Genetics: From Genes to Genomes, 5th edition
Genetics: From Genes to Genomes, 5th edition
5th Edition
ISBN: 9780073525310
Author: Leland H. Hartwell, Michael L. Goldberg, Janice A. Fischer, Leroy Hood, Charles F. Aquadro
Publisher: McGraw-Hill Education
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Chapter 8, Problem 13P
Summary Introduction

To describe:

The explanation that helps in understanding the reason that inserted amino acids do not correspond to normal protein sequence.

Introduction:

Amino acids are vital molecules that produce proteins. The process by which mRNA is transformed into amino acids is known as translation. The amino acids join together with the help of polypeptide bonds to form proteins. A scientist observed that a natural enzyme is composed of 227 amino acids. However, the mutant form of this enzyme contains 312 amino acids. This indicates that the mutant form has 85 extra amino acids.

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After you feel comfortable with your counting method and identifying cells in the various stages of mitosis, use the four images below of whitefish blastula to count the cells in each stage until you reach 100 total cells, recording your data below in Data Table 1. (You may not need to use all four images. Stop counting when you reach 100 total cells.) After totaling the cells in each stage, calculate the percent of cells in each stage. (Divide total of stage by overall total of 100 and then multiply by 100 to obtain percentage.)   Data Table 1Stage    Totals    PercentInterphase        Mitosis:        Prophase        Metaphase        Anaphase        Telophase        Cytokinesis        Totals    100    100% To find the length of time whitefish blastula cells spend in each stage, multiply the percent (recorded as a decimal, in other words take the percent number and divide by 100) by 24 hours. (Example: If percent is 20%, then Time in Hours = .2 * 24 = 4.8) Record your data in Data…

Chapter 8 Solutions

Genetics: From Genes to Genomes, 5th edition

Ch. 8 - Before the technology existed to synthesize RNA...Ch. 8 - A particular protein has the amino acid sequence...Ch. 8 - Prob. 13PCh. 8 - How many possible open reading frames frames...Ch. 8 - a. In Fig. 8.3 on p. 257, the physical map the...Ch. 8 - Charles Yanofsky isolated many different trpA-...Ch. 8 - Remembering that the wobble base of the tRNA is...Ch. 8 - The sequence of a segment of mRNA, beginning with...Ch. 8 - You identify a proflavin-generated allele of a...Ch. 8 - In certain bacterial species, pyrrolysine Pyl,...Ch. 8 - Describe the steps in transcription that require...Ch. 8 - Chapters 6 and 7 explained that mistakes made by...Ch. 8 - The coding sequence for gene F is read from left...Ch. 8 - If you mixed the mRNA of a human gene with the...Ch. 8 - Describe the steps in translation that require...Ch. 8 - Locate as accurately as possible the listed items...Ch. 8 - Concerning the figure for Problem 26: a. Which...Ch. 8 - In prokaryotes, a search for genes in a DNA...Ch. 8 - The yeast gene encoding a protein found in the...Ch. 8 - The sequence of a complete eukaryotic gene...Ch. 8 - Using recombinant DNA techniques which will be...Ch. 8 - a. The genetic code table shown in Fig. 8.2 on p....Ch. 8 - Arrange the following list of eukaryotic gene...Ch. 8 - Concerning the list of eukaryotic gene elements in...Ch. 8 - The human gene for 2 lens crystallin has the...Ch. 8 - Do you think each of the following types of...Ch. 8 - Null mutations are valuable genetic resources...Ch. 8 - The following is a list of mutations that have...Ch. 8 - Considering further the mutations described in...Ch. 8 - When 1 million cells of a culture of haploid yeast...Ch. 8 - Why is a nonsense suppressor tRNATyr, even though...Ch. 8 - Prob. 42PCh. 8 - You are studying mutations in a bacterial gene...Ch. 8 - Another class of suppressor mutations, not...Ch. 8 - Yet another class of suppressor mutations not...Ch. 8 - There is at least one nonsense suppressing tRNA is...Ch. 8 - Prob. 47PCh. 8 - Brenners m mutant phages m1 m6 described in Fig....
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