生物化学英文版习题.docx

1、生物化学英文版习题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 Us

2、e the structure 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

3、one of the 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) Hyperventila

4、tion, leading 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 a

5、fter addition 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

6、 4 mM substrate. 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 su

7、bstrate (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

8、of 5 M for 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. T

9、he fumarate 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 g

10、lucose to 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 Vma

11、x under 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

12、 but does 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

13、 mol/min/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

14、noncompetitive 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：D

15、BC BBDD 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. Questio

16、ns 33-37 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 co

17、nverted 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 a

18、t its C-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

19、 its C- 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 affin

20、ity for 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 iro

21、n from the 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. Subst

22、rate mol- 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 K

23、m = 12 mM. 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 maxi

24、mal. 18-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% Vm

25、ax.20-A. 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 fe

26、d and fasting 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 i

27、ncreased, 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 t

28、his inhibitor, 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

29、hydroxylated 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

30、 hemoglobin 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 numb

31、ered 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. 1. In DNA, on a molar basis (A) adenine equals thymine (B) adenine equals uracil (C) guanine equals adenine (D) cytos

32、ine equals 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 (B)