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1、_I. Fuel MetabolismPART 1: Structure and Function of ProteinDirections: Each of the numbered items or incomplete statements in this section is followed by answers or by completions of the statement. Select the one lettered answer or completion that is best in each case. Questions 9 and 10 Use the st
2、ructure to answer questions 9 and 10： Asp-Ala-Ser-Glu-Val-Arg 9. The C-terminal amino acid of the hexapeptide shown is (A) alanine (B) asparagine (C) aspartate (D) arginine 10. At physiologic pH (7.4), this hexapeptide will contain a net charge of (A) -2 (B) -1 (C) 0 (D) +1 (E) +2 11. Which one of t
3、he following types of bonds is covalent? (A) Hydrophobic (B) Hydrogen (C) Disulfide (D) Electrostatic 13. Which one of the following conditions causes hemoglobin to release oxygen more readily? (A) Metabolic alkalosis (B) Increased production of 2,3-bisphosphoglycerate (BPG) (C) Hyperventilation, le
4、ading to decreased levels of CO 2 in the blood (D) Replacement of thesubunits with subunits 14. Production of which of the following proreins would be most directly affected in scurvy? (A) Myoglobin (B) Collagen (C) Insulin (D) Hemoglobin 15. The active site of an enzyme (A) is formed only after add
5、ition of a specific substrate (B) is directly involved in binding of allosteric inhibitors (C) resides in a few adjacent amino acid residues in the primary sequence of the polypeptide chain (D) binds competitive inhibitors 16. An enzyme catalyzing the reaction E + S = ES E + P was mixed with 4 mM su
6、bstrate. The initial rate of product formation was 25% of Vm. The Km for the enzyme is (A) 2 mM (B) 4 mM (C) 9 mM (D) 12 mM (E) 25 mM 17. The velocity (v) of an enzyme-catalyzed reaction (A) decreases as the substrate concentration increases (B) is lowest when the enzyme is saturated with substrate
7、(C) is related to the substrate concentration at lh Vm (D) is independent of the pH of the solution Questions 18 and 19 Refer to the following reaction when answering questions 18 and 19. Fumarate + H20 malate fumarase 18. Fumarase catalyzes the conversion of fumarate to malate. It has a Km of 5 M f
8、or fumarate and a Vmax of 50 mol/min/mg of protein when measured in the direction of malate formation. The concentration of fumarate required to give a velocity of 25 mol/min/mg protein is (A) 2 M (B) 5 M (C) 10 M (D) 20 M (E) 50 M 19. The Km for fumarase is approximately 5 M for fumarate. The fumar
9、ate concentration in mitochondria is approximately 2 mM. If the fumarate concentration dropped to 1 mM, the reaction rate would (A) increase slightly (B) decrease slightly (C) decrease by one half (D) stay exactly the same 20. Hexokinase and glucokinase both catalyze the phosphorylation of glucose t
10、o glucose 6-phosphate. The values of Km for the enzymes are 10 M and 0.02 M, respectively. If blood glucose is 5 mM under fasting conditions and 20 mM after a high-carbohydrate meal (A) hexokinase will function near its Vmax under fasting conditions (B) glucokinase will function near its Vmax under
11、fasting conditions (C) hexokinase will function at less than one-half Vmax after a high-carbohydrate meal (D)glucokinase will function at less than one-half Vmax after a high-carbohydrate meal 21. A competitive inhibitor of an enzyme (A) increases Km but does not affect Vmax (B) decreases Km but doe
12、s not affect Vmax (C) increases Vmax but does not affect Km (D) decreases Vmax but does not affect Km (E) decreases both Vmax and Km Questions 22-25 Refer to the graph when answering questions 22-25. 22. The value of Km for the enzyme depicted by curve A is (A) 0.5 mM (B) 1 mM (C) 2 mM (D) 1 mol/min
13、/mg (E) 10 mol/min/mg 23. The value of Vm for the enzyme depicted by curve A is (A) 0.1 mol/min/mg (B) 1 mol/min/mg (C) 10 mol/min/mg (D) 0.5 mM (E) 2 mM 24. Curve B depicts the effect of an inhibitor on the system described by curve A. This inhibitor (A) is a competitive inhibitor (B) is a noncompe
14、titive inhibitor (C) increases the Vmax (D) decreases the Km 25. Curve C depicts the effect of a different inhibitor of the system described by curve A. This second inhibitor (A) is a competitive inhibitor (B) is a noncompetitive inhibitor (C) increases the Vmax (D) decreases the Km answer：DBC BBDD
15、CBBAA ACABDirections: Each group of items in this section consists of lettered options followed by a set of numbered items. For each item, select the one lettered option that is most closely associated with it. Each lettered option may be selected once, more than once, or not at all. Questions 33-37
16、 Match each characteristic below with the protein it best describes. (A) Hemoglobin (B) Myoglobin (C) Collagen (D) Insulin 33. Requires vitamin C for its synthesis 34. Has one oxygen binding site and one polypeptide chain 35. Contains four molecules of heine per molecule of protein 36. Is converted
17、into a triple helix during its synthesis 37. Is composed of two polypeptide chains joined by disulfide bonds answer： CBACD9-D. By convention, peptides are drawn with the N-terminal amino acid on the left and the C- terminal amino acid on the right. Therefore, this peptide contains arginine at its C-
18、terminus. 10-B. The N-terminal aspartate contains a positive charge on its N-terminal amino group and a negative charge on the carboxyl group of its side chain. Glutamate contains a negative charge on the carboxyl group of its side chain. The C-terminal arginine contains a negative charge on its C-
19、terminal carboxyl group and a positive charge on its side chain. Thus, the overall charges are +2 and -3, which gives a net charge of-1. 11-C. Disulfide bonds are covalent. 13-B. Increased H+, BPG, and CO2 decrease the affinity of HbA for O2. Fetal hemoglobin (HbF = (2Y2) has a greater affinity for
20、O2 than HbA (22). Increased BPG would cause O2 to be more readily released. 14-B. Scurvy is caused by a deficiency of vitamin C. The hydroxylation of proline and lysine residues in collagen requires vitamin C and oxygen. Globin synthesis might be indirectly affected because absorption of iron from t
21、he intestine is stimulated by vitamin C. Iron is involved in heine synthesis, which regulates globin synthesis. 15-D. The active site is formed when the enzyme folds into its three-dimensional configuration and may involve amino acid residues that are far apart in the primary sequence. Substrate mol
22、- ecules bind at the active site. Competitive inhibitors compete with the substrate. (Both bind at the active site.) Allosteric inhibitors bind at a site other than the active site. 16-D. In the Michaelis-Menten equation, v = (V m x S)/(Km + S). In this case, 1/4 Vm = (Vm x 4)/(Km + 4), or Km = 12 m
23、M. 17-C. The velocity of an enzyme-catalyzed reaction increases as the substrate concentration increases. It is highest when the enzyme is saturated with substrate. Then, v equals Vm, the maximum velocity. The velocity depends on Km. Enzymes have an optimal pH at which their activity is maximal. 18-
24、B. A velocity of 25 is 1/2 Vm, which is 50. Km = S at 1/2 Vm. Km. = 5 M. 19-B. The velocity decreases slightly when the concentration of the substrate drops from 2 mM to 1 mM. At 2 mM, v = (Vmax x 2,000 M)/(5 M + 2,000 M) = 99.8% Vmax. At 1 mM, v = (Vmax x 1,000 M)/(5pM + 1,000 M) = 99.5% Vmax.20-A.
25、 During fasting, for hexokinase, v = (5 x Vm)/(0.01 + 5) = 99.8% Vm; for glucokinase, v = (5 x Vm)/(20 + 5) = 20% Vm. In the fed state, for hexokinase, v = (20 x Vm)/(0.01 + 20) = 99.9% Vm; for glucokinase, v = (20 x Vm)/(20 + 20) = 50% Vm. Hexokinase will function near its Vm in both the fed and fa
26、sting states. Glucokinase (a liver enzyme) is more active in the fed than the fasting state. At 20 mM glucose, its velocity is 50% Vm. 21-A. A competitive inhibitor competes with the substrate for the active site of the enzyme, in effect increasing the Km. As the substrate concentration is increased
27、, the substrate, by competing with the inhibitor, can overcome its inhibitory effects, and eventually the normal V is reached. 22-A. The intercept on the x axis is -1/Km = -2. Therefore, Km = 0.5 mM. 23-C. The intercept on the y axis is 1/V m = 0.1. Therefore, V = 10 mol/min/mg. 24-A. With this inhi
28、bitor, V is the same (the y intercept is the same), but Km is larger (the x intercept is less negative). Therefore, this is a competitive inhibitor. 25-B. With this inhibitor, Vm is lower but Km is the same. It is a noncompetitive inhibitor. 33-C. Proline and lysine residues in collagen are hydroxyl
29、ated in a reaction that requires tamin C. 34-B. Each myoglobin molecule contains one polypeptide chain and one heine molecule that binds one O2 molecule. Each hemoglobin molecule contains four polypeptide chains, four mole- cules of heine, and four molecules of oxygen. 35-A. Each molecule of hemoglo
30、bin contains four molecules of heine. Each myoglobin molecule contains one molecule of heine. 36-C. Collagen forms a triple helix during its synthesis. 37-D. Insulin is composed of an A chain and a B chain, which are linked by disulfide bonds. PART 2: nucleic acidDirections: Each of the numbered ite
31、ms or incomplete statements in this section is followed by answers or by completions of the statement. Select the one lettered answer or completion that is best in each case. 1. In DNA, on a molar basis (A) adenine equals thymine (B) adenine equals uracil (C) guanine equals adenine (D) cytosine equa
32、ls thymine (E) cytosine equals uracil 2. Which of the following sequences is complementary to the DNA sequence 5-AAGTCCGA-3? (A) 5-AAGUCCGA-3 (B) 3-TTCAGGCT-5 (C) 5-TTCAGGCT-3 (D) 3-TCGGACTT-5 3. DNA contains which one of the following components? (A) Nitrogenous bases joined by phosphodiester bonds
33、 (B) Negatively charged phosphate groups in the interior of the molecule (C) Base pairs stacked along the central axis of the molecule (D) Two strands that run in the same direction (E) The sugar ribose 4. Which RNA contains 7-methylguanine at the 5 end? (A) 5S RNA (B) rRNA (C) hnRNA (D) tRNA 5. Thy
34、mine is present in which type of RNA? (A) mRNA (B) rRNA (C) hnRNA (D) tRNA answer：ABCCD1-A. On a molar basis, DNA contains equal amounts of adenine and thymine and of guanine and cytosine. Uracil is not found in DNA. 2-B. Complementary sequences base-pair with each other. The strands run in opposite
35、 direc- tions. 5-AAGTCCGA-3 base-pairs with 3-TTCAGGCT-5 (or with the RNA sequence 3-UUCAGGCU-5). 3-C. DNA chains are composed of nucleosides joined by 3,5-phosphodiester bonds. Each nucleoside consists of a nitrogenous base linked to deoxyribose. Two DNA chains, oriented in opposite directions, bas
36、e-pair with each other and are twisted to form a double helix. The bases are stacked on top of each other, forming a spiral staircase in the center of the molecule, and the sugar-phosphate backbone, in which the phosphates are negatively charged, is wrapped around the outside. 4-C. hnRNA is capped a
37、t its 5 end during transcription. The cap, which contains 7-methylgua- nine, is retained when hnRNA is converted to mRNA. 5-D. Each tRNA contains one thymine residue in the ribothymidine that is found in the TC loop.PART 3 oxidationDirections: Each of the numbered items or incomplete statements in t
38、his section is followed by answers or by completions of the statement. Select the one lettered answer or completion that is best in each case. 5. Of the total energy available from the oxidation of acetate, what percentage is transferred via the TCA cycle to NADH, FADH2, and GTP? (A) 38% (B) 42% (C)
39、 82% (D) 86% (E) 100%6. What percentage of the energy available from the oxidation of acetate is converted to ATP? (A) 3% (B) 3O% (C) 4O% (D) 85% (E) 100%7. If the enzyme concentration for a biochemical reaction is increased 100-fold, the equilibrium constant for the reaction will (A) decrease twofo
40、ld (B) remain the same (C) increase in proportion to the enzyme concentration (D) change inversely with the enzyme concentration8. All of the following are electron carriers in the electron transport chain EXCEPT (A) cytochromes (B) coenzyme Q (C) Fe-S centers (D) hemoglobin (E) riboflavin9. In the
41、tricarboxylic acid cycle, thiamine pyrophosphate (A) accepts electrons from the oxidation of pyruvate and -ketoglutarate (B) accepts electrons from the oxidation of isocitrate (C) forms a covalent intermediate with the-carbon of -ketoglutarate (D) forms a thioester with the sulfhydryl group of CoASH
42、 (E) forms a thioester with the sulfhydryl group of lipoic acid 10. Each of the following vitamins is required for reactions in the oxidation of pyruvate to CO2 and H20 EXCEPT (A) pantothenate (B) niacin (C) thiamine (D) biotin (E) riboflavin12. The reactions of the TCA cycle oxidizing succinate to
43、oxaloacetate (A) require coenzyme A (B) include an isomerization reaction (C) produce one high-energy phosphate bond (D) require both NAD+ and FAD (E) produce one GTP from GDP + Pi13. Each of the following statements concerning pyruvate dehydrogenase is true EXCEPT (A) it is an example of a multienz
44、yme complex (B) it requires thiamine pyrophosphate as a cofactor (C) it produces oxaloacetate from pyruvate (D) it is converted to an inactive form by phosphorylation (E) it is inhibited when NADH levels increase 14. The principl function of the TCA cycle is to (A) generate CO2 (B) transfer electron
45、s from the acetyl portion of acetyl CoA to NAD+ and FAD (C) oxidize the acetyl portion of acetyl CoA to oxaloacetate (D) generate heat from the oxidation of the acetyl portion of acetyl CoA (E) dispose of excess pyruvate and fatty acids 15. During exercise, stimulation of the TCA cycle results princ
46、ipally from (A) allosteric activation of isocitrate dehydrogenase by increased NADH (B) allosteric activation of fumarase by increased ADP (C) a rapid decrease in the concentration of four-carbon intermediates (D) product inhibition of citrate synthase (E) stimulation of the flux through a number of
47、 enzymes by a decreased NADH/NAD+ ratio 16. CO2 production by the TCA cycle would be increased to the greatest extent by a geneticabnormality that resulted in (A) a 50% increase in the concentration of ADP in the mitochondrial matrix (B) a 50% increase in the oxygen content of the cell (C) a 50% decrease in the Vm of a-ketoglutarate dehydrogenase (D) a 50% increase in the Km of isocitrate dehydrogenase17. A man presents to the emergency department after ingesting an insecticide.