n e n o u g h t o d r i v e t h e u n f a v o r a b l e m o v e m e n t o f o n e m o le o f g lu co se a t + 1 7 .1 kJ/m o le (￿ 2 ∆G Na > ∆G glucose). (4e) 3 pts. The properties of the Na+ and K+ gradients are different. You could design a transporter for this cell that is capable of using the energy of the K+ GRADIENT to drive the formation of the GLUCOSE GRADIENT. Fill in its properties in the boxes below: Symporter or antiporter: A n tip o rte r Direction of K+ movement through the transporter: o u t o f c y to p la s m ( o r o u t o f c e ll o r o u tw a r d ) Stoichiometry of the transporter, in moles K+ to moles glucose: ￿ 4 :1 [Although the energetically favorable outward movement one mole of K+ (4.8kJ) is not adequate to drive the unfavorable inward movement of one mole of glucose (need 17.1kJ), an antiporter that coupled the movement of four or more moles of K+ (= 19.2 kJ or more) per mole of glucose would work under these conditions.] -9- NAME: ________KEY_________________ MIDTERM EXAM II BIOLOGY 231 18 October 2010 ________________________________________________________________________________ (5) 8 p ts. W h e n yo u d ep o la rize th e n e u ro n in th e p re v io u s p ro b le m to its th re sh o ld V m, it h a s a c h a r a c t e r i s t i c a c t i o n p o t e n t i a l , g r a p h e d b e l o w i n a V m v s t i m e p l o t . A s yo u k n o w , V m c h a n g e s w ith a tim e c o u r s e th a t is d e te r m in e d b y th e o p e n in g a n d c lo s in g o f v o lta g e - g a te d io n c h a n n e ls . E x a m in e th e p lo t a n d m a tc h e a c h d e s c r ip tio n in th e le f t c o lu m n o f th e ta b le w ith th e O N E lo c a tio n a lo n g th e p lo t th a t it d e s c r ib e s b e s t. L e tte rs c a n b e u s e d m o r e th a n once. Event ONE location on plot (A-F) that matches Vm is heading toward the equilibrium potential of K E many voltage-gated Na channels open D resting membrane potential A many voltage-gated K channels open E threshold Vm for an action potential C Vm is heading toward the equilibrium potential of Na D membrane depolarizing but not yet at the threshold level B both Na and K channels are in their refractory period F -10- NAME: ________KEY_________________ MIDTERM EXAM II BIOLOGY 231 18 October 2010 ________________________________________________________________________________ ( 6 ) W i t h t h e r i g h t e q u i p m e n t , i t i s p o s s i b l e t o p e r f o r m a n e x p e r i m e n t i n w h i c h yo u s e t o r “ c l a m p ” t h e m e m b r a n e p o t e n t i a l o f i s o l a t e d m i t o c h o n d r i a . L e t ’ s s a y t h a t yo u c l a m p t h e v o lta g e a c r o s s th e in n e r m ito c h o n d r ia l m e m b r a n e s o th a t th e m a tr ix is 1 1 8 m V m o r e n e g a t i v e t h a n t h e c yt o p l a s m . I n a d d i t i o n , yo u a d j u s t t h e c o m p o s i t i o n o f t h e s o l u t i o n t h a t b a th e s th e is o la te d m ito c h o n d r ia s u c h th a t it h a s a p H 2 .0 u n its g r e a te r th a n th a t o f th e m a trix . ( a ) 6 p t s . U n d e r t h e s e c o n d i t i o n s , yo u f i n d t h a t yo u r i s o l a t e d m i t o c h o n d r i a w i l l N O T m a k e A T P . U s in g a c le a r c a lc u la tio n a n d b rie f e x p la n a tio n , d e m o n s tr a te W H Y th is is th e c a s e . F o r a p r o t o n g r a d i e n t, ∆G = F (∆p ) , w h e r e ∆p = ψ ￿ (5 9 m V ￿ ∆p H ) , ψ = V m o f th e IM M , a n d ∆ p H = (p H I M S ￿ p H m a tr ix) M a tr ix is 1 1 8 m V m o r e n e g th a n c y to p la s m , s o V m fo r I M M = + 1 1 8 m V S o ∆ p = 1 1 8 m V ￿ ( 5 9 m V ￿ ∆ p H ) , a n d ∆p H = 2 = 1 1 8 m V ￿ (5 9 m V × 2 ) = 0 mV U n d e r t h e s e c o n d i t io n s p r o t o n m o t i v e f o r c e = 0 . T h e r e f o r e ∆G = 0 a n d t h e m i to c h o n d r i a c a n n o t m a k e a n y A T P . [ y o u h a v e b a l a n c e d ∆p H w i t h a n exa ctly o p p o sed ψ] ( b ) 2 p t s . H o w c o u l d yo u c h a n g e t h e c o n d i t i o n s i n t h e s o l u t i o n t h a t b a t h e s t h e m ito c h o n d r ia s o th a t th e y c o u ld s ta r t m a k in g A T P ? E X P L A I N in a s e n te n c e b e lo w : L o w e r t h e p H (i n c r e a s e th e [ H + ] ) o f t h e b a t h i n g s o l u t i o n , s o a s t o in c r e a s e th e fa v o r a b ility o f H + flo w in w a r d a c r o s s th e I M M . -11- NAME: ________KEY_________________ MIDTERM EXAM II BIOLOGY 231 18 October 2010 ________________________________________________________________________________ (7 ) E x a m in e th is p ic tu re o f a rib o s o m e tra n s la tin g a stra n d o f m R N A , a n d u s in g it a s th e s t a r t i n g p o i n t , a n s w e r t h e q u e s t i o n s b e l o w a b o u t t h e s t e p - b y- s t e p p r o c e s s t h r o u g h w h i c h t h e r i b o s o m e t r a n s l a t e s m R N A i n t o a p o l yp e p t i d e c h a i n . (a ) 2 p ts . W h a t w ill h a p p e n n e x t to th e tR N A w ith th e a n tic o d o n 5 'C A U 3 ' ? I t w ill e x it th e E s ite o f th e r ib o s o m e . (b ) 1 p ts . W h a t a m in o a c id s h o u ld b e c o u p le d to th e tR N A in th e A s ite ? N (a s p a ra g in e ) (c ) 2 p ts . E x a c tly h o w d id th e tR N A in th e A s ite g e t th e re ? I t w a s p la c e d in to th e s ite w ith th e h e lp o f e lo n g a tio n f a c to r T u , u s in g th e e n e rg y o f G T P h y d ro ly s is . -12- NAME: ________KEY_________________ MIDTERM EXAM II BIOLOGY 231 18 October 2010 ________________________________________________________________________________ ( d ) 3 p t s . W h a t w i l l b e t h e n e x t a c t i o n s o f t h e l a r g e s u b u n i t a n d i t s p e p t i d yl t r a n s f e r a s e a c t i v i t y? T h e la rg e s u b u n it w ill s h if t in th e 3 ' d ire c tio n . T h e tra n s f e ra s e w ill b re a k th e b o n d b e tw e e n S a n d its tR N A , a n d c a ta ly z e th e f o rm a tio n o f a p e p tid e b o n d b e tw e e n th e S o f th e g r o w in g c h a in a n d th e [ ? ] a m in o a c id . (e ) 2 p ts . W h a t is th e N E X T C O D O N th a t w ill b e in th e A s ite ? W rite it b e lo w a n d b e su re to L A B E L th e 5 ' a n d 3 ' e n d s. 5 ' C A G 3 ' ( o r 3 'G A C 5 ') ( f ) 2 p t s I f t h e 5 ' U C U 3 ' c o d o n a b o v e w e r e m u t a t e d t o 5 ' U C A 3 ', w h a t w o u l d b e t h e e f f e c t o n th e a m in o a c id s e q u e n c e ? N o ch an g e – b o th co d o n s en co d e S (serine) (8) 5 pts. A “sizes” question. Number each of the following molecules so that 1 is the LARGEST and 5 is the SMALLEST. ___3___ A n N G F re c e p to r ___1___ A p o l ys o m e ___2___ A rib o s o m a l la rg e s u b u n it ___4___ A codon ___5___ D e o x yr i b o s e -13- NAME: ________KEY_________________ MIDTERM EXAM II BIOLOGY 231 18 October 2010 ________________________________________________________________________________ -14- NAME: ________KEY_________________ MIDTERM EXAM II BIOLOGY 231 18 October 2010 ________________________________________________________________________________ -15- ...
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This note was uploaded on 03/12/2012 for the course BIO 231 taught by Professor Hollenbeck during the Fall '11 term at Purdue University-West Lafayette.

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