BIOL0280 First Midterm Examination 2012

Draw the structure of Ala–Glu-Gly in the ionic form that predominates at pH 7. Draw all atoms (including hydrogens) as well as any charges as necessary. O O O H H H N CH C N CH C N CH C O 3 CH3 CH2 H CH2 C O O 2. [2 points] Titration of isoleucine by a strong base, for example NaOH, reveals two pK ’s. The titration reaction occurring at pK (pK = 2.4) is: 1 1 − − A) —COOH + OH → —COO + H O. 2 + − B) —COOH + —NH2 → —COO + —NH2 . + − C) —COO + —NH → —COOH + —NH . 2 2 + − D) —NH + OH → —NH + H O. 3 2 2 − − E) —NH + OH → —NH + H O. 2 2 Circle the correct answer. 3. [4 points] Pauling and Corey showed that in small peptides, six atoms associated with the peptide bond all lie in a plane. Draw a dipeptide of two amino acids in trans linkage (side-chains must be shown as —R), and indicate which six atoms are part of the planar structure of the peptide bond.

An octapeptide composed of four repeating alanylglycyl units has: A) one free amino group on an glycyl residue. B) one free amino group on an glycyl residue and one free carboxyl group on a alanyl residue. C) one free amino group on a alanyl residue and one free carboxyl group on an glycyl residue. D) two free amino and two free carboxyl groups. E) two free carboxyl groups, both on glycyl residues. Circle the correct answer. 5. [2 points] The diagram to the right illustrates: A) the polypeptide chain of an alpha-helix B) two polypeptides of a beta-pleated sheet running in a parallel fashion C) two polypeptides of a beta-pleated sheet running in an antiparallel fashion D) a flexible coil segment of a polypeptide E) two polypeptides of a coiled coil Circle the correct answer. 6. [2 points] A sequence of amino acids in a certain part of protein is -S-G-P-G-. The sequence is most probably part of a(n): A) random coil B) parallel beta sheet C) beta turn D) alpha helix E) antiparallel beta sheet Circle the correct answer. 7. [2 points] Roughly how many amino acids are there in one turn of an α helix? A) 1 B) 2.8 C) 3.6 D) 4.2 E) 10 Circle the correct answer. 8. [4 points] Describe (succinctly) the experimental design of the experiment conducted by Dr. Anfinsen that led to the conclusion: “The primary sequence of a protein determines its three-dimensional shape and thus its function.” 2 points for the word “denatured” 1 point for concept of fold back into native configuration 1 point for concept of native configuration has enzymatic activity

In hemoglobin, the transition from R state to T state (high to low affinity) is triggered by: A) Fe2+ binding. B) heme binding. C) oxygen binding D) subunit dissociation. E) 2,3 BPG binding Circle the correct answer. 10. [2 points] Which of the following statements about allosteric control of enzymatic activity is false ? A) Allosteric effectors give rise to sigmoidal V vs. [S points] kinetic plots. 0 B) Allosteric proteins are generally but not always composed of several subunits. C) An effector may either inhibit or activate an enzyme. D) Binding of the effector changes the conformation of the enzyme molecule. E) Heterotropic allosteric effectors compete with substrate for binding sites. Circle the correct answer. 11. [2 points] When oxygen binds to a heme-containing protein, the two open coordination bonds of Fe2+ are occupied by: A) one O molecule and one amino acid atom. 2 B) one O molecule and one heme atom. 2 C) two O atoms. D) two O molecules. 2 Circle the correct answer. 12. [2 points] The following values were determined for alcohol dehydrogenase in the absence and in the presence of acetaldehyde, an inhibitor: K (mM) V (micromoles/min) m max In the absence of acetaldehyde 0.10 750 In the presence of acetaldehyde 0.10 500 Acetaldehyde is a(n) __________________ inhibitor. A) competitive B) noncompetitive C) uncompetitive D) mixed E) irreversible Circle the correct answer. 13. [2 points] The proximal histidine residue in myoglobin coordinates Fe2+ and acts to (choose one): A) prevent oxidation of the heme Fe2+ B) lower the relative affinity for CO C) assist in the binding of O 2 D) prevent binding of N 2 E) decrease Θ Circle the correct answer.

The steady state assumption, as applied to enzyme kinetics, implies: A) K = K . m D B) the enzyme is regulated. C) the ES complex is formed and broken down at equivalent rates. D) the K is equivalent to the cellular substrate concentration. m E) the maximum velocity occurs when the enzyme is saturated. Circle the correct answer. 15. [6 points] (No calculators allowed on this question) An enzyme catalyzes the reaction A → B. The initial rate of the reaction was measured as a function of the concentration of A. The following data were obtained: [A], micromolar V , nmoles/min 0 0.05 0.08 0.1 0.16 0.5 0.79 1 1.6 5 7.3 10 13 50 40 100 53 5,000 79 10,000 80 20,000 80 A) What is the K of the enzyme for the substrate A (expressed as a number and explain how you m determined it to get credit for the answer)? K =[S] when V =1/2 * Vmax or 50 micromolar m 0 B) What is the value of V when [A] = 43 (Setup but do not solve the calculation)? 0 V =Vmax [S]/(K +[S])=80 * 43 / (50+43) or 37 nmoles/min 0 m The above data was plotted as 1/ V vs. 1/[A] , and a straight line was obtained. 0 C) What is the value of the x-intercept of the line (expressed as a fraction)? -1 -1 x intercept: -1/km=-1/50 micromolar or -0.02 micromolar

An enzyme catalyzes a reaction at a velocity of 10 µmol/min when all enzyme active sites -5 are occupied with substrate. The K for this substrate is 1 × 10 M. Assuming that Michaelis- m Menten kinetics are followed, what will the reaction velocity be when the concentration of S is (a) 1 -5 × 10 M . (a)= 5 umol/min 17. [2 points] One of the enzymes involved in glycolysis, aldolase, requires Zn2+ for catalysis. Under conditions of zinc deficiency, when the enzyme may lack zinc, it would be referred to as the: A) apoenzyme. B) coenzyme. C) holoenzyme. D) prosthetic group. E) substrate. Circle the correct answer. 18. [2 points] The scheme S → T → U → V → W → X → Y represents a hypothetical pathway for the metabolic synthesis of compound Y. The pathway is regulated by feedback inhibition. Assuming S → T is the rate-limiting step, indicate where the inhibition is most likely to occur and what the most likely inhibitor is. Draw an arrow at where the inhibition occurs and circle the likely inhibitor: S → T → U → V → W → X → Y 19. [4 points] a) What is the effect of increased concentration of CO2 on the binding of oxygen to hemoglobin? [2 points] The affinity decreases with increasing CO2. (b) Briefly describe the mechanism of this effect. At higher [CO2], CO2 reacts with the n terminus of the polypeptide (1point), which alters charge-charge interactions between the subunits(1 point), which stabilizes the low affinity conformation of the protein subunits.

Provide the correct answers below in the spaces given A) In the active site of chymotrypsin, a ___His[1 points]_____ residue increases the nucleophilicity of a nearby ____Ser[1 points]______ residue, which makes a covalent bond to the substrate. B) A hydrophobic pocket provides_________ substrate specificity [2 points]________ C) The oxyanion hole provides________ transition state stabilization[2 points]____________ 21. [2 points] Which of the following contains an ether-linked alkyl group? A) Oleoyl oleate B) Triacylglycerol C) Phosphatidyl serine D) Platelet-activating factor E) Sphingomyelin Circle the correct answer. 22. [2 points] An example of a glycerophospholipid that is involved in cell signaling is: A) cholesterol. B) triacylglycerol. C) phosphatidylinositol. D) testosterone. E) vitamin A (retinol). Circle the correct answer. 23. [2 points] Which of these statements about facilitated diffusion across a membrane is true? A) A specific membrane protein lowers the activation energy for movement of the solute through the membrane. B) It can increase the size of a transmembrane concentration gradient of the diffusing solute . C) It is impeded by the solubility of the transported solute in the nonpolar interior of the lipid bilayer. D) It is responsible for the transport of gases such as CO across biological membranes. 2 E) It requires the highly exergonic hydrolysis of ATP. . Circle the correct answer. 24. [4 points] Explain how a biochemist might discover that a certain enzyme is allosterically regulated in the absence of a crystal structure. The enzyme would show kinetics that do not fit the Michaelis-Menten equation; the plot of V0 vs. [S] would be sigmoidal, not hyperbolic and the plot of 1/v vs 1/[S] would be nonlinear. The 0 enzyme kinetics would be affected by molecules other than the substrate(s). 4 points for sigmoidal v vs. [S points] plot or nonlinear 1/v vs 1/[s] plot 4 points for enzyme kinetics affected by molecules other than substrate for a max of 4 points

Which of the sugars shown in the figure are L sugars? A) A and B B) B and C C) C and D D) A and D E) None of the above Circle the correct answer. 26. [2 points] Which sugar is the enantiomer of sugar A? A) B B) C C) D D) B and D E) none of the above Circle the correct answer. 27. [2 points] Which two sugars shown in the figure are epimers? A) A and B B) B and C C) C and D D) A and D E) None of the above Circle the correct answer. 28. [2 points] Which of the following monosaccharides is not an aldose? A) erythrose B) fructose C) glucose D) glyceraldehyde E) ribose Circle the correct answer. 29. [4 points] Show the basic structure of all glycerophospholipids. Indicate the charge on the molecule if any at pH 7. All glycerophospholipids have two fatty acids in ester linkage with C-1 and C-2 of glycerol; often the fatty acid at C-1 is saturated, and that at C-2 is unsaturated. C-3 of glycerol is joined to an alcohol-containing head group through a phosphodiester linkage, which is negatively charged at neutral pH.

The fluidity of the lipid side chains in the interior of a bilayer is generally increased by: A) a decrease in temperature. B) an increase in fatty acyl chain length. C) an increase in the percentage of phosphatidyl ethanolamine D) a decrease in the number of unsaturated fatty acids. E) an increase in the number of double bonds in fatty acids. Circle the correct answer. 31. [2 points] Which of following is an anomeric pair? A) D-glucose and D-fructose B) D-glucose and L-fructose C) D-glucose and L-glucose D) α-D-glucose and β-D-glucose Ε) α-D-glucose and β-L-glucose Circle the correct answer. 32. [2 points] Indicate whether the following two statements are true or false (True / False) 2 protons are transferred in every redox reaction (True / False) Standard reduction potentials, E’ º, are relative to that of oxygen 33. [2 points] Redox reactions require an electron _____donor____________ and an electron _______acceptor_________ Answer can be inverted and still be correct 34. [3 points] Explain how homopolysaccharides of glucose can have either helical or planar structures. Helical confirmation results from alpha1-4 linkages (1.5 pts) between the glucose monomers and planar structures result from beta1-4 (1.5 pts) linkages between the glucose monomers. Hydrogen bonding stabilizes these structures

Once inside a cell, glucose is rapidly phosphorylated to glucose-6-phosphate. What is the main purpose of this phosphorylation? A) to form a low-energy compound B) to activate PFK-1 C) to trap glucose inside the cell D) to prevent mutarotation Circle the correct answer. 36. [2 points] Hamsters love to run on exercise wheels. Prolonged running at a high rate of speed requires ATP. Could a hamster with a defective gene for the enzyme lactate dehydrogenase meet the extra ATP demand for prolonged, fast wheel-running by maintaining a high rate of glycolysis? Why or why not? A) No, not enough NAD+ can be regenerated for glycolysis to continue at a high rate. B) No, the defective gene will cause a rapid decline in pH in the muscles used for running. C) Yes, the defective gene has no effect on the glycolytic pathway. D) Yes, the enzyme alcohol dehydrogenase will supply the needed NAD+ if the lactose dehydrogenase cannot. Circle the correct answer. 37. [2 points] Indicate whether the following statement is true or false (True/False) In the preparatory phase of glycolysis, a glucose molecule is converted to two pentose phosphate molecules, at the expense of (driven by) the hydrolysis of two ATP molecules. 38. [2 points] Indicate whether the following statements are true or false by circling T or F for each question. T / F Phosphofructokinase 1 requires ATP as a substrate. T / F Phosphofructokinase 1 is allosterically activated by AMP. 39. [2 points] The ultimate electron acceptor in the fermentation of glucose to ethanol is: A) acetaldehyde. B) acetate. C) ethanol. + D) NAD . E) pyruvate. Circle the correct answer.

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