BIOCHEMISTRY 460 FIRST HOUR EXAMINATION FORM A (yellow) ANSWER KEY February 11, 2008

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1 WRITE YOUR AND I.D. NUMBER LEGIBLY ON EVERY PAGE PAGES WILL BE SEPARATED FOR GRADING! CHECK TO BE SURE YOU HAVE 6 PAGES, (print): ANSWERS INCLUDING COVER PAGE. I swear/affirm that I have neither given nor received any assistance with this exam. Signature: BIOCHEMISTRY 460 FIRST HOUR EXAMINATION FORM A (yellow) ANSWER KEY February 11, 2008 Date: A NON-PROGRAMMABLE CALCULATOR MAY BE USED ON THIS EXAM. No programmable calculators are permitted, and no sharing of calculators. We have a couple of spare calculators to lend in an emergency. SHOW YOUR WORK FOR ALL CALCULATIONS, AND BE SURE TO STATE UNITS OF ANY NUMERICAL ANSWERS. If the reasoning, calculations, or answer are shown anywhere other than in the space provided, make a note in the space provided and put answer on BACK OF SAME PAGE so the grader for that page will have it. USEFUL CONSTANTS: R (gas constant) = J mol 1 K 1 = x 10 3 kj mol 1 K 1 If temperature = 25 C, absolute temperature T = 298 K (Assume this temperature unless problem states otherwise.) Use these generic pk a values only when no precise pk a for a specific group is given. Ionizable group in peptides and proteins Approximate ("generic") pk a in peptides & proteins (from Berg, Tymoczko & Stryer, Biochemistry, 6th ed., 2007) α-carboxyl 3.1 side chain carboxyl 4.1 imidazole 6.0 α-amino 8.0 thiol 8.3 aromatic hydroxyl 10.9 ε-amino 10.8 guanidino 12.5 Potentially Useful Equations ΔG' = ΔH' TΔS' ΔG' = ΔG ' + RTln(mass action ratio)

2 1. (13 pts) The catalytic activity of an enzyme is being measured at ph 7.5, and requires that the side chain of a specific Cys (C) residue be in its unprotonated (conjugate base) form for the enzyme to be active. That specific Cys side chain in the protein has a pk a of 7.8. To receive credit, you must show your work for all calculations. A. (5 pts) For a Cys side chain with a pk a of 7.8, what is the base/acid ratio at ph 7.5? ph = pk a + log [base]/[acid] log [base]/[acid] = ph pk a = = 0.3 [base]/[acid] ratio = = 0.5 / 1 B. (3 pts) What fraction of the total enzyme molecules in a solution of the enzyme at ph 7.5 have that Cys side chain in the active form? (Express the answer as a per cent if you prefer. Show your work.) Active/total = [base] / ([acid] + [base]) = 0.5 / ( ) = 0.33 or 33% active (base form) C. (2 pts) The name of the Cys side chain functional group is thiol (or sulfhydryl). D. (3 pts) The active form of the Cys side chain in this enzyme would be 1) positively (+) charged 2) uncharged (neutral) *3) negatively ( ) charged 2. (7 pts) Consider the equilibrium between unfolded (denatured) myoglobin and folded (native) myoglobin, shown at right. For the reaction shown (folding direction) for myoglobin under at 25 C (298K), ΔH' = 0 kj/mol ΔS' = kj / (K mol). A. (3 pts) Calculate ΔG' for the folding of myoglobin under these conditions. Show your work and state units. ΔG' = ΔH' TΔS' ΔG' = 0 kj/mol (298 K){0.170 kj/(k mol)} ΔG' = 50.7 kj/mol B. (4 pts) Under these conditions, would the majority of myoglobin molecules in solution be in the 1) unfolded, or *2) folded state? Briefly explain your answer. Because ΔG' for folding is < 0 /20 p. 2 (20 points) p. 3 (28 points) p. 4 (17 points) p. 5 (17 points) p. 6 (18 points) TOTAL: (100 points) page 2

3 3. (8 pts) Consider a peptide with the sequence Lys-Ala-Arg-Gln-Asp-His (K-A-R-Q-D-H). At ph 5.0, what would be the approximate charges (if any) on the terminal groups and the side chain (R group) of each residue in the peptide: positively charged (+), negatively charged ( ), or uncharged (0), and what would be the approximate net charge on this peptide at ph 5.0? Show your work by filling out the table. functional group Charge on group, if any, at ph 5 (, +, or 0 ) N-terminal amino group of chain + C-terminal carboxyl group of chain R group (side chain) of Lys + R group (side chain) of Ala 0 R group (side chain) of Arg + R group (side chain) of Gln 0 R group (side chain) of Asp R group (side chain) of His + NET charge on peptide at ph 5: (9 pts) A certain protein can bind either of two ligands, A or B. On the axes on the graph paper below, a plot is shown for fractional saturation (Y) vs. [ligand] for ligand A and ligand B. A. (3 pts) What is K d for ligand A? (Remember to specify units, if any.) _4 µm_ Show where you got your answer. (see graph) B. (3 pts) What is K d for ligand B? (Remember to specify units, if any.) _10 µm_ Show where you got your answer. (see graph) C. (3 pts) Which ligand, A or B, binds to the protein with higher affinity (tighter binding)? Briefly explain your reasoning. Ligand A binds more tightly (higher affinity) ligand A has a lower K d, i.e., 50% saturation (half the sites occupied) occurs at a lower concentration of ligand A. 5. (6 pts) Circle T (true) or F (false) to indicate if each of the following statements is true or false. T F Increasing the concentration of protons (reducing the ph) decreases the binding affinity of hemoglobin for O 2. T F 2,3-bisphosphoglycerate (2,3-BPG) inhibits O 2 binding of human hemoglobin by binding to the distal His residue in the O 2 binding site, preventing O 2 from binding. (2,3-BPG binds in central cavity, not to distal His or in any of the 4 O 2 binding sites.) T F Increasing the concentration of carbon dioxide (CO 2 ) shifts more hemoglobin into the R state. 6. (5 pts) For a hemoglobin mutant in which the T-R equilibrium is shifted toward the T state, A. (2 pts) would P 50 of the mutant be higher or lower than for normal hemoglobin? HIGHER B. (3 pts) Does that mean that the O 2 binding affinity of the mutant is higher or lower than normal the affinity of normal hemoglobin? Briefly explain your reasoning. lower: If P 50 for mutant is higher, that means it takes a higher concentration of O 2 to halfsaturate the mutant Hb, which means the mutant s O 2 binding affinity is LOWER than that of normal Hb. (T state has a lower affinity than R state, so increasing T state reduces affinity.) /28 page 3

4 7. (10 pts) Suppose that a membrane protein is folding in the interior of a membrane lipid bilayer, a very hydrophobic environment, out of contact with H 2 O. Protein folding in a membrane (very hydrophobic environment): A. (3 pts) If the interior of the membrane lipid bilayer is very hydrophobic, the dielectric constant in the interior of the lipid bilayer would be than the dielectric constant of water. 1) somewhat lower *2) much lower 3) somewhat higher 4) much higher 5) Dielectric constants of H 2 O and the interior of the lipid bilayer would be very similar. B. (3 pts) Ionic bonds (charge-charge interactions) would be in the lipid bilayer than in water. 1) somewhat weaker 2) much weaker 3) somewhat stronger *4) much stronger 5) The strength of ionic interactions should be very similar in the two solvents. C. (4 pts) In any protein folding reaction, whether in a membrane or in water, one source of entropy change comes from the change in conformation of the polypeptide chain. Which state of the protein is favored by the difference in conformational entropy between the two states? *1) the unfolded state 2) the folded state Briefly justify your answer based on your understanding of entropy. (This asked about conformational entropy, not about free energy change or enthalpy.) The unfolded state has a much greater degree of disorder (its entropy is much higher), so the unfolded state is favored. (Conversion to a much more ordered folded structure would give a negative ΔS, a decrease in entropy for folding, which is unfavorable. Unfolding the protein would have a positive ΔS, an increase in entropy, which is favorable.) 8. (3 pts) What type of axis of cyclic symmetry is apparent in the protein structure at right? A. There is no apparent symmetry in the structure. B. a 2-fold symmetry axis *C. a 3-fold symmetry axis D. a 4-fold symmetry axis E. a 10-fold symmetry axis 9. (4 pts) In certain microorganisms, the enzyme sucrose phosphorylase catalyzes the cleavage of sucrose by inorganic phosphate (P i ), yielding glucose-1-phosphate and fructose: (1) sucrose + P i glucose-1-phosphate + fructose ΔG =? Use the following information for reactions 2 and 3 to calculate the standard free energy change for the cleavage of sucrose (reaction 1 above). Show your work and state units. (2) glucose + P i glucose-1-phosphate + H 2 O ΔG = kj/mol (3) sucrose + H 2 O glucose + fructose ΔG = 29.3 kj/mol Reaction (1) = Reaction (2) + Reaction (3) overall ΔG '(1) = ΔG '(2) + ΔG '(3) = kj/mol + ( 29.3 kj/mol) ΔG '(1) = 8.4 kj/mol /17 page 4

5 10. (13 pts) In the pathway for glucose degradation to pyruvate (the glycolytic pathway), glucose 6- phosphate is isomerized to fructose 6-phosphate in a reaction catalyzed by phosphoglucose isomerase: glucose 6-phosphate (G-6-P) fructose 6-phosphate (F-6-P) ΔG = +1.7 kj/mol A. (5 pts) Calculate the ratio of [F-6-P]/[G-6-P] at equilibrium for this reaction at 25 C (298 K). As always, show your work and state units. (The value of R is on cover page of exam.) Equilibrium ratio of [product]/[reactant] is K eq '. ΔG' = ΔG ' + RTln(m.a.ratio), so when reaction is at equilibrium: 0 = ΔG ' + RT lnk eq ' ΔG ' = RTlnK eq ' K eq ' = e ΔG '/RT = e (1.7 kj/mol) /(2.478 kj/mol) K eq ' = 0.5 (unitless) = e B. (5 pts) Suppose that in a cell, the concentration of glucose-6-phosphate is 3.9 x 10 4 M, and the concentration of fructose-6-phosphate is 1.5 x 10 4 M. Calculate the actual free energy change (ΔG') for this reaction at 298K. (Show work and state units.) ΔG' = ΔG ' + RT ln(actual mass action ratio) (The actual mass action ratio is NOT K eq ' unless the reaction is at equilibrium, in which case ΔG' = 0.) ΔG' = 1.7 kj/mol + (8.315 x 10 3 kj mol 1 K 1 )(298K) ln (1.5 x 10 5 M / 3.9 x 10 4 M) ΔG' = 1.7 kj/mol + (2.478 kj/mol) ln (0.385) = 1.7 kj/mol + (2.478 kj/mol)( 0.956) ΔG' = 1.7 kj/mol + ( 2.4 kj/mol) = 0.7 kj/mol C. (3 pts) Under the cellular conditions in part B, would the reaction go *1) forward (as written, glucose 6-phosphate fructose 6-phosphate), or 2) in reverse (fructose 6-phosphate glucose 6-phosphate), and WHY? The reaction would go forward (glucose-6-phosphate fructose-6-phosphate) because the actual ΔG' under those conditions is < 0 (a negative number) (exergonic) 11. (4 pts) Consider the tertiary structure of a protein in which two β sheet structures stack one above the other. Contact interactions at the interface between the two sheets primarily involve A. hydrogen bonding between main chain groups (backbone N-H and C=O groups) of strands in one β sheet with main chain groups of strands on the other sheet. B. polar interactions between the dipoles of the main chains of β strands on one sheet with main chains of β strands on the other sheet. *C. close van der Waals packing interactions between side chains on the two β sheets, burying hydrophobic groups between them. D. burying hydrophobic groups of side chains of residues 3-4 residues apart in the polypeptide sequence, forming a spiral structure with hydrophobic groups up the middle of the spiral. E. hydrogen bonds between side chains from one β sheet with the main chain groups of strands in the other sheet. /17 page 5

6 12. (4 pts) State the most important (Nobel-Prize-winning) conclusion resulting from Anfinsen s experiments on the refolding of ribonuclease. (One sentence would be enough here.) The primary structure of a protein has all the information necessary to specify the tertiary structure. (The amino acid sequence of a protein dictates its 3-dimensional structure.) 13. (8 pts) The reaction below has the following rate constants for the uncatalyzed (uncat.) and enzymecatalyzed (cat.) reactions: k F(uncat) = 3 sec 1 k R(uncat) = 1.2 x 10 3 sec 1 k F(cat) = 6 x 10 9 sec 1 A. (3 pts) What is K' eq for the uncatalyzed reaction? (Show how you got your answer.) K' eq = k F(uncat) / k R(uncat) = 3 sec 1 / 1.2 x 10 3 sec 1 K' eq = or 2.5 x x 10 3 (no units) B. (3 pts) What is the rate enhancement that results from the presence of the enzyme? (Show how you got your answer.) Rate enhancement is the factor by which the catalyst increases the RATE (or rate constant) of the same reaction, over the uncatalyzed rate. Rate enhancement = k F(cat) / k F(uncat) = 6 x 10 9 sec 1 / 3 sec 1 = 2 x 10 9 C. (2 pts) What is K' eq for the enzyme-catalyzed reaction? (Show how you got your answer.) K' eq = (from the answer to part A, because the equilibrium constant does not change in the presence of a catalyst) or you could (more work) calculate k R(cat), which wasn t given, using the rate enhancement, which is the same for the reverse reaction as for the forward reaction. Then K' eq = k F(cat) / k R(cat), which comes out to be the same as for the uncatalyzed reaction. 14. (6 pts) Consider the following side chain functional groups and their hydrogen bonding capabilities. A. (2 pts) At ph 6.0, how many hydrogen bonds could the side chain functional group of a Glu residue donate? (Circle one answer.) (unprotonated carboxyl group has no O H (and of course, no N H), so can t donate any hydrogen bonds) B. (2 pts) At ph 8.0, how many hydrogen bonds could the side chain functional group of a His residue donate? (Circle one answer.) (unprotonated imidazole ring has 1 N H group; the other N is in the form =N, with a lone pair, so can t donate H bond) C. (2 pts) At ph 3.0, how many hydrogen bonds could the side chain functional group of an Asn residue donate? (Circle one answer.) (has 2 N H bonds, independent of ph) /18 page 6