Concepts of Genetics Plus Mastering Genetics with Pearson eText -- Access Card Package (12th Edition) (What's New in Genetics)
12th Edition
ISBN: 9780134811390
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino, Darrell Killian
Publisher: PEARSON
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Chapter 6, Problem 21ESP
During the analysis of seven rII mutations in phage T4, mutants 1, 2, and 6 were in cistron A, while mutants 3, 4, and 5 were in cistron B. Of these, mutant 4 was a deletion overlapping mutant 5. The remainder were point mutations. Nothing was known about mutant 7. Predict the results of complementation (+ or −) between 1 and 2; 1 and 3; 2 and 4; and 4 and 5.
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Complementation tests of recessive mutants a through f produced the data shown in the table below. "+" means wild-type phenotype; "-" means mutant phenotype.
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A new mutant, g, is isolated and it fails to complement c. Which other mutant(s) would g also fail to complement? (Select all correct answers.)
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In recombination studies of the rII locus in phage T4, what is the significance of the value determined by calculating phage growth in the K12 versus the B strains of E. coli following simultaneous infection in E. coli B? Which value is always greater?
In E. coli, the gene bioD+ encodes an enzyme involved in biotin synthesis, and galK+ encodes an enzyme involved in galactose utilization. An E. coli strain that contained wild-type versions of both genes was infected with P1 phage, and then a P1 lysate was obtained. This lysate was used totransduce (infect) a strain that was bioD− and galK−. The cellswere plated on a medium containing galactose as the sole carbonsource for growth to select for transduction of the galK+ gene.This medium also was supplemented with biotin. The resultingcolonies were then restreaked on a medium that lacked biotin tosee if the bioD+ gene had been cotransduced. The following resultswere obtained:What topic in genetics does this question address?
Chapter 6 Solutions
Concepts of Genetics Plus Mastering Genetics with Pearson eText -- Access Card Package (12th Edition) (What's New in Genetics)
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 - A 4-month-old infant had been running a moderate...Ch. 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 - 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. 23ESPCh. 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...
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- E. coli cells are simultaneously infected with two strains of phage λ. One strain has a mutant host range, is temperature sensitive, and produces clear plaques (genotype h st c); another strain carries the wildtype alleles (genotype h+ st+ c+). Progeny phages are collected from the lysed cells and are plated on bacteria. The following numbers of different progeny phages are obtained: Progeny phage genotype Number of plaques h+ c+ st+ 321 h c st 338 h+ c st 26 h c+ st+ 30 h+ c st+ 106 h c+ st 110 h+ c+ st 5 h c st+ 6 a. Determine the order of the three genes on the phage chromosome. b. Determine the map distances between the genes. c. Determine the coefficient of coincidence and the interferencearrow_forwardTwo mutations that affect plaque morphology in phages (a and b ) have been isolated. Phages carrying both mutations (a b) are mixed with wild-type phages (a* b*) and added to a culture of bacterial cells. Once the phages have infected and lysed the bacteria, samples of the phage lysate are collected and cultured on plated bacteria. The following numbers of plaques are observed: Plaque phenotype Number a* b* 2043 a* b- 320 a b* 357 2134 What is the frequency of recombination between the a and b genes?arrow_forwardIn E. coli, the gene bioD+ encodes an enzyme involved in biotin synthesis, and galK+ encodes an enzyme involved in galactose utilization. An E. coli strain that contained wild-type versions of both genes was infected with P1 phage, and then a P1 lysate was obtained. This lysate was used totransduce (infect) a strain that was bioD− and galK−. The cellswere plated on a medium containing galactose as the sole carbonsource for growth to select for transduction of the galK+ gene.This medium also was supplemented with biotin. The resultingcolonies were then restreaked on a medium that lacked biotin tosee if the bioD+ gene had been cotransduced. The following resultswere obtained:What information do you know based onthe question and your understanding of the topic?arrow_forward
- The following recombinants are recovered when conjugation occurs between an a*d*g+ donor and an adg recipient. at d+ g+ = 84% a d g+ = 6% at d g+ = 10% a dt g+ = less than 1% What is the map distance between the a and d genes? 10 map units 74 map units less than 1 map unit 84 map units 6 map unitsarrow_forwardIn an analysis of rII mutants, complementation testing yielded thefollowing results:Mutants Results (+/- lysis)1, 2 +1, 3 +1, 4 -1, 5 -Predict the results of testing 2 and 3, 2 and 4, and 3 and 4 together.arrow_forwardTime mapping is performed in a cross involving the genes his, leu, mal, and xyl. The recipient cells were auxotrophic for all four genes. After 25 minutes, mating was interrupted with the following results in recipient cells. Diagram the positions of these genes relative to the origin (O) of the F factor and to one another. (a) 90% were xyl+ (b) 80% were mal+ (c) 20% were his+ (d) none were leu+arrow_forward
- Shown below are the complementation test results involving 4 independently isolated lethal mutants in a bacteriophage. Complementation was assayed by simultaneouly infecting bacteria with two phage strains, each with a different mutation, neither of which could alone lyse the cells. In the table below, a "+" indicates the strains complemented each other and therefore lysed open the bacteria. A "0" indicates no complementation and therefore no cell lysis occurred. Test pair Results 1___2___3___4 1,2 + 1 0 + + 0 1,3 + 2 0 + + 1,4 0 3 0 + 2,3 + 4 0 2,4 + 3,4 + How many genes are there? a. 3 b.1 c. 2 d. 4arrow_forwardTwo mutations that affect plaque morphology in phages (a− and b −) have been isolated. Phages carrying both mutations (a− b−) are mixed with wild-type phages (a+ b+) and added to a culture of bacterial cells. Once the phages have infected and lysed the bacteria, samples of the phage lysate are collected and cultured on plated bacteria. The following numbers of plaques are observed: Plaque phenotype Number a+ b+ 2043 a+ b− 320 a− b+ 357 a− b− 2134 What is the frequency of recombination between the a and b genes?arrow_forwardNine rII− mutants of bacteriophage T4 were used inpairwise infections of E. coli K(λ) hosts. Six of themutations in these phages are point mutations; theother three are deletions. The ability of the doubly infected cells to produce progeny phages in large numbers is scored in the following chart.1 2 3 4 5 6 7 8 91 − − + + − − − + +2 − + + − − − + +3 − − + − + − −4 − + − + − −5 − − − + +6 − − − −7 − + +8 − −9 −The same nine mutants were then used in pairwise infections of E. coli B hosts. The production of progenyphages that can subsequently lyse E. coli K(λ) hosts isnow scored. In the table, 0 means the progeny do notproduce any plaques on E. coli K(λ) cells; − meansthat only a very few progeny phages produce plaques;and + means that many progeny produce plaques(more than 10 times as many as in the − cases).1 2 3 4 5 6 7 8 91 − + + + + − − + +2 − + + + + − + +3 0 − + 0 + + −4 − + − + + +5 − + − + +6 0 0 − +7 0 + +8 − +9 −a. Which of the mutants are the three deletions? Whatcriteria did…arrow_forward
- By conducting conjugation experiments between Hfr and recipientstrains, Wollman and Jacob mapped the order of many bacterialgenes. Throughout the course of their studies, they identified severaldifferent Hfr strains in which the F-factor DNA had been integratedat different places along the bacterial chromosome. A sample of theirexperimental results is shown in the following table:Analyze data. Compare and contrast. Make a drawing.arrow_forwardHow does homologous recombination with transfected disruption constructs can inactivate specific target genes in yeast? Explain this with the help of figure.arrow_forwardAustin Taylor and Edward Adelberg isolated some new strains of Hfr cells that they then used to map several 200 mal+ genes in Escherichia coli by using interrupted conjugation. 150 In one experiment, the researchers mixed cells of Hfr strain AB-312, which were xyl* mtl* mal* met* and sensitive to phage T6, with F strain AB-531, which was xyl mtl mal met and resistant to phage T6. The 100 mt/+ cells were allowed to undergo conjugation. At regular intervals, the researchers removed a sample of cells and 50 met+ interrupted conjugation by killing the Hfr cells with phage T6. The F cells, which were resistant to phage T6, survived and were then tested for the presence of 0. 20 40 60 80 100 genes transferred from the Hfr strain. The results of this experiment are shown in the graph. Time of sampling (minutes) On the basis of these data, give the order of the xyl, mtl, mal, and met genes on the bacterial chromosome and the minimum distances between them in minutes. The origin of transfer is…arrow_forward
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Mitochondrial mutations; Author: Useful Genetics;https://www.youtube.com/watch?v=GvgXe-3RJeU;License: CC-BY