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b. 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 individual’s 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 individual’s 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. Denote the point in the pathway in which each individual is blocked.

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Human Heredity: Principles and Iss...

11th Edition
Michael Cummings
Publisher: Cengage Learning
ISBN: 9781305251052

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BuyFindarrow_forward

Human Heredity: Principles and Iss...

11th Edition
Michael Cummings
Publisher: Cengage Learning
ISBN: 9781305251052
Chapter 10, Problem 10QP
Textbook Problem
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  1. b. 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 individual’s 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 individual’s 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.

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

Denote the point in the pathway in which each individual is blocked.

Summary Introduction

To determine: The block points in the pathway for synthesis of protein E where each individual is blocked.

Introduction: Production of proteins is controlled by the genetic code for that particular protein, present in the genome of that organism. Different metabolic pathways are controlled by different proteins. Synthesis of these metabolic pathway controlling proteins can be interrupted by many different reasons that are present in the pathway.

Explanation of Solution

To determine the block points, the complete pathway for the synthesis of protein needs to be determined. This pathway can be determined as given below:

To find the pathway for the production of protein E, some assumptions are made as follows:

Z is the initial substance that is converted into protein E by the enzymatic activity of different enzymes.

Enzyme 1, enzyme 2, enzyme 3, enzyme 4, and enzyme 5 are used in the conversion of compound Z to A, A to B, B to C, C to D, and D to E respectively.

It can be seen that in individual 1 and 6, individual 2 and 7, and individual 3 and 8 shows a similar pattern of production of protein E even after additions of the intermediates.

In individual 4, the addition of any of the compounds did not form protein E. Hence, it can be said that individual 4 has null mutation for all the enzymes involved in the pathway.

In individual 8, the addition of compound D only forms protein E. This shows that individual 8 has only enzyme 5 as functional and all other enzymes are nonfunctional for the production of protein E. It also shows that compound D and enzyme 5 is necessary for the production of protein E. Hence, this necessary step can be shown as:

DEnzyme5E

In individual 1 (and individual 6), addition of intermediate A and B did not form protein E, but addition of intermediate C and D formed protein E. This shows that individual 1 and 6 have faulty enzyme 1, enzyme 2, and enzyme 3 but they have properly working enzyme 4, and enzyme 5. Based on the pathway obtained in individual 8 and conversion pathway by these properly working enzymes in individual 1, it can be shown as:

CEnzyme4DEnzyme5E

In individual 2 (and individual 7), it can be seen that the addition of compound A did not form protein E but addition of compound B, C, and D led to formation of protein E. This shows that individual 2 and 7 have a faulty enzyme 2 and properly working enzyme 2, enzyme 3, enzyme 4, and enzyme 5

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