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 7, Problem 26P
a. You have a test tube containing 5 ml of a solution of bacteriophages, and you would like to estimate the number of bacteriophages in the tube. Assuming the tube actually contains a total of 15 billion bacteriophages, design a serial dilution experiment that would allow you to estimate this number. Ideally, the final plaque-containing plates you count should contain more than 10 and fewer than 1000 plaques.
b. When you count bacteriophages by the serial dilution method as in part (a), you are assuming a plating efficiency of 100%; that is, the number of plaques on the petri plate represents exactly the number of bacteriophages you mixed with the plating bacteria. Is there any way to test the possibility that only a certain percentage of bacteriophage particles can form plaques (so that the plating efficiency would be less than 100%)? Conversely why is it fair to assume that any plaques are initiated by one rather than multiple bacteriophage particles?
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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 7 Solutions

Genetics: From Genes to Genomes, 5th edition

Ch. 7 - Remember that balancer chromosomes prevent the...Ch. 7 - In the experiment shown in Fig 7.13 on p. 218, H....Ch. 7 - Figure 7.14 on pp. 219-220 shows examples of base...Ch. 7 - So-called two-way mutagens can induce both a...Ch. 7 - In 1967, J. B. Jenkins treated wild-type male...Ch. 7 - Aflatoxin B1 is a highly mutagenic and...Ch. 7 - When a particular mutagen identified by the Ames...Ch. 7 - Prob. 18PCh. 7 - The Ames test uses the reversion rate his-/i to...Ch. 7 - In human DNA, 70 of cytosine residues that are...Ch. 7 - Albinism in animals is caused by recessive...Ch. 7 - Imagine that you caught a female albino mouse in...Ch. 7 - Plant breeders studying genes influencing leaf...Ch. 7 - In humans, albinism is normally inherited in an...Ch. 7 - Prob. 25PCh. 7 - a. You have a test tube containing 5 ml of a...Ch. 7 - Prob. 27PCh. 7 - The rosy ry gene of Drosophila encodes an enzyme...Ch. 7 - Prob. 29PCh. 7 - In a haploid yeast strain, eight recessive...Ch. 7 - The pathway for arginine biosynthesis in...Ch. 7 - In corn snakes, the wild-type color is brown. One...Ch. 7 - In a certain species of flowering plants with a...Ch. 7 - Prob. 34PCh. 7 - In each of the following cross schemes, two...Ch. 7 - Prob. 36PCh. 7 - The following complementing E. coli mutants were...Ch. 7 - In 1952, an article in the British Medical Journal...Ch. 7 - Mutations in an autosomal gene in humans cause a...Ch. 7 - Antibodies were made that recognize six proteins...Ch. 7 - Prob. 41PCh. 7 - Prob. 42PCh. 7 - In addition to the predominant adult hemoglobin,...Ch. 7 - Prob. 44PCh. 7 - Humans are normally trichromats; we have three...
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