2005级七年制生化A答案.doc

试卷A答案PART ONE: (40 marks, 1 mark/each)A型题1.A 2.E 3.A 4.A 5.D 6.B 7.E 8.B 9.A 10.C11. E 12. D 13. B 14. C 15. A 16.D 17. A 18. E 19. A 20.C X型题21. BCDE 22. ABCD 23. ABCDE 24. BCE 25. BCE 26. CDE 27. ABC 28. ACD 29. ADE 30.ABD31. AB 32. ABC 33. ABC 34. BD 35. BCD 36. ACE 37. BC 38. BDE 39. ACD 40. ABDPART TWO: Explain the following terms (20 marks, 4 marks/each)1．Allosteric enzyme oligomer, allosteric effector binding regulatory site, conformation and activity change2. in one tertiary structure, few parts, folding tightly and shape, specific function, pr-pr or pr-DNA binding 3. in liver, kindney, noncarbohydrate, pyruvate, lactate, aa, glycerol, transfer to G or Gn4. in aa metabolism, 1 carbon chemical group, 5 kinds, FH4 carry, pu or py synthesis5. all bile acids enter intestine, 95% reabsorption, back to liver and resecretion to intestine, fully use for lipid digestionPART THREE: Answer the following questions (40 marks)1. 10 marksin cytosol, glycolysis pathway 3 key steps, 2 pyruvate;(4)enter mito and oxidative decarboxylation 2 acetyl CoA; (2) enter TAC, 3 key steps 2CO2 and 3NADH, 1 FADH2, enter respiratory chain form H2O and ATP(4)2. 6 markszymogen, protect secrete tissue and as enzyme storage form, zymogen activation, (2)give physiological effect in suitable site and condition(2), trypsin and coagulation factor (2)3. 8 markstransport for endogenetic triglyceride and cholesterol from liver(2), in liver TG ,CH and apoB100,E to form VLDL(1), get apo C,E from HDL, apo C II (1) activate LPL in blood capillary and breakdown the TG to FFA to use,(2) remove apoC and PL, CH, to IDL, liver take IDL(B100,E)by LRP, IDL breakdown and to LDL(B100) (2)4. 10 marksFA activation form acyl-CoA reaction(2); enter into mito, carnitine acyl transferase;(2)βoxidation 4 steps, form acetyl CoA(2); enter TAC ; NADH and FADH2 enter 2 res.(2) chain produce 3 or 2 ATP.(2) 5. 6 marksGluconeogenesis: glucose (2)Oxidation: H2O + CO2 (1)FA synthesis (2)Transamination: Ala (1)Ch synthesisLactate。