BIS 102-Theg-Final (1)

BIS 102-Theg-Final (1) - BileZ Steven Theg Final Exam...

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Unformatted text preview: BileZ Steven Theg Final Exam December 13, 2001 Your name Last, First Your social security number Instructions: Write your name and social security number on the cover page, and your name on subsequent pages, where indicated. It is important to show all your work for calculations. Full credit will not be given for calculated answers that are not justified. O uestion Value 0 uestion Value .—u p—d .—u ._. _. y—d DON b—d ._.. .._. O i . r t w > F p—d ,... J; I u—A I N II Total 100 Here are some of the notable equations we encountered while covering the material. They are not necessarily needed for this exam. Pa) = 30(1— 6”) r_ PA _ A PA'I'P A+Kd vo=—-—-——— I AG =-RT 111K” C. AG = RTln-C—+ 2mg; Also, you might like to know: 4V R = 3.314 meolxK T = 298 K F= 96,485 JNxmol Use the following values for amino acid pKas rewrote waw 9.4 9.2 8.2 9.2 8.8 n - - - - - - m - - - - - - “- - - m 97 T‘PE‘JE‘J comm“: 1. (3 pts) Answer the following questions concerning the phospholipid drawn below: 4.; The three carbon molecule making the backbone to which the two fatty acids are attached is glycerol . A nomenclature indicating the number of carbon atoms and the number and position of double bonds was described in class. Using that nomenclature, what fatty acid is attached at the uppermost backbone carbon atom is C16:1A9 The group attached to the phosphate can change to form different lipids. The group shown here is serine This phospholipid goes by the name phosphatidylserine The charge of this lipid at neutral pH is -1 Name a phospholipid with a different charge than the one drawn phosphatidylcholine or phosphatidylethanolamine 2. (3 pts) In class an argument was presented indicating that a protein must follow a folding pathway to achieve its final conformation. Very generally, what is that argument? The time required to randomly sample all conformations of even a small protein is far greater than the time it takes a real protein to fold. Therefore it can’t fold by random sampling of conformations but must have a specific pathway. 3. (12 pts) Glycine is commonly used as a buffer . Preparation of a 0.1 M glycine buffer starts with 0.1 M solutions of glycine hydrochloride (HO0C-CHz—NH3+ + Cl”) and isoelectric glycine ("OOC—CHz-Ni-ig’). What volumes of these two solutions must be mixed to prepare 1 liter of 0.1 M glycine buffer having a pH of 3.1 ? 1.5 9.5 “*9.- “I-_ h“ fl "tn?" ch. cf i: “at” E“..- pH = pK. + log/la, with the equilibrium about the pK. = 2.3. a =10(3.l-2.3) = 6.31 A'=6.31HA Total=CT=0J=A‘+HA=6.31HA+HA=7.31HA Hit-"LEI— =0.014M 7.31 A" = 0.1 — 0.014 = 0.086 M 0.014 molJ’L = W, so need 0.14 L of HOOC-CHrNH3+ 1L and 0.86 L of ‘OOC-CHr-Nflg‘. 4. (4 pts) There was a time when scientists thought that perhaps living systems operated by a different set of chemical rules that those governing inanimate reactions. One argument for such a viewpoint was that cells appear to decrease their entropy as they live, thereby violating the second law of thermodynamics. How was this resolved? That is, how do cells live and not violate the second law of thermodynamics? Start this way: The second law of thermodynamics states entropy increases in any process in a closed system. But cells are not a closed system - the system includes the surroundings. Cells exchange heat with the surroundings, so that the entropy of the entire system = cells + surroundings increases. 5. (6 pts) The graph below shows v0 as a function of [S] in the presence and absence of an inhibitor. On the axes below draw the Lineweaver—Burke plot that you would expect to see with this set of data, and identify the type of inhibitor used (competitive, etc.). What features of the graphs identify the kind of competition? 1m] —in|ihilor E 80 2 on E +inhibitnr “a 40 :II 20 ll [Slum 41 N*°”P‘¥‘fiu Esau-$0.; 0"- m MoMiW-M VS} 6. (4 pts) The coat for Tobacco Mosaic Virus (TMV) is made up of 2000 identical subunits of a 158 amino acid protein. Why does this structural arrangement lead to an economy of genetic material? The answer should look something like “The TMV genetic material is this way instead of that way.” Make a calculation to support your statement. The gene for a 158 amino acid protein is 3X 158 = 474 nucleotides long. One gene to encode the entire TMV coat would be 3 X 158 X2000 = 948,000 nucleotides long. More economical to use subunits. 7. (4 pts) You are purifying a 40 kDa protein. In a previous experiment it was bound to a cation exchange resin and eluted between 300 and 350 mM NaCl. You have at your disposal a cation exchange column, an anion exchange column, and two gel filtration (= size exclusion) columns filled with beads with size cutofi‘s at 20 and 60 kDa, respectively. How would you use one of these to lower the salt concentration of your protein solution? Why would that work? You would put the protein over the 20 kDa cutoff gel filtration column. The protein is too large to enter the beads and would come off in the void volume = flow through. The salts would enter the beads and be retained. You would be using the gel filtration column as a desalting column. 8. (12 pts) Draw the structure of the following tetrapeptide at pH 7.5. H—E-L—P Calculate its isoelectric point. Show your work and reasoning. 0 ° I? - n .. .. ., .. ..¢H.-L~M_cl-\—c<>o ml? - CH " c "3 c“ t T: ‘ | \ z o l ‘4 ‘ a cut, c We ‘3’ cfi C“; ‘ k / I 6.0 ' C a“ c .- N C“; I . mu ‘ 6 at; CH C- N / cor- H 4" f; 919M44- z.. ‘1' '- 6.0 9 V .5- -—=' - .3- : ’4. q C— e. f— " pl = pH at which Po dominates = ‘A (4.2 + 6.0) = 5.1 9. (8 pts) A 140 amino acid protein resides in a lipid vesicle with the topology noted in the drawing. The protein is drawn with the correct proportions. Answer the following questions about this protein: ._ c What secondary structure is likely to be found in the membrane-spmming regions? when: Would glutamic acid or isoleucine be more liker to be found in the out-of-the- membrane loops? Glutamic acid Would lysine or vaIine be more likely to be found within the membrane? Vallne Antibody #1 was raised against an epitope created by amino acids 1 — 10, antibody #2 was raised against an epitope created by amino acids 45 — 55, and antibody #3 was raised against an epitope created by amino acids 130 —- 140. In the table below put a + or — in the b0x to indicate whether the antibody would react with the protein in the vesicle as drawn. — a lied withde- ent + Antibod #l a . lied without deter ent — Antibod #2 _5_ Antibod #3 —5— Sketch the possible hydropathy plot below that would be generated using the amino acid sequence of this protein. 5 P 5‘ i o 00 120 arr-Ina and amber (4]. by 10. (l 2 pts) An enzyme catalyzes a reaction at an initial velocity of 20 pmoir‘min when the concentration of the substrate is 0.01 M. The KM for this substrate is 10'5 M. Assuming Michaelis-Menten kinetics are followed, what will the initial velocity be when the concentration of S is: (Show yetn' work and reasoning.) a. 1x10" M? b. 1x10‘M? At [S] = 0.01, [S] >> KM, and v.I = V...“ = 20 umolhnin a. At [8] = 1 x 10“, [s1= KM, so v.= 54 vm= 1o pmob’min b. At [S] = 1 x 10“, use the M-M equation: V -[S] 21mm" v =—'““—-=-——— =1.82 mollmin ° [S]+KM 10"+10-‘ ” 11. (6 pts) Fetal hemoglobin differs from adult hemoglobin by 3 His to Ser substitution in the B subunit. How does this translate into a protein with increased affinity for oxygen? That is, why does fetal hemoglobin have a lower P50 than does adult hemoglobin? Please be specific in terms of mechanism. be does not bind as well to BPG as does aHb. Since the binding of BPG stabilizes the deoxy form of Hb, it has the effect of lowering the affinity of Hb for 0;. Since be binds BPG poorly, it undergoes the BPG-mediated decrease in affinity for 0; less well than does aHb. By comparison then, be has a higher affinity for 02 than does aHb. 12. (4 pts) In lecture KM was described three times, as a ratio of rate constants, as the affinity of the enzyme for the substrate and as the substrate concentration giving v0 = V: Vm. Derive the latter relationship algebraically. v0 = vmts] [swim V S afifl 2 mg”:MI lel l=s f; %+ M [81% +KM =2[S]}é 13. (10 pts) Given Phosphocreatine —> creatine + P; AG‘” = -43.0 lemol ATP —> ADP + Pi A6“ = -30.5 ldz’mol What is the overall change in Gibbs free energy of the reaction Phosphocreatine + ADP —> creatine + ATP What is the equilibrium constant for this latter reaction? SHOW YOUR WORK. Numbering the reactions above 1, 2 and 3, rm 3 is the sum of rm 1 plus the reverse of mm 2. So AG” = 43.0 — (30.5) = -12.5 kamol From AG‘” = —-RT anq, —ao°' -{-12.5) Km =6 RT =80£D8314x293 =155_3 lO we, 14. (2 pts) Deseribe in a simple drawing the difi‘erence between a parallel and anti-parallel [3 sheet. N. null] t llll ‘ 15. (10 pts) You have separated tow connected compartments with a piece of plastic. You place KC] at 15 mM and 205 mM into compartments #1 and #2, respectively. Then you poke a very small hole in the plastic, such that the ions can diffuse from one compartment to the other. Because K-I- diffuses faster than Cl-, a charge imbalance develops between the compartments that can be read with a volt meter. What is the voltage difference that you will read between the two compartments? Which compartment is more negative? What is the name of this voltage difference? Since AG = RT lng- + ZFAV and AG = [I when ion movement stops, we have J —s.314 x298K Aly=fl1nfl=——m°l°K—J__ln—l—5—=—O.026x(—2.61)=0.067V ZF Ce Ix96,485 205 mol~V =67mV Compartment #2 is negative with respect to compartment #1. The voltage difference is called a diffusion potential or a Nernst potential. 16. (0 pts) Fill in the blank, writing out the number, to make a true statement. The letter ‘t’ appears in this sentence exactly times. Eight, no nine, no eight - hey! wait a minute ...I Have a nice break. ll ...
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This note was uploaded on 01/31/2012 for the course BIS 102 taught by Professor Hilt during the Winter '08 term at UC Davis.

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BIS 102-Theg-Final (1) - BileZ Steven Theg Final Exam...

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