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Unformatted text preview: University of Houston BCHS 3304 Final Exam Instructor Dr. Briggs May 11, 2006 Name___ANSWER KEY______________ Answer all questions. LAST, FIRST Multiple choice 2 pts each (100 pts total) – Choose the most correct answer and fill it in on the Scantron sheet. Also circle your answer on the exam and show ALL work. 1). What is the only compound that can phosphorylate the unphosphorylated catalytic histidine residue in phosphoglycerate mutase (PGM)? a. 2-PG b. 3-PG c. 1,3-BPG d. 2,3-BPG e. PEP 2). In the citric acid cycle, electrons are transferred from succinate to FAD in which of the following ways: a. electrons are transferred directly to NAD+ b. as hydrogen atoms c. as a hydride ion d. as molecular oxygen e. as a phosphate group 3). Which enzyme in the citric acid cycle is allosterically activated by calcium? a. Citrate synthase b. α-ketoglutarate dehydrogenase c. Fumarase d. Succinate dehydrogenase e. Succinyl-CoA synthetase 4). Which of the following enzymes is responsible for the production of the CO2? a. Succinyl-CoA synthetase b. Aconitase c. α-ketoglutarate dehydrogenase d. Succinate dehydrogenase e. Fumarase 5). Identify the MOST IMPORTANT interaction in stabilizing the secondary structures of proteins. a. Hydrogen-bonding interactions b. Hydrophobic interactions c. Covalent bonds d. Ionic interactions e. van der Waals interactions 6). Which of the following describes the pI, or isoelectric point? a. The concentration of H+ ions in solution b. The pH at which a protein molecule is most soluble c. The pH at which a protein has a net charge of zero d. The negative log of [H+] e. The log of 1/[A] 1 7). You have created a mutant form of chymotrypsin using site-directed mutagenesis. Which of the following effects would you expect to observe if you replaced the catalytic serine 195 with an alanine. a. No effect or a slight increase in affinity for substrate coupled to a complete loss of enzyme activity b. A decrease in the affinity for substrate coupled to a decrease in enzyme activity c. An increase in the rate of peptide bond cleavage due to an increase in the rate of acid-base catalysis d. An increase in the rate of peptide bond cleavage due to an increase in the rate of covalent catalysis e. A complete loss of enzyme activity due to the inability to bind substrate 8). Which of the following glycolysis enzymes is NOT used during gluconeogenesis? a. Hexokinase b. Aldolase c. Triose phosphate isomerase d. Phosphoglycerate mutase e. Phosphoglucose isomerase 9). The common disaccharide, maltose, contains which of the following monosaccharide residues? a. glucose and galactose b. glucose and mannose c. glucose and fructose d. glucose only e. mannose and galactose 10). In the uncompetitive inhibition of an enzyme, the a. inhibitor binds with enzyme-substrate complex b. inhibitor binds in the substrate binding pocket preventing substrate binding c. inhibitor displaces the substrate from the enzyme-substrate complex to form an enzymeinhibitor complex d. inhibitor reduces the activation free energy e. None of the above 11). All of the following citric acid cycle intermediates participate in other metabolic pathways EXCEPT: a. α-ketoglutarate b. citrate c. fumarate d. succinate e. oxaloacetate 12). What is the maximum number of moles of ATP that can be produced from the aerobic breakdown of one mole of glucose? a. 1 b. 2 c. 4 d. 24 e. 38 2 13). What is the pH of a 55 mM NaOH solution? (Sodium hydroxide completely dissociates in H20) [H+]=1x10-14/55x10-3=1.818x10-13, 12.7 a. 12.7 b. 12.0 c. 11.7 d. 5.0 e. 1.8x10-13 14). What is the fate of the carboxyl group that is alpha to the carbonyl of oxaloacetate in the citric acid cycle? Be VERY specific. a. It is released as CO2 via the enzyme citrate synthase b. It is released as CO2 via the enzyme pyruvate dehydrogenase c. It is released as CO2 via the enzyme α-ketoglutarate dehydrogenase d. It is released as CO2 via the enzyme isocitrate dehydrogenase e. It is retained in oxaloacetate for the next pass through the citric acid cycle 15). Compute the ΔG° (25 °C) for a reaction with a Keq of 3.52 x 10-3. a. 14 kJ/mol b. -14 kJ/mol c. 30.5 kJ/mol d. -30.5 kJ/mol e. -61.5 kJ/mol ΔG°' = -RT ln Keq 14 kJ/mol = -(8.3145 J/K mol) (298 K) ln Keq 14000 J/mol/[-(8.3145 J/K mol) (298 K)] = ln Keq -5.65 = ln Keq Keq = 3.52 x 10-3 16). A septapeptide containing one of each of the following amino acids, Arg, Asp, Ile, Lys, Met, Val, and Phe, was subjected to the following degradative techniques resulting in polypeptide fragments with the indicated amino acid compositions. The N-terminal amino acid is Asp. What is the amino acid sequence of the entire polypeptide? Draw arrows pointing to the bonds in your sequence cleaved by trypsin and cyanogen bromide (CNBr). Note that the data below are NOT sequences but are compositions!! I. Trypsin hydrolysis 1. Arg, Asp 2. Lys, Val 3. Ile, Met, Phe II. Cyanogen bromide treatment 4. Arg, Asp, Ile, Lys, Met, Val 5. Phe a. b. c. d. e. Arg-Asp-Val-Lys-Ile-Met-Phe Asp-Arg-Lys-Val-Ile-Met-Phe Asp-Arg-Val-Lys-Ile-Met-Phe Asp-Arg-Val-Lys-Met-Ile-Phe Asp-Arg-Val-Lys-Ile-Phe-Met Asp-Arg-Val-Lys-Ile-Met-Phe Trypsin Trypsin CnBr 17). A competitive inhibitor alanine racemase was added in a concentration of 2.5x10-3 M to a solution containing alanine racemase and L-alanine. The Vmax, and KM for the uninhibited enzyme at 3 25 oC are 245 M/s and 0.4x10-4 M, respectively, and the substrate concentration was 2.5x10-4M. The KI of this inhibitor is 7.5x10-6 M. What is the initial velocity (v0) of this inhibited enzyme? a. 0.39 M/s b. 0.61 M/s c. 34.3 M.s d. 37.8 M/s e. 211.4 M/s ANSWER IS NOT ABOVE The correct answer is: 4.51 M/s v0 = (Vmax*[S])/(alpha*KM + [S]) alpha = 1 + (2.5x10-3/7.5x10-6) = 333.3 v0 = (245 M/s*2.5x10-4M)/(333.3*0.4x10-4M + 2.5x10-4M) v0 = (6.13x10-2M2/s) / (1.36x10-2M) v0 = 4.51 M/s v0 = (Vmax*[S])/(KM + [S]) v0 = (245 M/s*2.5x10-4M)/(0.4x10-4M + 2.5x10-4M) v0 = (6.13x10-2M2/s) / (2.9x10-4M) v0 = 211.4 M/s 18). What is a Brønsted acid? a. A molecule that can form a hydrogen-bond b. A molecule that has a net charge c. A molecule that can accept a proton d. A molecule that can donate a proton e. A molecule that cannot form a hydrogen-bond 19). How many nm are there in 3.21 m? a. 32.1x10-9 nm b. 3.21x10-9 nm c. 0.321x109 nm d. 3.21x109 nm e. 32.1x109 nm 1 nm = 1x10-9 m; conversion factor = 1nm/1x10-9 m 1= 1nm 1nm ; x3.21m = 3.21x109 nm −9 −9 1x10 m 1x10 m 3.21 m = 3.21x109 nm 20). The enthalpy change (ΔH) for a reaction is –6.5 kJ/mol and the entropy change (ΔS) is 45.0 cal/mol/K, what is the free energy (ΔG) for the reaction in kJ/mol at 35 oC? a. –64.49 kJ/mol b. –57.99 kJ/mol c. –51.49 kJ/mol d. –62.61 kJ/mol e. –6.5 kJ/mol ΔG = ΔH – TΔS = –6.5 kJ/mol – (273+35)K*(45.0 cal/mol/K*4.184 kJ/kcal / (1000 J/kJ)) = –64.49 kJ/mol 21). Calculate the pH of a 1L solution containing 150mL of 5.04M acetic acid and 150mL of 1.2M sodium acetate? The pKa of acetic acid is 4.76. a. 7.00 4 b. c. d. e. 4.24 4.76 4.14 3.24 Acetic acid: (0.15L)(5.04M)/1L = 0.756M Acetate: (0.15L)(1.2M)/1L) = 0.180M pH = pKa + log([acetate]/[acetic acid]) = 4.76 + log(0.180/0.756) = 4.76 + (–0.62) = 4.14 22). What enzyme catalyzes the conversion of fructose to fructose-6-phosphate (in muscle cells) which is the mechanism for the metabolism of fructose via the glycolysis pathway? a. Pyruvate Kinase b. Hexokinase c. Phosphofructokinase d. Phosphoglucose isomerase e. Triose phosphate isomerase 23). What form of this amino acid cannot exist in the pH range 1-14? H H2N C COO- H H2N C COO- H NH3+ C COO- (CH2)4 a. (CH2)4 (CH2)4 NH2 NH3+ NH3+ b. H H2N C COOH H NH3+ C COOH (CH2)4 d. c. (CH2)4 NH2 NH3+ e. 24). Calculate the pI of the amino acid Arginine. a. 10.94 pI=0.5*(12.48+9.4) b. 12.48 c. 10.54 d. 9.97 e. None, it is always positively charged. 5 25). What is the stereochemistry of the following chiral molecule (R or S)? a. R b. S c. Molecule is not chiral 4 3 CH2OH HOOC OH HS 1 2 Use Cahn-Ingold-Prelog rules – rank order is based on atomic mass. Remember also that the lowest ranked group is supposed to be pointing away from you – it is facing you here… 26). Identify the sequence of the following hexapeptide. COOO NH3+ N H OH a. b. c. d. e. H N O N H O H N O HN H2N TIFERG SLPDKG TIFEKG SIFERG SIYDRG N H (CH2)3 NH2+ 27). What percentage of the histidine imidazole is protonated at pH 6.7? a. 100.0% b. 17.95% c. 14.8% d. 5.75% e. 0.0% Henderson-Hasselbalch pH = pKa + log [ApH - pKa = log [A-]/[HA] [A-]/[HA] = 10(pH-pKa) = 10(6.7-6.04) = 4.57 %HA = 100 x [HA] ([HA] + [A-]) %HA = 100 x [HA]/[HA] ([HA]/[HA] + [A-]/[HA]) %HA = 100 x 1 = 17.9% (1 + 4.57) 6 H O COO- 28). Given the amino acid sequence FKAGKTMDAHGH, what reagent will cut the peptide into two equal length pieces? a. Trypsin – cleaves to the right of Arg and Lys b. Phenyl isothiocyanate Cleaves disulfide bonds c. Cyanogen bromide Cleaves to the right of Met: THIS IS THE MOST CORRECT ANSWER d. Chymotrypsin – cleaves to the right of Phe, Trp, and Tyr e. Iodoacetate Reacts with Cys sidechains 29). What is the ionic strength, I, of a 60 mM solution of CaCl2? [Cl-] a. 60 mM [Ca2+] 2 2 I=0.5*Sum(ci*Zi )=0.5*(1*[60mM]*2 +2*[60mM]*(-1)2)=0.5(240+120) = b. 180 mM 180mM c. 240 mM d. 360 mM e. 720 mM 30). The Φ dihedral angle of a peptide bond is described by rotation about the bond a. that is between the N and C atoms. b. that is between the Cα and Cβ atoms. c. that is between the Cα and N atoms. d. that is between the Cα and C atoms. e. none of the above. 31). SDS-PAGE is a. a dehydrating gel technique. b. an anaerobic gel technique. c. a hydrophilic gel technique. d. a denaturing gel technique. e. none of the above. 32). Enzymes increase the rates of chemical reactions by a. lowering the pH required for the reaction. b. reducing the activation free energy of the reaction. c. making them diffusion limited. d. sequestering the catalytic residues from protons. e. reducing the gap between the reactants and products. They decrease the activation free energy by preferentially stabilizing the transition state. (see figure below) Book, pg. 289 7 Book, pg. 167 33). The P50 of hemoglobin is 26 torr. What is the fractional saturation of hemoglobin (YO2) at 34.5 torr? Assume that n=3 for this calculation. a. 0.26 b. 0.50 c. 0.70 d. 0.73 e. 0.95 YO2 = ( pO2 ) n 34.53 = 3 = 0.70, if you assume n = 3 ( p50 ) n + ( pO2 ) n 26 + 34.53 34). Affinity chromatography is used to separate proteins by a. size b. ligand binding c. mass d. salting out e. salting in 35). Given the following φ and ψ angles, what is the secondary structure in which this residue is involved? Phi(φ, deg) -120 a. b. c. d. e. Psi(ψ, deg) 110 Alpha helix (right handed) Anti-parallel beta strands Alpha helix (left handed) Parallel beta strands Collagen coil 8 36). In what order would you expect the amino acids Asp, Arg, and Ile to be eluted from a carboxymethyl column at pH 7? a. Asp, Ile, Arg b. Arg, Ile, Asp c. Ile, Arg, Asp d. Ile, Asp, Arg e. Asp, Arg, Ile Since this is a cation exchange column (it is anionic), the column will bind most strongly those molecules that are most positively charged. Related to: Book, Ch 5, problem 1b. 37). What specific portion of the IgG antibody is responsible for binding directly with antigens? a. Light chain. b. Constant region. c. Variable region. d. Hypervariable loops. e. Heavy chain. The hypervariable loops in the variable portion of the light chain. 38). Fully identify the subunit composition of a protein from the following information. Molecular mass by gel filtration: 400 kD Molecular mass by SDS-PAGE: 150 kD and 50 kD Molecular mass by SDS-PAGE with 2-mercaptoethanol: 80 kD, 70 kD, and 50 kD a. One 400 kD polypeptide. b. Two 200 kD polypeptides. c. One 200 kD polypeptide, one 150 kD polypeptide, and one 50 kD polypeptide. d. Two 50 kD polypeptides and two 150 kD polypeptides. e. Two 50 kD polypeptides, two 70 kD polypeptides, and two 80 kD polypeptides. The 400 kD protein consists of four chains (150 kD, 150 kD, 50 kD, and 50 kD) that are not crosslinked. The 150 kD chains consist of two polypeptides, one of 70 kD and another of 80 kD. In the entire protein, there are two 80 kD chains, two 70 kD chains, and two 50 kD chains. 39). How many residues per turn do the coiled-coil helices of α-keratin contain? a. 1.54 b. 2.0 c. 3.5 d. 3.6 e. 5.4 9 Book, pg. 133 40). What is the identity of the catalytic triad in serine proteases? a. Asp-Tyr-Ser b. His-Trp-Ser c. Lys-His-Ser d. Arg-Trp-Ser e. Asp-His-Ser, book, pg. 311 41). The Michaelis constant, KM, is a. The concentration at which the v0 is half maximal b. The rate at which the Michaelis complex goes on to form product c. The rate at which the Michaelis complex is formed d. The concentration at which the v0 is maximal e. The concentration at which the initial velocity, v0, is rate limiting 42). The lines on a Lineweaver-Burk plot for a mixed inhibitor, with the variation of α and α’ (i.e. =1.0, 1.5, 2.0, etc.) are characteristically a. Parallel b. Perpendicular c. Coincident on the left of the Y-axis d. Antiparallel e. Approaching Vmax in an asymptotic fashion 43). What is the half-life of a first order reaction (t½) with a rate constant of 5.616 x 10-4 sec-1? a. 2.06 min b. 206.0 min c. 1234 sec, t½ = 0.693/(5.616x10-4 sec-1) = 1234 sec d. 1234 min e. 0.34 min 44). What is the half-life of a second order reaction (t½) with a rate constant of 1.155 x 10-2 M-1s-1 and an initial substrate concentration of 7.016 x 10-2 M? a. 2.06 min b. 206.0 min c. 1234 sec., t½ = 1/(k[A]0) = 1/(1.155 x 10-2 M-1s-1 * 7.016 x 10-2 M) d. 1234 min e. 0.34 min 45). At what concentration of S (expressed as a multiple of KM) will v0 = 0.75 Vmax? a. 0.25 KM b. 0.33 KM c. 0.75 KM d. 1.50 KM e. 3.00 KM v0 = Vmax [S] V [S] [S] ; 0.75Vmax = max ; 0.75 = K M + [S] K M + [S] K M + [S] 0.75(K M + [S]) = [S]; 0.75K M = 0.25[S]; [S] = 3.0K M 10 46). What is the name of the enzyme in the glycolysis pathway that is responsible for converting 1,3bisphosphoglycerate to 3-phosphoglycerate? a. phosphoglycerate mutase b. triose phosphate isomerase c. hexokinase d. phosphoglycerate kinase e. fructose-1,6-bisphosphatase 47). For a Michaelis-Menton reaction, k1 = 5x102M-1s-1, k KM for this reaction. a. 6.5 M b. 4.0 M c. 2.0 M d. 0.5 M e. 0 M -1 = 4x102s-1, and k2 = 6x102s-1. Calculate KM = (k-1 + k2)/k1 KM = [(4x102 sec-1)+(6x102 sec-1)]/(5x102 M-1sec-1) KM = 2.0 M 48). Compute the free energy for the overall reaction below using the half reactions near the end of the exam: phosphoenolpyruvate + ADP ⇆ pyruvate + ATP a. 31.4 kJ/mol b. –31.4 kJ/mol c. –30.5 kJ/mol d. –61.9 kJ/mol e. –92.4 kJ/mol Phosphoenolpyruvate + H2O ADP + Pi ⇆ ⇆ Pyruvate + Pi ATP + H2O 61.9 30.5 Phosphoenolpyruvate + ADP ⇆ Pyruvate + ATP 31.4 49). Calculate KM from the following data. It might be helpful to plot the data [S] (mM) 0.1 0.2 0.4 0.8 1.6 v0 (mM/s) 0.34 0.53 0.74 0.91 1.04 a. 0.01 mM b. 0.025 mM c. 0.25 mM d. 2.5 mM e. 25.0 mM [S] (mM) v0 (mM/s) 1/[S] 1/v0 0.1 0.34 10 2.94 0.2 0.53 5 1.89 0.4 0.74 2.5 1.35 0.8 0.91 1.25 1.10 1.6 1.04 0.625 0.96 11 Problem 6 y = 0.2111x + 0.83 4 1/v0 3 2 1 0 -5 0 5 10 15 1/[S] • KM=-1/x-intercept=-1/(-4 mM-1)=0.25 mM • Vmax=1/y-intercept=1/(0.83 mM-1-s) • Vmax=1.20 mM-s –1 • Slope=KM/Vmax 50). What is the equilibrium constant (K) for the hydrolysis of glucose–1–P at 25oC. a. 0 b. 8.4 c. 46 d. 272.4x106 e. 4606 Glucose-1-P + H2O ⇆ Glucose + Pi Glucose–3–phosphate + H2O ⇆ glucose + Pi; ΔG = –9.2 kJ/mol ΔG = –RTlnK; K = e–ΔG/RT K = e–(–20,900J/mol)/(8.3145J/K-mol)(298K) = 4606 12 -20.9 Standard free energies (ΔG0) of some hydrolysis and phosphorolysis reactions. (R = 8.3145x10-3 kJ/Kmol) Reactants Products ⇆ ΔG0’(kJ/mol) Pi + Glucose Glucose-6-P + H2O 13.8 ⇆ ATP + H2O AMP + PPi -32.2 ⇆ ATP + H2O ADP + Pi -30.5 ⇆ Phosphoenolpyruvate + H2O Pyruvate + Pi -61.9 ⇆ Glucose-1-P + H2O Glucose + Pi -20.9 ⇆ Glucose + Pi Glucose-3-P + H2O 9.2 ⇆ 1,3-bisphosphoglycerate + H2O 3-bisphosphoglycerate + Pi -49.4 ⇆ Acetate + Pi Acetylphosphate + H2O 43.1 ⇆ Phosphocreatine + H2O Creatine + Pi -43.1 ⇆ H2O + PPi 2Pi -33.5 ⇆ Fructose-6-P + H2O Fructose + Pi -13.8 ⇆ Acetyl-CoA + H2O Acetate + CoA -31.5 ⇆ 13 14 15 16 ...
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This note was uploaded on 03/10/2009 for the course BCHS 3304 taught by Professor Johnson during the Spring '08 term at University of Houston.

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