After solving the crystal structure of a novel enzyme and looking at the active site, a researcher makes the following point mutations to test which residues are involved in the catalysis, binding, or both. The total enzyme concentration is 4 uM in each test. Here are the results: Wild Type (unmutated enzyme) His64 Ala64 Arg12- Ala12 Asp121- Alat4 18015 3500 50 5501 10 8015 18117 3465 45 1000 20 35501 30 Calculate the catalytic efficiency for the wild type enzyme.

After solving the crystal structure of a novel enzyme and looking at the active site, a researcher makes the following point mutations to test which residues are involved in the catalysis, binding, or both. The total enzyme concentration is 4 uM in each test. Here are the results: Wild Type (unmutated enzyme) His64 Ala64 Arg12- Ala12 Asp121- Alat4 18015 3500 50 5501 10 8015 18117 3465 45 1000 20 35501 30 Calculate the catalytic efficiency for the wild type enzyme.

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 10QP: b. Compounds A, B, C, and D are known to be intermediates in the pathway for production of protein...

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a. Compounds A, B, C, and D are known to be intermediates in the pathway for production of protein E. To determine where the block in protein-E production occurred in each individual, the various intermediates were given to each individuals cel Is in culture. After a few weeks of growth with the intermediate, the cells were assayed for the production of protein E. The results for each individuals cells are given in the following table. A plus sign means that protein E was produced after the cells were given the intermediate listed at the top of the column. A minus sign means that the cells still could not produce protein E even after being exposed to the intermediate at the top of the column. Draw the pathway leading to the production of protein E.
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