Biochemistry Exam I Hydrogen Bonds

EXAM I I. __________________/36 II. __________________/21 III. __________________/18 IV. __________________/11 V. __________________/14 TOTAL /100 I. MULTIPLE CHOICE (36 points; 3 pts each) Choose the BEST answer to the question by circling the appropriate letter. 1. All of the following compounds are capable of forming hydrogen bonds with water EXCEPT ______. A. B. C. D. E. 2. Titration of valine by a strong base, for example NaOH, results in two pKs. The titration reaction occurring at pK2 (pK2 = 9.62) is: A. -COOH + OH- B. -COOH + -NH2 -COO- + -NH3+ -COOH + -NH2. -NH3+ + OH-NH2 + H2O. -NH2 + OH-NH- + H2O. C. D. E. 3. -COO- + H2O. -COO- + -NH3+. The pK of the carboxyl group of glycine (pK = 2.3) is less than that for acetic acid (pK=4.8) due to ________. A. B. C. D. E. 4. methanol acetamide methyl acetate ethanol hexane the the the the the electronegativity of the carbonyl oxygen electrostatic stabilization of the adjacent amino group electronegativity of the nitrogen in the adjacent amino group electronic nature of amino acids fact that glycine can be a zwitterion and acetic acid cannot Which of the following is true about the change in enthalpy ( H) of a reaction that is spontaneous (favorable) at room temperature? A. B. C. D. E. It It It It It is equal to T S. is positive and the reaction is exothermic. is negative and the reaction is endothermic must be equal to zero. can be either positive or negative. Page 1 5. The peptide Ala-Glu-Gly-Ala-Leu has: A. B. C. D. E. 6. An enzyme that catalyzes the reaction A <---> B changes the ____. A. B. C. D. E. 7. heat of reaction equilibrium constant equilibrium concentration of A entropy of the reaction rate of both forward and reverse reactions An amino acid contains at least two functional groups: a carboxylic acid and an amine. Condensation of an amino group and a carboxyl group with loss of water forms a(n): A. B. C. D. E. 8. four peptide bonds. two free amino groups. a disulfide bond. no free carboxyl group. five peptide bonds. carbonyl. ester. imidazole. guanidine. amide. The reaction shown below is conducted in a closed system containing gaseous CO2 and a buffered aqueous solution. After the reaction is allowed to reach equilibrium, the concentration of bicarbonate ions in the aqueous phase would most likely be increased by _____. H2O + CO2 A. B. C. D. E. 9. + adding water while keeping the partial pressure of CO2 constant adding carbonic anhydrase, a catalyst for the reaction increasing the pH of the buffered aqueous solution reducing the partial pressure of CO2 reducing the temperature of the reaction solution A water-soluble globular protein is most likely to have the highest proportion of which of the following amino acid residues buried within its core? A. B. C. D. E. 10. HCO -3+ H Ser His Ile Glu Lys Which of the following is not a biological role for amino acids? A. B. C. D. E. components of proteins components of peptide hormones function as neurotransmitters components of cell walls All of the above are biological uses of amino acids. Page 2 11. An acid or base that dissociates less than 100% in water is described as being __________. A. B. C. D. E. 12. a Lewis acid or base a weak acid or base a strong acid or base a little acid or base hydrophobic acid or base The condition called _________ results when a person's blood pH becomes lower than normal. A. B. C. D. E. F. metabolic acidosis vitamin deficiency beri-beri strong acid weak acid gastric acid II. STRUCTURES. (21 points) 13. Draw the structures of the following functional groups,pH 7.0: (5 pts) (a) hydroxyl (b) carboxylate (c) amino (d) phosphoryl (e) indole 14. Draw the amino acid asparagine as you would find it at pH 7. Identify the hydrogen bond donor and acceptor groups using arrows. (6 pts) 15. Draw the peptide ATLDAK. (a) Calculate is approximate pI. the net charge at pH 7.0? (10 pts) Page 3 (b) What is III. SHORT ANSWER. (18 points) Give a brief answer or diagram to each problem or question. 16. If all living organisms use the same amino acids, nucleotides, and monosaccharides to form proteins, nucleic acids, and polysaccharides, how can there be so many different kinds of living organisms? (2 pts) 17. Draw the titration curve for a weak acid, HA, whose pKa is 3.2. Label the axes properly. Indicate with an arrow where on the curve the ratio of salt (A-) to acid (HA) is 10:1. What is the pH at this point? (5 pt) 18. Suppose you have just added 100 mL of a 1.5 M solution of acetic acid to 400 mL of 0.5 M NaOH. What is the final pH? (The pKa of acetic acid is 4.7.) (3 pts) 19. What is the uniquely important acid-base characteristic of the histidine R group? (2 pts) Page 4 T 20. A beaker of distilled water sitting out on your lab bench is slightly acidic. Explain. (2 pts) 21. For the reaction A to B at 300 °K, the change in enthalpy is -7kJ/mol and the change in entropy is -30 J/°K-mol. Is the reaction favorable? If not, should the temperature be increased or decreased to make the reaction favorable (spontaneous)? (4 pts) IV. TRUE/FALSE. (11 points) CIRCLE the appropriate answer. F 22. The O-H bonds of water are polar due to the high electronegativity of hydrogen. T F 23. Water is a polar molecule (has a dipole moment) because it is V-shaped. T F 24. In a medium of pH 2.0, aspartic acid has a net positive charge. T F 25. All 20 of the amino acids used to make natural proteins are optically active. T F 26. The isoelectric point for most of the amino acids that have non-polar side chains is about 6. T F 27. The hydrophilic side chains of amino acid residues normally locate themselves toward the exterior of globular proteins. T F 28. For life to exist, the -T S term in the Gibbs free energy equation must be more negative than H is positive. T F 29. The gecko utilizes van der Waals interactions to walk on ceilings. T F 30. Hyperventillation will lower the CO2 concentration in the blood by changing the pKa of the bicarbonate reaction. T F 31. Morphine must have the same configuraton as Met-enkephalin because both can bind to the same brain receptor proteins T F 32. Enzymes can make a reactions more or less "spontaneous". Page 5 V. MATCHING. (14 points) 33. _______ 1. _______ 2. _______ 3. _______ 4. _______ 5. _______ 6. 34. _ 1. _ 2. _ 3. _ 4. The choices on the left refer to the substances on the right. Select the one substance that is most closely related to the choice on the left. A substance can be used ONLY once and all have a best choice. (6 pts) Largely insoluble in aqueous solutions Highly soluble in water and solvated largely by hydrogen bonds Forms hydrated ion pairs in aqueous solutions Exists largely as a zwitterion in aqueous solution at pH 7 Structure is largely determined by the hydrophobic effect van der Waals interactions help stabilize the bonding A. NaCl B. the R-group of Glutamine C. Valine D. the R-group of Phenylalanine E. a ligand with its receptor protein SDS micelles F. For the bonds or interactions in the left column, indicate all the characteristics in the right coluumn that are appropriate. Items on right can be used more than once. (8 pts) A. Charge-charge electrostatic interactions ___________________________ Hydrogen bond ___________________________ van der Waals interactions ___________________________ Hydrophobic interactions ___________________________ D. Has an energy of around 0.2-0.5 kcal/mole per bond Has an energy between 3 and 7 kcal/mole per bond Is optimal at or near the van der Waals contact distance Requires nonpolar species E. F. G. H. I. J. Is also called a salt-bridge Involves polarizable atoms Involves O and H or N and H atoms Requires polar or charged species Exists only in water Involves charged species only B. C. Page 6 Answer Sheet for Test Chapter/ Question Test Correct Question Answer The following questions are in section “Multiple Choice” 1 E (-,b,-) 2 D 3 B 4 E (-,a,-) 5 A 6 E (-,b,-) 7 E 8 C 9 C (-,c,-) 10 E 11 B 12 A The following questions are in section “Structures” 3-41 13 O O + (a) OH; (b) C O ; (c) NH 2; (d) P 3 3-38 5-8 5-59 3-37 5-13 3-35 3-3 3-36 3-39 3-14 4-37 4-38 OH O(e) N H 14 acceptors + COO- O = 5-60 H3N-CH-CH2-C-NH2 donors O O O = O = O = + = 15 = 5-61 H3N-CH-C-NH-CH-C-NH-CH-C-NH-CH-C-NH-CH-C-NH-CH-COO CH3 H-C-OH CH2 CH3 H-C- CH3 CH2 CH3 COO- (CH2)4 NH + 3 CH3 Ala(A) Thr(T) Leu(L) Asp(D) Ala(A) Lys(K) (a) pI is about 6; C-term and D have pK ~4 and ~3, respectively; N-term and K have pK ~8 and ~11, respectively. The pI is midway between the pK's of the ionizations involving the neutral species; 4+8/2 = 6 (b) zero; at 7 both C-term and D are negative and both N-term and K are positive. Although, it might be slightly negative due to some N-term deprotonation and loss of plus charge. Page 1 Answer Sheet for Test Chapter/ Question Test Question Correct Answer The following questions are in section “Short Answer” 16 The diversity is achieved through the nearly limitless variety of sequences that can be achieved when amino acids are linked to form proteins, and nucleotides are linked to form nucleic acids. 4-21 (-,c,-) 17 The plot of pH vs. added base will look like this, with the midpoint of the titration (inflection point) at pH 3.2. 1-12 (-,b,-) 5.2 [A-]/[HA]=10:1 4.2 pKa = 3.2 3.2 2.2 pH 0.5 0.9 1.0 equivalents NaOH added The ratio of A- to HA is 10 when about 0.9 equivalents of base have been added. From the Henderson-Hasselbalch equation, the pH at this point can be calculated: pH = pKa + log [salt] = 3.2 + log 10 = 4.2 4-27 (-,c,-) 18 [acid] -Addition of 200 mmol of NaOH (400 mL x 0.5 M) to 150 mmol of acetic acid (100 mL x 1.5 M) completely titrates the acid so that it can no longer act as a buffer and leaves 50 mmol of NaOH dissolved in 500 mL, an [OH-] of 0.1 M. Given [OH-], [H+] can be calculated from the water constant: [H+][OH-] = 10-14 M2 [H+] = 10-14 M2/0.1 M = 10-13 M pH is, by definition, -log [H+] 5-31 (-,b,-) 19 pH = -log (10-13 M) = 13. Only the imidazole ring of the histidine R group has a pKa near physiological pH (pKa = 6.0), which suggests that histidine may participate in proton exchanges (acid-base chemistry) in physiological systems. Page 2 Answer Sheet for Test Chapter/ Question 4-40 3-42 3-29 3-30 3-31 3-32 3-33 3-34 3-44 3-45 3-46 3-47 3-48 3-40 4-39 (a,c,-) Test Question Correct Answer 20 Atomospheric carbon dioxide reacts with the distilled water in the beaker to form carbonic acid. The carbonic acid dissociates into protons and bicarbonate because of its pK of about 6. These protons make the water slightly acidic. 21 G = H -T S, so G = -7 kJ/mol - (300)(-0.03 kJ/°K-mol); this is -7 kJ/mol + 9 kJ/mol = +2 kJ/mol. It is not favorable. Since the -T S term is more positive, the reaction would be favorable if the temperature were decreased. The following questions are in section “True/False” 22 F 23 T (the shape is asymetric) 24 T (COOH not charged for either -carboxyl or the R-group carboxyl, thus only the positive -amino is charged) 25 F (Gly is not) 26 T; they only have a carboxylate (pKa=2.5) and an aminogroup (pKa=9.5). pI = (pKa(acid) + pKa (base))/2 27 T 28 F ( H must be more negative than the positive (-T S) term ( S is negative). 29 T 30 F (it lowers [CO2], pKa does not change. 31 F (conformation) 32 F (enzymes change the rate(kinetics) not the thermodynamics). The following questions are in section “Matching” 33 D, B, A, C, F, E 34 1. B, E, J (C) (H) 2. B, G, H (F) 3. A, C, F (G) 4. D, I, (A) [Note: letters in () are correct, but not counted as wrong] Page 3