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Fundamentals of Biochemistry: Life at the Molecular Level
5th Edition
ISBN: 9781118918401
Author: Donald Voet, Judith G. Voet, Charlotte W. Pratt
Publisher: WILEY
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Chapter 15, Problem 30CQ
Summary Introduction
To explain: The reason why galactose catabolism is slower than glucose catabolism in yeast.
Concept introduction: In human diet, the source of galactose is dairy products. In dairy products, lactose is present. Lactose gets hydrolyzed in the human intestine to form its constituent monosaccharide called glucose and galactose. Galactose
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Chapter 15 Solutions
Fundamentals of Biochemistry: Life at the Molecular Level
Ch. 15 - Prob. 1ECh. 15 - Prob. 2ECh. 15 - 3. The reversible reaction shown here is part of...Ch. 15 - 4. Step 4 of the pentose phosphate pathway...Ch. 15 - 5. The aldolase reaction can proceed in revere as...Ch. 15 - Prob. 6ECh. 15 - 7. Identify the intermediate in the...Ch. 15 - 8. The compound you identified in Problem 7 is a...Ch. 15 - 9. The pyruvate ? lactate reaction in animals is...Ch. 15 - Prob. 10E
Ch. 15 - 11. Why is it possible for the ?G values in Table...Ch. 15 - 12. If a reaction has a ?G0' value of at least...Ch. 15 - 13. Although it is not the primary flux-control...Ch. 15 - 14. What is the advantage of activating pyruvate...Ch. 15 - 15. Tumor cells, which tend to grow rapidly,...Ch. 15 - 16. The pyruvate kinase isozyme in cancerous cells...Ch. 15 - 17. Compare the ATP yield of three glucose...Ch. 15 - 18. If G6P is labeled at its C2 position, where...Ch. 15 - Prob. 19ECh. 15 - Prob. 20ECh. 15 - Prob. 21ECh. 15 - 22. Describe the products of the transketolase...Ch. 15 - Prob. 23CQCh. 15 - 24. The enzyme phosphoglucomutase interconverts...Ch. 15 - 25. You combine 0.2 g of yeast. 0.2 g of sucrose...Ch. 15 - 26. Nerve cells require a source of free energy to...Ch. 15 - Prob. 27CQCh. 15 - Prob. 28CQCh. 15 - 29. Consider the pathway for catabolizing...Ch. 15 - 30. Yeast take up and metabolize galactose, using...Ch. 15 - 31. (a) Describe how glycerol enters the...Ch. 15 - 32. Some organisms can anaerobically convert...Ch. 15 - 33. Explain why some tissues continue to produce...Ch. 15 - 34. Some bacteria catabolize glucose by the...Ch. 15 - Prob. 35CQCh. 15 - 36. For enzymes a–e in Problem 35, identify their...Ch. 15 - Prob. 1MTE
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- Problem 15 of 15 Submit Using the following reaction data points, construct Lineweaver-Burk plots for an enzyme with and without an inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Using the information from this plot, determine the type of inhibitor present. 1 mM-1 1 s mM -1 [S]' V' with 10 μg per 20 54 10 36 20 5 27 2.5 23 1.25 20 Answer: |||arrow_forward12:33 CO Problem 4 of 15 4G 54% Done On the following Lineweaver-Burk -1 plot, identify the by dragging the Km point to the appropriate value. 1/V 40 35- 30- 25 20 15 10- T Км -15 10 -5 0 5 ||| 10 15 №20 25 25 30 1/[S] Г powered by desmosarrow_forward1:30 5G 47% Problem 10 of 15 Submit Using the following reaction data points, construct a Lineweaver-Burk plot for an enzyme with and without a competitive inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. 1 -1 1 mM [S]' s mM¹ with 10 mg pe 20 V' 54 10 36 > ст 5 27 2.5 23 1.25 20 Answer: |||arrow_forward
- Problem 14 of 15 Submit Using the following reaction data points, construct Lineweaver-Burk plots for an enzyme with and without an inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Using the information from this plot, determine the type of inhibitor present. 1 mM-1 1 s mM -1 [S]' V' with 10 μg per 20 54 10 36 20 5 27 2.5 23 1.25 20 Answer: |||arrow_forward12:36 CO Problem 9 of 15 4G. 53% Submit Using the following reaction data points, construct a Lineweaver-Burk plot by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Based on the plot, determine the value of the catalytic efficiency (specificity constant) given that the enzyme concentration in this experiment is 5.0 μ.Μ. 1 [S] ¨‚ μM-1 1 V sμM-1 100.0 0.100 75.0 0.080 50.0 0.060 15.0 0.030 10.0 0.025 5.0 0.020 Answer: ||| O Гarrow_forwardProblem 11 of 15 Submit Using the following reaction data points, construct a Lineweaver-Burk plot for an enzyme with and without a noncompetitive inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. 1 -1 1 mM [S]' 20 V' s mM¹ with 10 μg per 54 10 36 > ст 5 27 2.5 23 1.25 20 Answer: |||arrow_forward
- Problem 13 of 15 Submit Using the following reaction data points, construct Lineweaver-Burk plots for an enzyme with and without an inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Using the information from this plot, determine the type of inhibitor present. 1 mM-1 1 s mM -1 [S]' V' with 10 μg per 20 54 10 36 20 5 27 2.5 23 1.25 20 Answer: |||arrow_forward12:33 CO Problem 8 of 15 4G. 53% Submit Using the following reaction data points, construct a Lineweaver-Burk plot by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Based on the plot, determine the value of kcat given that the enzyme concentration in this experiment is 5.0 μM. 1 [S] , мм -1 1 V₁ s μM 1 100.0 0.100 75.0 0.080 50.0 0.060 15.0 0.030 10.0 0.025 5.0 0.020 Answer: ||| Гarrow_forward1:33 5G. 46% Problem 12 of 15 Submit Using the following reaction data points, construct a Lineweaver-Burk plot for an enzyme with and without an uncompetitive inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. 1 -1 1 mM [S]' 20 V' s mM¹ with 10 μg per 54 10 36 > ст 5 27 2.5 23 1.25 20 Answer: |||arrow_forward
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