BioChemistry > QUESTIONS & ANSWERS > Virginia Commonwealth University BIOC 403 Biochemistry_BIOCHEM 403 Test2 . 25 Q&A (All)

Virginia Commonwealth University BIOC 403 Biochemistry_BIOCHEM 403 Test2 . 25 Q&A

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BIOC 403 BIOCHEM 403 Test2 Biochemistry 403 Exam #2-3 Multiple Choice (2 points each) 1. The demonstration that RNA can have catalytic activity suggests that: a. DNA may be more catalytic tha... n RNA b. Catalytic RNAs were present early in evolution c. RNA cannot contain genetic information d. DNA preceded RNA in evolution e. None of the above 2. Km= a. k1+k2/k-1 b. k1+k-1/k2 c. k2+k-2/k1 d. k-1+k2/k1 e. none of the above 3. vmax= a. k2[ET] b. k1[ET] c. k-1[ET] d. k1k2 4. For the Michaelis –Menten equation, k-2 can be ignored because: a. [EF]=[ET] b. We measure v0 c. [ES] is in steady state d. k1=k2 e. all of the above 5. The steady state assumption is that: a. ES formation=ES breakdown b. We can ignore k-2 c. T and pH are constant d. 1 substrate reactions e. All of the above 6. One cofactor that is not tightly bound to an enzyme is: a. Heme groups b. Cations c. NAD d. Coenzymes e. None of the above 7. Enzymes catalyze reactions by all except: a. Increasing reaction rates b. Decreasing the activation energy c. Decreasing the energy of the transition state d. loweringDG0 e. none, all of the above are true 8. Bringing together multiple active sites is an advantage conferred by: a. Multiple subunits b. Alpha helices c. Beta sheets d. Fibrous proteins e. None of the above 9. The amino acid sequence of a protein is derived directly from: a. mRNA b. tRNA c. DNA d. A, B and C e. A and C 10. Urea denatures proteins by: a. Increasing H bonds b. Disrupting H bonds c. Disrupting hydrophobic interactions d. Increasing ionic interactions e. Breaking covalent bonds 11. Energy stabilization of 3o structures comes mainly from: a. Hydrophobic interactions b. H bonds c. ionic interactions d. Covalent bonds e. None of the above 12. Chaperones are thought to function by: a. Forming H bonds b. Increasing hydrophobic interactions c. Lowering EA between energy minima d. Increasing energy stabilization e. All of the above 13. Protein domains have: a. Specific functions b. Independent, stable structures c. Predictable functional motifs d. Structures stabilized by weak forces e. All of the above 14. The first stage in protein folding is: a. Formation of secondary structures b. Interaction of secondary structures c. Subunit binding d. Hydrophobic collapse e. None of the above 15. Molecular chaperones were first identified as: a. Heat shock proteins b. Secondary structures c. Regulatory enzymes d. Membrane proteins e. None of the above 16. The active tertiary structure of an enzyme is usually: a. Non-specific b. Unfolded c. The lowest energy conformation d. The highest energy conformation e. None of the above 17. The statement that all of the information needed for proper protein sequencing is contained in the amino acid sequence is: a. True for all proteins b. False for all proteins c. True for most proteins d. True for fibrous proteins e. None of the above 18. A region of a polypeptide that has a specific function such as DNA binding is called: a. A secondary structure b. A tertiary structure c. A subunit d. A domain e. All of the above 19. Enzyme interactions with substrate are best described by: a. Lock and key b. Rigid structure interactions c. Induced fit d. Ionic bonding e. All of the above 20. What is the catalytic power of an enzyme if the rate of the catalyzed reaction is 1X105 molecules/sec and the uncatalyzed rate is 2X10-6 molecules/sec? a. 1X1011 b. 1X10-11 c. 5X1010 d. 5X10-10 e. 8X102 21. (15 points) For an enzymatic reaction that obeys Michaelis-Menten kinetics: a. What is kcat for the enzyme if vmax is 4X103 nanomoles/ml sec and there are 2 nanomoles enzyme per ml? K2 = kcat vmax = k2[Et] = (4 X 103 nanomoles/ml sec)/(2 nanomoles/ml) = 2 X 103 sec-1 b. If the catalytic efficiency of the enzyme is 4X108, what is the Km for the enzyme? Catalytic efficiency = kcat/Km Km = kcat/ catalytic efficiency = (2 X103 sec-1)/(4 X 108 sec-1M-1) = 5 X 10-6M c. What is k1 for the enzyme if k-1 is 4X102? Km = k-1 + k2/K1 K1 = k-1 + k2/Km = (4 X 102 sec-1) + (2 X 103 sec-1)/(5 X 10-6 M) = 4.8 X 108 sec-1 22. (15 points) A. Draw a double reciprocal plot for pure noncompetitive inhibition. (8 points) B. How are Km and vmax affected by the inhibitor? (3 points) C. What forms of the enzyme are bound by inhibitor? (2 points) D. Where on the enzyme does I bind? (2 points) A. See book B. Km does not change, Vmax decreases C. I binds both free E and ES D. Away from active site 23. (16 points) Consider the following 4 enzymes: Enzyme A Enzyme B Enzyme C Enzyme D k1 (1/M sec) 2 X 107 1 X 105 2 X 106 1 X106 k-1 (1/sec) 1 X 102 4 X 103 3 X 103 1 X 103 k2 (1/sec) 5 X 103 2 X 104 2 X 102 5 X 104 A. (4 points) Order the 4 enzymes (A-B) from highest to lowest turnover number. D B A C B. (4 points) Order the 4 enzymes from highest to lowest substrate affinity. A C D B C. ( 4 points) Order the 4 enzymes from highest to lowest catalytic efficiency. A D C B D. (4 points) Which enzyme has the highest catalytic power? Cannot determine 24. (12 points) With reference to the intermediate structure of the substrate protein shown below, fill in the following table. Short answers please. Serine Protease Aspartic Acid Protease 1. What does X represent? Ser-O- HO- 2. Where did the second H on -NH2+- directly come from? His Asp 3. What happens to the substrate protein as a result of the next step? Cleavage of peptide bond Cleavage of peptide bond 25. [Show More]

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Virginia Commonwealth University BIOC 403 Biochemistry_BIOCHEM 403 Test2 . 25 Q&A Virginia Commonwealth University BIOC 403 Summer Test2 Key Biochemistry 403 Exam #2 Virginia Commonwealth Univer...

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