Concepts of Genetics (12th Edition)
12th Edition
ISBN: 9780134604718
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino, Darrell Killian
Publisher: PEARSON
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Chapter 6, Problem 20PDQ
Using mutants 2 and 3 from Problem 19, following mixed infection on E. coli B, progeny viruses were plated in a series of dilutions on both E. coli B and K12 with the following results.
- (a) What is the recombination frequency between the two mutants?
- (b) Another mutation, 6, was tested in relation to mutations 1 through 5 from Problems 18–20. In initial testing, mutant 6 complemented mutants 2 and 3. In recombination testing with 1, 4, and 5, mutant 6 yielded recombinants with 1 and 5, but not with 4. What can you conclude about mutation 6?
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Chapter 6 Solutions
Concepts of Genetics (12th 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 - 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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- Austin Taylor and Edward Adelberg isolated some new strains of Hfr cells that they then used to map several genes in E. coli by using interrupted conjugation . In one experiment, they 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 cells were allowed to undergo conjugation. At regular intervals, the researchers removed a sample of cells and interrupted conjugation by killing the Hfr cells with phage T6. The F− cells, which were resistant to phage T6, survivedand were then tested for the presence of genes transferred from the Hfr strain. The results of this experiment are shown in the accompanying graph. On the basis of these data, give the order of the xyl, mtl, mal, and met genes on the bacterial chromosome and indicate the minimum distances between them.arrow_forwardIn an analysis of five rII mutants, complementation testing yielded the following results:arrow_forwardT. 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?arrow_forward
- 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.arrow_forwardT. 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_forwardConsider the five E. coli merodiploid strains listed here. Strain #1 I+P+O+Z-Y+/I+P+OcZ+Y+ Strain #2 I+P+O+Z+Y+/I-P+OcZ+Y- Strain #3 I+P+O+Z-Y+/I-P+OcZ+Y- Strain #4 I-P-O+Z+Y-/I+P+OcZ-Y+ Strain #5: ISP+O+Z+Y+/I-P+O+Z+Y- Which of these strains will be inducible for expression of b-galactosidase? Select all correct answers. A.) Strain #2 B.) Strain #3 C.) Strain #1 D.) Strain #4 E.) None of these F. )Strain #5arrow_forward
- Austin Taylor and Edward Adelberg isolated some new strains of Hfr cells that they then used to map several genes in Escherichia coli by using interrupted conjugation. 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 cells were allowed to undergo conjugation. At regular intervals, the researchers removed a sample of cells and 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 genes transferred from the Hfr strain. The results of this experiment are shown in the graph. 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 represented by the red triangle. The distances between genes are not…arrow_forwardWhen the interrupted mating technique was used with five different strains of Hfr bacteria, the following orders of gene entry and recombination were observed. On the basis of these data, draw a map of the bacterial chromosome. Do the data support the concept of circularity?Hfr Strain Order1 T C H R O2 H R O M B3 M O R H C4 M B A K T5 C T K A Barrow_forwardWhat is the cotransduction coefficient of histidine and valine? For consistency, standardize on histidine. In E. coli, the genes for histidine (his), arginine (arg), tyrosine (tyr), and valine (val) synthesis are closely linked. A wild-type strain is used as a donor for transducing viruses. Various recipient strains were treated with these viruses. Wild-type recombinant colonies were then counted, with the results shown in the following table. Genotype of recipient Number of wild-type colonies Genotype of recipient Number of wild-type colonies his- arg+ val+ tyr+ 254 his+ arg- val- tyr+ 132 his+ arg+ val- tyr- 36 his+ arg- val+ tyr+ 240 his- arg- val+ tyr+ 28 his- arg+ val- tyr+ 102 his- arg+ val+ tyr- 100 his+ arg- val+ tyr- 2 What is the cotransduction coefficient of histidine and valine? For consistency, standardize on histidine. 0.38 0.537 0.402 None of thesearrow_forward
- An Hfr strain that is hisE + and pheA + was conjugated to a strain that is hisE − and pheA −. The conjugation was interrupted at different times, and the percentage of recombinants for each gene was determined by streaking on media that lacked either histidine or phenylalanine. The following results were obtained: A. Determine the map distance (in minutes) between these twogenes.B. In a previous experiment, it was found that hisE is 4 minutesaway from pabB and that PheA is 17 minutes from pabB. Drawa genetic map showing the locations of all three genes.arrow_forwardFour E. coli strains of genotype a+ b- are labeled 1, 2, 3, and 4. Four strains of genotype a- b+ are labeled 5, 6, 7, and 8. The two genotypes are mixed in all possible combinations and (after incubation) are plated to determine the frequency of a+ b+ recombinants. The following results are obtained, where M = many recombinants, L = low numbers of recombinants, and 0 = no recombinants:On the basis of these results, assign a sex type (either Hfr, F+, or F-) to each strain.arrow_forwardThe DNA of a deletion of alpha bacteriophage has a length of 15 micrometers instead of 17 micrometers? How many base pairs are missing from this mutant?arrow_forward
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genetic recombination strategies of bacteria CONJUGATION, TRANSDUCTION AND TRANSFORMATION; Author: Scientist Cindy;https://www.youtube.com/watch?v=_Va8FZJEl9A;License: Standard youtube license