Biochemistry 3304 Practice Exam 1 2015 PDF

Biochemistry 3304, Spring 2015 Practice Exam1 1. Biochemistry is a broad field with overlap across many other scientific disciplines. Which of the following would not have significant overlap with Biochemistry? a. Cell Biology b. Medicinal Chemistry c. Immunology d. None of the above because all ov...

Biochemistry 3304, Spring 2015 Practice Exam1 1. Biochemistry is a broad field with overlap across many other scientific disciplines. Which of the following would not have significant overlap with Biochemistry? a. Cell Biology b. Medicinal Chemistry c. Immunology d. None of the above because all overlap with Biochemistry. 2. How can functional groups enable interactions between biological molecules? a. Complementarity of charge b. Complementarity of reactivity c. Complementarity of structure d. All of the above can be important. 3. What is the following functional group? O R H a. Carboxylic acid b. Aldehyde c. Thiol d. Alcohol 4. What is the following functional group? O R2 R1 N H a. Carboxylic acid b. Amide c. Thiol d. Alcohol 5. What is the following functional group? R SH a. Thiol b. Thioester c. Amide d. Alcohol 6. What is the following functional group? O R O P O R OH a. Amide b. Thiol c. Phosphate diester d. Phosphate ester 7. Compartmentation conferred what kind of advantage for early cellular organisms? a. Concentrated the molecules and reactants necessary for important biological processes. b. Diluted the molecules and reactants necessary for important biological processes. c. Separated the inside of the cell from the surrounding environment. d. None of the above. 8. What is the underlying theory behind evolutionary processes? a. Evolution is not a goal-directed process. All the choices are correct. b. Random variation is exhibited among individuals. c. Evolution is a continual process. d. Organisms with beneficial traits (mutations) will survive and pass on these traits. 9. Conservation of Energy is which Law of Thermodynamics? a. First Law b. Second Las c. Third Law d. None of the above. 10. The Second Law of Thermodynamics defines what thermodynamic principle? a. Enthalpy b. Entropy c. Gibbs Free Energy d. Standard Free Energy. 11. In a first aid heat pack, a salt solution is mixed with a solvent and undergoes crystallization in an exothermic reaction. What would you observe considering only this reaction? a. The reaction would become hot and the entropy decreases. b. The reaction would become cold and the entropy decreases. c. The reaction would become hot and the entropy increases. d. The reaction would become cold and the entropy increases. 12. If you want to drive the reaction in Question 11 in the backward direction most effectively, what would you do? a. Increase the temperature b. Decrease the temperature c. Try to maintain the same temperature d. Temperature is not important 13. What is the equilibrium ∆Go at standard temperature for a unimolecular reaction A B where A = 10mM and B = 1 mM? a. -5.7 kJ/mol b. 5.7 kJ/mol c. -5.2 kJ/mol d. 5.2 kJ/mol ∆Go =-RT ln Keq Keq = [B]/[A] = 1mM / 10 mM = 0.1 ∆Go = - (8.314 J/mol K)(298K) ln 0.1 ∆Go = - (8.314 J/mol K)(298K) (-2.303) = -2478 J/mol * -2.303 = 5,700 J/mol or 5.7 kJ/mol 14. The Gibbs free energy for a reaction is 8.7 kJ/mol and proceeds with ∆H = -15.2 kJ/mol. What is the entropy for the reaction at standard temperature? a. 22.2 J/mol K b. 81.6 J/mo K c. -81.6 J/mol K The numbers in this and the next problem are not accurate. d. -87.5 J/mol K Please choose the closest if this happens in your tests. ∆G = ∆H - T∆S 8,700 J/mol = -15,200 J/mol - T∆S 8,700 J/mol + 15,200 J/mol = -T∆S 23,900 J/mol = -T∆S 23,900 J/mol / -293K = -80.2 J/mol K 15. At what temperature does the reaction in Question 14 become non-spontaneous? a. 100 K b. 186 K c. 298 K d. 1000 K ∆G = ∆H - T∆S = 0 at equilibrium, which is where the reaction goes from spontaneous to non- spontaneous 0 = ∆H - T∆S T∆S = ∆H T = ∆H/ ∆S = -15,200 J/mol / -81.6 J/mol K = 186 K 16. The ∆Go for a reaction A B at standard temperature is 2.5 kJ/mol. If the starting concentration of B is 4 and A is 2, what is the ∆G for the reaction as written at standard temperature? a. 782 J/mol b. 927 J/mol c. 4,217 J/mol d. -4,217 J/mol ∆G = ∆Go + RT ln K = 2,500 J/mol + (8.314 J/mol K)(298K) ln [4/2] ∆G = 2,500 J/mol + 2477.572 J/mol * (0.693) = 2,500 J/mol + 1717.3 J/mol ∆G = +4,217 J/mol (This is really an estimate based on rounding) 17. From the ∆Go for the reaction A B in Question 16, what direction would the equilibrium ratio of A and B be? a. A > B by about 1 to 3 b. A < B by about 1 to 3 c. A = B d. Can’t determine ∆Go =-RT ln Keq and Keq = [B]/[A] 2,500 J/mol =-(8.314 J/mol K) (298 K) ln Keq 2,500 J/mol / -(8.314 J/mol K) (298 K) = ln Keq -1.0 = ln Keq Keq = e(-1.0) = 0.36 = [B]/[A] 18. The equilibrium constants for a reaction were measured at two temperatures. From the following data, what is the ∆Ho for this reaction? HINT, the slope of a line is defined as the rise/run. Temperature = 50K, Keq = 403.43 Temperature = 30K, Keq = 54.60 a. -1.3 kJ/mol b. 1.3 kJ/mol c. -2.2 kJ/mol d. 2.2 kJ/mol Two Points of a van’t Hoff plot ln Keq = -∆Ho/ R (1/T) + ∆So/R (van’t Hoff Equation) y = mx + b for a line Therefore, m = slope = rise over run = -∆Ho/ R T1 = 50K, Keq1 = 403.43 T2 = 30K, Keq2 = 54.60 y = ln Keq and x = 1/T The points are (x, y): 0.020, 6 and 0.033, 4 The slope m = ∆y/∆x = 6-4/0.020-0.033 = 2/(-0.013) = - 153.85 NOTE: 1/T is in the denominator, so units are K m = -∆Ho/ R = -153.85 ∆Ho= - (-153.85K * R) = -(-153.85K * 8.314 J/mol K) = 1279.1 J/mol Makes sense because the slope of the line is negative. Since m = -∆Ho/ R, the only way for a negative slope is to have a positive ∆Ho! 19. What physical property of water is most responsible for its remarkable properties as a solvent? a. It can accept and donate protons in supporting biological catalytic processes and structures of biological molecules. b. It has a polar bent structure with both hydrogens and lone-pair electrons that can form hydrogen bonding networks. c. The density of water decreases upon freezing. d. All of the above 20. What is the fundamental difference between hydrogen bonding in ice versus liquid water? a. The number of hydrogen bonds per water molecule is different in ice versus liquid water b. The orientation of water molecules in ice is different than in liquid water c. The structure of the hydrogen bonding network in water is much more dynamic than in ice. d. The density of ice is lower than that for water. 21. Which of the following amino acid sidechains would not participate in hydrogen bonding? a. Tyrosine b. Phenylalanine c. Lysine d. Serine 22. What would be the order of bond length from shortest to longest between covalent bonds, ionic bonds, hydrogen bonds, and van der Waals bonds? a. Ionic Bond > Hydrogen Bond > van der Waals > Covalent Bond b. Ionic Bond > Covalent Bond > Hydrogen Bond > van der Waals c. Covalent Bond > Hydrogen Bond > Ionic Bond > van der Waals d. Covalent Bond > Ionic Bond > Hydrogen Bond > van der Waals WHILE IT MIGHT BE A LITTLE CONFUSING BECAUSE OF THE ARROW DIRECTIONS, ONLY D MAKES SENSE AT ALL WITH ANY ORDER. 23. Why do hydrophobic molecules cluster together and separate from water in aqueous environments? a. Non-polar substances like oil interact with each other more strongly than with water. b. Polar water molecules prefer to hydrogen bond with other water molecules. c. Water molecules organize around non-polar molecules to release as many water molecules into bulk solvent as possible and increase entropy. d. Clustering of non-polar molecules together in an aqueous environment is favored by enthalpy. 24. How does the hydrophobic effect impact the structure of proteins? a,b,c can all be considered correct. a. Proteins form a globular structure to maximize the attraction between non-polar sidechains in an aqueous environment. b. Exclusion of water molecules from the hydrophobic interior of the protein releases water molecules into bulk water and contributes to the globular structure. c. The protein globular structure maximizes the presence of hydrophilic sidechains on the surface of the protein. d. Protein globular structure requires an organized water “shell” around the non-polar sidechains. 25. Dialysis can be best described as which of the following? a. Movement of solvent molecules from regions of high concentration to regions of lower concentration. b. Movement of soluble molecules from regions of high concentration to regions of lower concentration. c. Movement of solvent molecules across a semipermeable membrane from regions of high concentration to regions of lower concentration. d. Movement of soluble molecules across a semipermeable membrane from regions of high concentration to regions of lower concentration. 26. In the earliest phase of dialysis of a protein and salt solution inside a dialysis membrane bag, osmosis is responsible for what effect? a. The movement of water molecules into the protein and salt solution through the dialysis membrane. b. The movement of salt molecules out of the protein and salt solution through the dialysis membrane. c. The retention of the protein inside the dialysis membrane. d. None of the above. 26. What is the pH of a solution of 0.0001M NaOH? a. 14 b. 10 c. 8 d. 4 [H+][OH-] = 10-14 M2 [OH-] = 10-4 M [H+] = 10-14 M2 /[OH-] = 10-14 M2 / 10-4 M = 10-10 M pH = -log [H+] = -log 10-10 = 10 27. A solution of formic acid is 10% ionized at a pH of 2.7. Calculate the pKa for this acid. a. 1.7 b. 2.7 c. 3.7 d. 4.7 HCOOH yields HCOO- + H+ You can use the Henderson-Hasselbalch Equation. 10% of HCOOH is actually in H3COO- form, which means 90% is in HCOOH form. The ratio is ~1/10. 2.75 = pKa + log (1/10) = pKa – 1.00 Therefore, pKa = 2.75 + 1.00 = 3.75 You can also remember that the log term in the Henderson-Hasselbalch Equation is in groups of 10, so that the pH where 10% of the conjugate base exists is mostly in the protonated state and is 1 pH unit. 28. You have 100ml of 0.1M MES buffer at the pH = 5.09. The pKa for MES is 6.09. How much 1M NaOH would you need to add to change the pH from 5.09 to 7.09? Do not concern yourself with dilution effects. a. 2 ml This level of difficulty may not appear in your tests. b. 4 ml c. 8 ml d. 10 ml We are adding NaOH to remove protons, deprotonate MES, and to generate a new pH = 7.09 from 5.09. Using the pKa = 6.09 and the the Henderson-Hasselbalch Equation, we can see that at pH = 5.09 10% of the MES is deprotonated. At pH = 7.09, the Henderson-Hasselbalch Equation shows that 90% of the MES is deprotonated. Therefore, there is a change of 80% of the MES from protonated at pH = 5.09 to deprotonated at pH = 7.09. Given that we have 100ml of 1M MES, that makes for 100 millimoles of MES (or 0.1 moles). We therefore need to add 80 millimoles (or 0.08 moles) of NaOH. Since 1M NaOH is 1 millimole/ml (or 1mole/liter), we need to add 80 milliliters (or 0.08 liters) of 1M NaOH. 29. H2CO3 is a diprotic acid. The pKa for H2CO3 is 6.35 and the pKa for HCO3- = 10.33. At pH = 5.35, what is the approximate relative quantities of H2CO3, HCO3-, and CO32-? a. 90 H2CO3, 10 HCO3-, and negligible CO32- b. 99 H2CO3, 1 HCO3-, and negligible CO32- c. 1 H2CO3, 90 HCO3-, and 10 CO32- d. Negligible H2CO3, 75 HCO3-, and 25 CO32- THIS QUESTION IS SIMILAR TO QUESTION 28. I require you to know this one, as it is much simpler than 28. 30. What would be the approximate pKa for an acid with the following titration curve? 13 12 11 10 9 8 7 0.0 0.25 0.50 0.75 1.0 H+ ions Dissociated/Molecule H+ Ions Dissociated/Molecule a. 12 b. 11 c. 10 d. 9 31. If this titration curve were for an amino acid sidechain, what amino acid(s) could be represented? a. Cysteine and Tyrosine b. Tyrosine and Lysine c. Lysine and Arginine d. Tyrosine and Arginine 32. Three different amino acids can be post-translationally modified by a kinase enzyme on an alcohol on the sidechain? Which amino acids are these? a. His, Ser, Tyr b. Thr, Tyr, Lys c. Ser, Thr, Lys d. Ser, Thr, Tyr 33. Identify the following amino acids? a. V, H, S, K b. T, K, H, V c. H, L, R, F d. I, C, T, K N HN O O + - - H 3N C O + H 3N C O H H + NH3 HO O O + - + - H 3N C O H 3N C O H H 34. What amino acids exhibit protonation/deprotonation states that are significant at physiological pH? a. C, K b. D, C c. H, E d. H, C 35. What is the approximate pI for the peptide MEKVDRP. Because this is a peptide, assume the pKa for the amino terminus is 9.0 and the pKa for the carboxyl terminus is 3.5. a. 4.5 b. 6.5 c. 7.5 d. 9.5 pKa’s are 3.5 (C-terminus), 4.0 (Glutamic Acid), 3.9 (Aspartic Acid), 9.0 (N-terminus), 10.5 (Lysine), 12.5 (Arginine) I will solve a similar problem in class.