Concepts of Genetics (11th Edition)
11th Edition
ISBN: 9780321948915
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Textbook Question
Chapter 6, Problem 28ESP
For the experiment in Problem 26, another gene, g, was studied. It demonstrated positive cotransformation when tested with gene f. Predict the results of testing gene g with genes a, b, c, d, and e.
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So if you see growth on a histidine-deficient medium in a yeast 1 hybrid and you assume that it is a false positive. What could have caused this? The yeast cells are normal and this wasn't caused by other molecules.
T. Miyake and M. Demerec examined proline-requiring mutations in the bacterium Salmonella typhimurium (). On the basis of complementation testing, they found four proline auxotrophs: proA, proB, proC, and proD. To determine whether proA, proB, proC, and proD loci were located close together on the bacterial chromosome, they conducted a transduction experiment. Bacterial strains that were proC+ and had mutations at proA, proB, or proD were used as donors. The donors were infected with bacteriophages, and progeny phages were allowed to infect recipient bacteria with genotype proC− proA+ proB+ proD+. The recipient bacteria werethen plated on a selective medium that allowed only proC+ bacteria to grow. After this, the proC+ transductants were plated on selective media to reveal their genotypes at the other three pro loci. The following results were obtained:
Q.Which genotypes represent single transductants and which represent cotransductants?
T. Miyake and M. Demerec examined proline-requiring mutations in the bacterium Salmonella typhimurium (). On the basis of complementation testing, they found four proline auxotrophs: proA, proB, proC, and proD. To determine whether proA, proB, proC, and proD loci were located close together on the bacterial chromosome, they conducted a transduction experiment. Bacterial strains that were proC+ and had mutations at proA, proB, or proD were used as donors. The donors were infected with bacteriophages, and progeny phages were allowed to infect recipient bacteria with genotype proC− proA+ proB+ proD+. The recipient bacteria werethen plated on a selective medium that allowed only proC+ bacteria to grow. After this, the proC+ transductants were plated on selective media to reveal their genotypes at the other three pro loci. The following results were obtained:
Q.Is there evidence that proA, proB, and proD are located close to proC? Explain your answer.
Chapter 6 Solutions
Concepts of Genetics (11th Edition)
Ch. 6 - When the interrupted mating technique was used...Ch. 6 - In a transformation experiment involving a...Ch. 6 - In complementation studies of the rII locus of...Ch. 6 - Prob. 1CSCh. 6 - Prob. 2CSCh. 6 - Prob. 3CSCh. 6 - Prob. 4CSCh. 6 - HOW DO WE KNOW? In this chapter, we have focused...Ch. 6 - Review the Chapter Concepts list on p. 123. Many...Ch. 6 - With respect to F+ and F bacterial matings, answer...
Ch. 6 - List all major differences between (a) the F+ F...Ch. 6 - Describe the basis for chromosome mapping in the...Ch. 6 - In general, when recombination experiments are...Ch. 6 - Why are the recombinants produced from an Hfr F...Ch. 6 - Describe the origin of F bacteria and merozygotes.Ch. 6 - In a transformation experiment, donor DNA was...Ch. 6 - Describe the role of heteroduplex formation during...Ch. 6 - Explain the observations that led Zinder and...Ch. 6 - Prob. 12PDQCh. 6 - Prob. 13PDQCh. 6 - Two theoretical genetic strains of a virus (abc...Ch. 6 - The bacteriophage genome consists of many genes...Ch. 6 - If a single bacteriophage infects one E. coli cell...Ch. 6 - A phage-infected bacterial culture was subjected...Ch. 6 - In recombination studies of the rII locus in phage...Ch. 6 - In an analysis of rII mutants, complementation...Ch. 6 - If further testing of the mutations in Problem 18...Ch. 6 - Using mutants 2 and 3 from Problem 19, following...Ch. 6 - During the analysis of seven rII mutations in...Ch. 6 - In studies of recombination between mutants 1 and...Ch. 6 - Prob. 24ESPCh. 6 - An Hfr strain is used to map three genes in an...Ch. 6 - A plaque assay is performed beginning with 1 mL of...Ch. 6 - In a cotransformation experiment, using various...Ch. 6 - For the experiment in Problem 26, another gene, g,...Ch. 6 - Bacterial conjugation, mediated mainly by...Ch. 6 - A study was conducted in an attempt to determine...Ch. 6 - Prob. 31ESP
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- T. Miyake and M. Demerec examined proline-requiring mutations in the bacterium Salmonella typhimurium (). On the basis of complementation testing, they found four proline auxotrophs: proA, proB, proC, and proD. To determine whether proA, proB, proC, and proD loci were located close together on the bacterial chromosome, they conducted a transduction experiment. Bacterial strains that were proC+ and had mutations at proA, proB, or proD were used as donors. The donors were infected with bacteriophages, and progeny phages were allowed to infect recipient bacteria with genotype proC− proA+ proB+ proD+. The recipient bacteria werethen plated on a selective medium that allowed only proC+ bacteria to grow. After this, the proC+ transductants were plated on selective media to reveal their genotypes at the other three pro loci. The following results were obtained: Q.Why are there no proC− genotypes among the transductants?arrow_forwardCompared to the normal A allele, the disease-causing allele in sickle cell anemia (S allele) is missing an MstII restriction site. On a Southern blot of genomic DNA cut with MstII and hybridized with the probe shown on the diagram below, a person with sickle anemia, carrying two S alleles, will show Choose an answer below: a single band at 1.1 kb. a single band at 1.3 kb. a single band at 0.2 kb. one band at 0.2 and one at 1.3 kb. one band at 1.1 and one at 1.3 kb.arrow_forwardWhich step of the Central Dogma is responsible for transmission of genetic information from generation to generation? Explain.arrow_forward
- In this gene order question, would the order be a/b -- kf -- nm since genes with shorter map unit distances are closely linked?arrow_forwardIf Gene A had a Ct of 23 and your NTC (no template control) had a Ct of 22-what, what would that tell you about Gene A?arrow_forwardA constitutive mutant is a strain that continuously makes aprotein that is inducible in the wild type. Describe two waysin which a change in a DNA molecule could lead to theemergence of a constitutive mutant. How could these twotypes of constitutive mutants be distinguished genetically?arrow_forward
- A bacterial transformation is performed with a donor strain that is resistant to four drugs, A, B, C, and D, and a recipient strain that is sensitive to all four drugs. The resulting recipient cell population is divided and plated onmedia containing various combinations of the drugs. The following table shows the results.a. One of the genes is distant from the other three, which appear to be closely linked. Which is the distant gene? b. What is the likely order of the three closely linked genesarrow_forwardIf you were to engineer some non-adhesive cells to express E-cadherin at either high levels or low levels and then mixed the E-cadherin high cells with the E-cadherin low cells, what do you predict will happen?arrow_forwardConsider the following types of cells: F+, F-, Hfr, and F’ cells. Which of these four types of cells are capable of acting as a donor during conjugation? What genes does each cell that is capable of acting as a donor donate to the recipient cell?arrow_forward
- Given a mutation in the promoter, coding region, and/or non-coding region of a gene, explain how these three scenarios create new alleles of the same gene.arrow_forwardTwo methods for functionally inactivating a gene without altering the gene sequence involve dominant negative alleles and RNA interference (RNAi). Describe how each method can inhibit expression of a gene.arrow_forwardYou can carry out matings between an Hfr and F strain by mixing the two cell types in a small patch on a plate and then replica plating to selective medium. This methodology was used to screen hundreds of different cells for a recombination-deficient recA - mutant. Why is this an assay for RecA function? Would you be screening for a recA mutation in the F or Hfr strain using this protocol? Explain.arrow_forward
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