Exam 1- 2010 - NAME: __________KEY_______________ MIDTERM...

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Unformatted text preview: NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 16 September 2010 ________________________________________________________________________________ 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 15 pages (plus two blank pages for scratch work) 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 1810, Mexican priest Miguel Hidalgo y Costilla issued his revolutionary tract, “Grito de Dolores,” launching the War of Independence from Spain. -1- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 16 September 2010 ________________________________________________________________________________ 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 16 September 2010 ________________________________________________________________________________ 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 16 September 2010 ________________________________________________________________________________ (1) 5 pts. A “sizes” question. Number each of the following molecules so that 1 is the LARGEST and 5 is the SMALLEST. _ __ 5 __ H 2O ___ 4 _ _ g ly cin e (G) ___ 2 __ hexokinase, a n enzym e _ __ 3 _ _ try p to p h an (W) ___ 1 __ a m itochondrion -4- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 16 September 2010 ________________________________________________________________________________ (2) The cell schematic we looked at in lecture is reproduced here. Note the following locations, designated with letters in the schematic: (a) mitochondrial matrix; (b) nucleus; (c) lysosome; (d) endosome; (e) cytoplasm; (f) outside cell. (A) 5 pts. For each pair of locations in the table below determine whether or not the small, hydrophilic dye molecule that we followed through the cell could move from the first to the second of the two locations WITHOUT breaking a membrane boundary. p a ir o f lo c a tio n s YE S o r N O from ( e) to (f) NO from (e) to (d) NO from (b) to (e) Y ES from (d) to (c) Y ES from (d) to (a) NO (B) 2 pts. The nucleus and the endosome labeled here are both spherical. Which one has the larger surface-to-volume ratio? E n d o so me (C) 2 pts. If the nucleus has a radius of 24 ìm and the endosome has a radius of 8 ìm, what is the ratio of the volume of the nucleus to the volume of the endosome? Answer in the box below. (24 ì m /8 ì m )3 = 3 3 = 2 7 -5- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 16 September 2010 ________________________________________________________________________________ (3) 12 pts. Consider each of the non-covalent interactions listed in the table below. In the middle column, briefly and clearly describe the basis of each. In the right column, choose the ONE set of groups from the list below (A-E) that is most likely to interact in each way and enter its letter. n on-c ov ale nt int eraction d esc rib e th e b a sis of th e in ter a ction o r b o n d l ik eliest to in te r a c t th is wa y (A- E) (1) h y d r o p h o b ic H yd rop ho bic grou ps associate w ith each o ther be c a us e e ntropic c ons ide ra tions fa vor this o ve r the o rderin g o f w ater B (2) io n ic T he attraction o f grou ps w ith o pp osite electrical charges A (3) hydrogen bonds A h y d ro g e n (o r p ro to n ) is s h a re d b e twee n two e le c tro n e g a tiv e a to ms s u c h a s O an d N . D (4) v a n d e r W a a ls A tom s ind uce snapsho t c harge asym m etries (dipo les) in each o ther a t a n op tim um d istance. C (A) a typical c arboxylic acid group and amine group at p hys iological p H (B) th e sid e c h a in s o f th e a m in o a c id re sid u e s L , W a n d V in th e fo ld o f a p ro tein (C) tw o proteins that h ave clos ely complim entary s hapes (D) a c arbon yl grou p like the on e on the b ase thy m ine, a nd a n am ine o n an other b ase (E) the C hicago W hite Sox -6- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 16 September 2010 ________________________________________________________________________________ (4) 12 pts. You have studied the properties of the four major classes of macromolecules that are important constituents of cells. They are listed below, followed by a list of properties. In the box next to each polymer, write the letters of the THREE properties that best correspond to it: M ACRO M OLECULE PROPE RTIES polysaccharide A, C , H n ucleic a cid B, D, F protein B, D, G lipid E, H, I PO SSIBL E PR O PER TI ES: (A ) o ften compr is ed of identical b uilding blocks, b ut b uilding blocks c an be linked together in d ifferent w ays to g ive p olym ers w ith d ifferent p roperties (B ) o nly form s a linear, u nbranched polym er (C ) ca n f orm a b ran ch ed p o lym er struc ture (D ) a p olym er b uilt u p of d ifferent b uilding block molecules, its p roperties d erive from the identity and sequence of thos e building blocks (E ) d o e s n o t fo rm lar g e c o v a len t p o ly m e rs, a ss o c iate s in to a lar g e r stru c tu re v ia h y d ro p h o b ic interaction s (F) c ontains the cell’s g enetic inform ation (G ) h a s m o st o f th e c e ll’s c a taly tic a c tiv ity (H ) m a jo r fo rm o f e n e rg y sto ra g e fo r c e lls (I) fo rm s m o st o f th e b o u n d a rie s fo r a n im a l ce lls -7- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 16 September 2010 ________________________________________________________________________________ (5) Consider the short polypeptide, a tripeptide, abbreviated as: N -term – g lu – a la – h is – C -term (A) 7 pts. DRAW THE STRUCTURE of this tripeptide in the box below. For each functional group that can be charged at any pH, SHOW IT IN THE CHARGED FORM in your drawing: (B) 2 pts. In your drawing, CIRCLE the peptide bonds (only). -8- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 16 September 2010 ________________________________________________________________________________ 5(C) 6 pts. Take into account all of the functional groups on this tripeptide that can be charged, and determine what the TOTAL CHARGE ON THE TRIPEPTIDE WILL BE AT EACH OF THE FOLLOWING pH values. Be sure to include the CORRECT SIGN (positive or negative). N-term glu his C-term What is the total charge at pH = 3? (pK 7.8) (4.3) (6.0) (3.6) +1 +0 +1 +0 +2 What is the total charge at pH = 7? +1 ￿1 +0 ￿1 0 ￿1 +0 ￿1 ￿1 What is the total charge at pH = 11? ￿2 5(D) 2 pts. This tripeptide is actually part of a protein whose quaternary structure involves forming a dimer with another protein. The tripeptide is in the protein-protein binding site, which binds to a basic region of the other protein. You can analyze the mechanism of binding by successively mutating each residue in the site and measuring the effects on binding. Of the possible changes to the tripeptide listed below, which do you think is most likely to disrupt binding with its protein partner? Which is least likely to disrupt binding? Check the appropriate box: C hange M O S T l ik ely to dis rupt (check ONE) X g lu to asp glu t o val g lu to g ly glu t o lys L E A S T l ik ely to dis rupt (check ONE) X glu to leu -9- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 16 September 2010 ________________________________________________________________________________ (6) You learned about an acetylcholinesterase (Achase) inhibitor, sarin, that is a deadly nerve agent. But other inhibitors of Achase may have therapeutic value. One example is huperzine A, originally found in a plant used in traditional Chinese medicine and purified by pharmacologists in the 1980s. Because huperzine A may help to relieve the symptoms of Alzheimer’s disease, you are studying its properties as an inhibitor in your laboratory. You carry out an in vitro assay of Achase activity over a range of substrate concentrations, with and without huperzine A, and get the data shown here: (A) 12 pts. Calculate the Vmax and Km of Achase with and without inhibitor. Show your work and circle your final answer. Be careful with units. V max , u ninhibite d: K m , u ninhibite d: V max = 1 /y-intc p = (1/1 (m m ol/m in)￿1) K m = ￿ (1/x-intc p) = ￿ (1/￿ 4 m M ￿1) = 1 .0 m m o l/m in = 0 .2 5 m M V max w ith h u p e rz in e A : K m w ith h u p e r z in e A : V max = 1 /y-intc p = (1/1 (m m ol/m in)￿1) K m = ￿ (1/x-intc p) = ￿ (1/￿ 2 m M ￿1) = 1 .0 m m o l/m in = 0 .5 m M -10- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 16 September 2010 ________________________________________________________________________________ (B) 2 pts. Which kind of inhibitor of Achase is huperzine A? Circle the correct answer below: COMPETITIVE NON-COMPETITIVE Briefly, explain how you deduced this: In the presence of huperzine A, Achase has the same Vm ax but a higher Km , indicating that it is a competitive inhibitor. (C) 2 pts. In your in vitro assay, could you overcome the effect of huperzine A by adding high concentrations of substrate (Ach)? Answer YES or NO and explain in a sentence or two: YES. Because huperzine is a competitive inhibitor, it competes with Ach for access to the active site. As [Ach]/[huperzine] increases, the probability that huperzine can get into an unoccupied active site diminishes and V approaches Vm ax. (D) 4 pts. Whether a Michaelis-Menten enzyme is inhibited by a competitive or a noncompetitive inhibitor, the slope of the reciprocal plot always INCREASES. In the box below, EXPLAIN this in 3 sentences or fewer. T he slop e o f the recipro cal (L ine w eav er-B urk ) plot = K M /V MAX . In com petitive inhibition, the apparent K M increases w hile the V MAX stays th e sam e, so th e slo p e, K M /V MAX, in cre ases as th e n u m era to r in cre ases. In non -com petitive inhibition the K M s tays the sam e w hile the V MAX g oes d o w n , so K M /V MAX in cre ases as th e d en o m in ato r d ecre ases. -11- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 16 September 2010 ________________________________________________________________________________ (7) We have talked about ATP hydrolysis driving the formation of reactive intermediates in glycolysis, but of course glycolysis does produce net formation of ATP. In one of the two reactions that do this, phosphoglycerate kinase catalyzes the conversion of 1,3-bisphosphoglycerate to 3phosphoglycerate, coupled to the synthesis of ATP from ADP: 1 ,3 -b is p h o s p h o g ly c e ra te + A D P < => 3 -p h o s p h o g ly c e ra te + A TP In the laboratory, you allow the reaction to proceed TO EQUILIBRIUM and then measure the concentrations of the reactants and products. They are: [1 ,3 - b is p h o s p h o g ly c e r a te ] = 1 m M [A D P] = 1 0 ì M [3 -p h o s p h o g ly c e ra te ] = 1 m M [ATP ] = 0 .1 m M (A) 4 pts. CALCULATE the Keq for this reaction. Show your work: K eq = [3-PG ][A TP] / [1,3-bisPG ][A D P] at equilibrium = (1 0 ￿3M)(1 0 ￿4M)/(1 0 ￿3M)(1 0 ￿5M) = 10 ￿7/10 ￿8 = 10 (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 ) ￿ 5.7 kJ/m ole ￿ 1 ￿ 5 .7 k J/m o le -12- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 16 September 2010 ________________________________________________________________________________ (C) 3 pts. You want to understand how far from equilibrium this reaction is in the intact cell, so you develop an assay for each of the reactants and products and find that in the intact cell their concentrations are: [1,3-bisphosphoglycerate] = 10 mM [ADP] = 0.1 mM [3-phosphoglycerate] = 0.1 mM [ATP] = 1.0 mM CALCULATE q, the ratio of products to reactants for these conditions. BE CAREFUL OF YOUR UNITS. Show your work: q = [P]/[R] = [3-P G ][A T P ] / [1,3-b isP G ][A D P ] = (1 0 ￿4M)(1 0 ￿3M)/(1 0 ￿2M)(1 0 ￿4M) = (10 ￿7 /10 ￿6) = 1 0 ￿1 (D) 3 pts. Do conditions in the cell favor the forward (L-to-R) or reverse (R-to-L) reaction? Answer forward or reverse, and in one sentence, explain HOW the cell keeps this reaction far from equilibrium: FO R W A R D o r R E V E R SE : _ ____ forw ard [q< K eq]______ H o w : T h is cell m aintains a v ery low ratio o f P /R for this reaction (3-PG to 1,3-bisPG ), p erhaps b y rapidly consum ing 3-PG in the n ext step o f glycolysis. (E) 3 pts. This coupled reaction looks favorable even without the enzyme – so why does the cell require the enzyme at all? EXPLAIN in 1 or 2 SENTENCES. T he enzym e vastly increases the rate of the reaction. It low ers the activ atio n en erg y b y p h y sically co u p lin g th e tw o re actio n s, efficien tly b rin g in g th e re actan ts in to a p o sitio n th at w o u ld o ccu r v ery in freq u en tly in so lu tio n . -13- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 16 September 2010 ________________________________________________________________________________ (8) (8 pts) On the following page, a protein called ribonucleotide reductase (PDB #1pj0) is shown using the Protein Data Bank WebMol viewer display modes that you have used to examine a number of proteins. The display modes are shown above each image. Review these images and answer the following questions: Which secondary structure element(s) does this protein contain? It contains both á helices and â strands (apparent in Color/Sec mode) Does it have disulfide bond(s)? YES (one is apparent in Back Bone / Color/Chain mode) Does it contain a detectable amount of unstructured, “random” coil? YES (apparent in Color/Sec mode) What can you tell about the quaternary structure of this protein? It is comprised of TWO subunits (apparent in Color/Chain mode, they have similar structures) -14- NAME: __________KEY_______________ MIDTERM EXAM I BIOLOGY 231 16 September 2010 ________________________________________________________________________________ R IB O N U C L E O T ID E R E D U C T 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 “M ain C hain” dis play mode, “C olor/C hain” mode “B ack Bone” dis play mode, “C olor/C hain” mode -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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