GENETIC ANALYSIS: INTEGRATED - ACCESS
GENETIC ANALYSIS: INTEGRATED - ACCESS
3rd Edition
ISBN: 9780135349298
Author: Sanders
Publisher: PEARSON
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Chapter 11, Problem 40P

Common baker’s yeast ( Saccharomyces cerevisiae ) is normally grown at 37°C , but it will grow actively at temperatures down to approximately 25°C . A haploidculture of wild - type yeast is mutagenized with EMS.Cells from the mutagenized culture are spread on a complete - medium plate and grown at 25°C . Six colonies ( 1 to 6 ) are selected from the original complete - medium plate and transferred to two fresh complete - medium plates. The new complete plates (shown below) are grown at 25°C and 37°C . Four replica plates are made onto minimal medium or minimal plus adenine from the 25°C complete - medium plate. The new plates are grown at either 25°C and 37°C and the growth results are shown.

Chapter 11, Problem 40P, 11.40 Common baker’s yeast () is normally grown at , but it will grow actively at temperatures down

a. Which colonies are prototrophic and which are auxotrophic? What growth information is used to makethese determinations?

b. Classify the nature of the mutations in colonies 1, 2, and 5 .

c. What can you say about colony 4 ?

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Two auxotrophic triple Escherichia coli strains (A: met- phe- ade- val+ bio+ thr+ and B: met+ phe+ ade+ val- bio- thr-) are mixed in LB liquid medium, diluted and then spread on LB solid rich medium. Six colonies are observed: Then, replicates are performed on 6 different media (minimum medium + glucose + indicated substances). The results are shown below. Determine the genotype of the 6 colonies observed. Which ones are from strain A? From strain B? Which hypotheses can explain these results and which one do you prefer? met phe val bio Abbreviations: met ade val thr phe ade bio thr Met: methionine; Phe: phenylalanine; Ade: adenine; Val: valine; Bio: biotin; Thr: threonine.
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Baker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. After crossing the F1 generation of the cross between mutant strains 1 and 3, you count and determine the phenotypes of 1,000 colonies (here a colony is equivalent to an individual):  563 colonies that can grow on minimal medium alone; 437 colonies that require adenine…

Chapter 11 Solutions

GENETIC ANALYSIS: INTEGRATED - ACCESS

Ch. 11 - 11.11 Two different mutations are identified in a...Ch. 11 - What is the phenotype effect of inserting a Ds...Ch. 11 - 11.13 Answer the following questions concerning...Ch. 11 - Several types of mutation are identified and...Ch. 11 - 11.15 A sample of the bacterium is exposed to...Ch. 11 - 11.16 A strain of is identified as having a null...Ch. 11 - Describe the difference between DNA transposons...Ch. 11 - 11.18 How are flanking direct repeat sequences...Ch. 11 - 11.19 Using the adeninethymine base pair in this...Ch. 11 - The partial amino acid sequence of a wild-type...Ch. 11 - Prob. 21PCh. 11 - 11.22 Many human genes are known to have homologs...Ch. 11 - The fluctuation test performed by Luria and...Ch. 11 - In this chapter, three features of genes or of DNA...Ch. 11 - Briefly compare the production of DNA double -...Ch. 11 - During mismatch repair, why is it necessary to...Ch. 11 - 11.27 Following the spill of a mixture of...Ch. 11 - 11.28 In an Ames test using Salmonella bacteria a...Ch. 11 - A wild - type culture of haploid yeast is exposed...Ch. 11 - A fragment of a wild - type polypeptide is...Ch. 11 - Prob. 31PCh. 11 - Alkaptonuria is a human autosomal recessive...Ch. 11 - 11.33 In an experiment employing the methods of...Ch. 11 - Using your knowledge of DNA repair pathways choose...Ch. 11 - 11.35 Ataxia telangiectasia is a human inherited...Ch. 11 - A geneticist searching for mutations uses the...Ch. 11 - 11.37 In a mousebreeding experiment a new mutation...Ch. 11 - 11.38 Considering the Dumbo mutation in a Problem,...Ch. 11 - 11.39 Thinking back to the discussion of...Ch. 11 - 11.40 Common baker’s yeast () is normally grown at...Ch. 11 - 11.41 The two gels illustrated below contain...
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