TAILIEUCHUNG - Biochemistry, 4th Edition P112

Biochemistry, 4th Edition P112. Continuing Garrett and Grisham's innovative conceptual and organizing framework, "Essential Questions," BIOCHEMISTRY guides students through course concepts in a way that reveals the beauty and usefulness of biochemistry in the everyday world. Streamlined for increased clarity and readability, this edition also includes new photos and illustrations that show the subject matter consistently throughout the text. New end-of-chapter problems, MCAT practice questions, and the unparalleled text/media integration with the power of CengageNOW round out this exceptional package, giving you the tools you need to both master course concepts and develop critical problem-solving skills you can draw upon. | Abbreviated Answers to Problems A-13 In a Vmax doubles but Km is constant. In b Vmax halves but Km is constant. In c Vmax is constant but the apparent Km increases. In d Vmax decreases but Km is constant. In e Vmax decreases Km decreases but the ratio Km Vmax is constant. 6. a. The slope is given by KmB Vmax KSA A 1 . b. y-intercept K A 1 1 Vmax. c. The horizontal and vertical coordinates of the point of intersection are 1 B KmA KSAKmB and 1 V 1 Vmax 1 - KmA KsA . 7. Top left 1 Competitive inhibition I competes with S for binding to E . 2 I binds to and forms a complex with S. Top right 1 Pure noncompetitive inhibition. 2 Random single-displacement bisubstrate reaction where A doesn t affect B binding and vice versa. Other possibilities include 3 Irreversible inhibition of E by I 4 1 v vs. 1 S plot at two different concentrations of enzyme E. Bottom left 1 Mixed noncompetitive inhibition. 2 Ordered single-displacement bisubstrate mechanism. Bottom right 1 Uncompetitive inhibition. 2 Doubledisplacement ping-pong bisubstrate mechanism. 8. Clancy must drink 694 mL of wine or about one 750-mL bottle. 9. a. KS 5 M b. Km 30 M c. kcat 5 X 103 sec 1 d. kcat Km X 108M 1sec 1 e. Yes because kcat Kmapproaches the limiting value of 109 M 1sec 1 f. Vmax 10 5mol mL sec g. S 90 h. Vmax would equal 2 X 10 5mol mL sec but Km 30 .j. 1 as before. 10. a. Vmax 132 rniol mL 1 sec 1 b. kcat 44 000 sec 1 c. kcat Km X 108 M 1 sec 1 d. Yes kcat Km actually exceeds the theoretical limit of 109 M 1 sec 1 in this problem e. The rate at which E encounters S the ultimate limit is the rate of diffusion of S. 11. a. Vmax jumol mL 1 sec 1 b. v miol mL 1 sec 1 c. kcat Km X 108 M 1 sec 1 d. Yes. 12. a. Vmax 6 jumol mL 1 sec 1 b. kcat X 106 sec 1 c. kcat Km 1 X 108 M 1 sec 1 d. Yes kcat Km approaches the theoretical limit of 109 M 1 sec 1 e. The rate at which E encounters S the ultimate limit is the rate of diffusion of S. 13. a. Vmax 64 jumol mL 1 sec 1 b. kcat X

• Sinh học
• Nền tảng của hóa sinh học
• trao đổi chất protein
• RNA trao đổi chất
• nguyên tắc của hóa sinh
• Axit amin
• công nghệ DNA