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 3NST
In complementation studies of the rII locus of phage T4, three groups of three different mutations were tested. For each group, only two combinations were tested. On the basis of each set of data (shown here), predict the results of the third experiment for each group.
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Bacteriophage P22 was used in generalised transduction experiments to infect the Salmonella
typhimurium donor strains described in the table below.
The resulting phage lysates were then used to infect the recipient strains of S. typhimurium
recipient strains listed in the table.
In each cross, a phenotype was selected for one of the selected for one of the three genetic
markers studied (str, aceA, thrA), and were made to select the recombinants corresponding to the
other two markers. markers.
The results are given in the following table:
Strain
I donor
str thrA
aceA thrA
str aceA+
Strain
recipient
strs thrA+
aceA thrA
str aceA
Phenotype
selected
Str
Ace+
Str
recombinants
selected
ThrA
ThrA
ThrA
ThrA
Ace
Ace
Number
60
40
95
5
10
90
str: gene involved in streptomycin resistance,
aceA: gene involved in the use of acetate as a carbon source,
thrA: gene involved in threonine biosynthesis.
1) What are the selective media used in these three transduction experiments? to obtain the
selected…
During the analysis of seven rII mutations in phage T4, mutants1, 2, and 6 were in cistron A, while mutants 3, 4, and 5 were incistron B. Of these, mutant 4 was a deletion overlapping mutant5. The remainder were point mutations. Nothing was knownabout mutant 7. Predict the results of complementation (+ or -)between 1 and 2; 1 and 3; 2 and 4; and 4 and 5.
In a generalized-transduction experiment, phages arecollected from an E. coli donor strain of genotype cys+leu+ thr+ and used to transduce a recipient of genotypecys- leu- thr-. Initially, the treated recipient populationis plated on a minimal medium supplemented with leucine and threonine. Many colonies are obtained.a. What are the possible genotypes of these colonies?b. These colonies are then replica plated onto threedifferent media: (1) minimal plus threonine only, (2)minimal plus leucine only, and (3) minimal. Whatgenotypes could, in theory, grow on these three media?c. Of the original colonies, 56 percent are observed togrow on medium 1, 5 percent on medium 2, and nocolonies on medium 3. What are the actual genotypes ofthe colonies on media 1, 2, and 3?d. Draw a map showing the order of the three genes andwhich of the two outer genes is closer to the middle gene
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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- a. Bacteriophage P22 was used in generalized transduction experiments to infect the Salmonella typhimurium donor strains described in the table below. The resulting phage lysates were then used to infect the S. typhimurium recipient strains listed in the table. In each cross, a phenotype was selected for one of the three genetic markers studied (str, aceA, thrA), and then replicates were performed to select the corresponding recombinants for the other two markers. The results are given in the following table: Recipient strain Selected phenotype Selected recombinants Donor strain str thrA aceAt thrA str aceA+ strs thrA+ aceA thrA+ str aceA Str Ace+ Str ThrA ThrA+ ThrA ThrA+ Ace Ace+ str: gene involved in streptomycin resistance, aceA: gene involved in the use of acetate as a carbon source, thrA: gene involved in the biosynthesis of threonine. Number 60 40 95 5 10 90 What are the selective media used in these three transduction experiments, on the one hand to obtain the selected…arrow_forwardBacteriophage P22 was used in generalized transduction experiments to infect the Salmonella typhimurium donor strains described in the table below. The resulting phage lysates were then used to infect the S. typhimurium recipient strains listed in the table. In each cross, a phenotype was selected for one of the three genetic markers studied (str, aceA, thrA), and then replicates were performed to select the corresponding recombinants for the other two markers. The results are given in the following table: Recipient strain Selected phenotype Selected recombinants Donor strain str thrA aceA+ thrA str aceA+ strs thrA+ aceA thrA+ str aceA Str Ace+ Str ThrA ThrA+ ThrA ThrA+ Ace Ace str: gene involved in streptomycin resistance, aceA gene involved in the use of acetate as a carbon source, thrA: gene involved in the biosynthesis of threonine. Number 60 40 95 5 10 90 Determine the order of the genes and draw a genetic map showing this orderarrow_forwardShown 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_forward
- Complementation tests of recessive mutants a through f produced the data shown in the table below. "+" means wild-type phenotype; "-" means mutant phenotype. PSCO a b с 0 0 0 0 0 a e a f b + C + d e + + + + + - d e f 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.) ++ f ++ + + -arrow_forwardBy 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_forwardIn an analysis of five rII mutants, complementation testing yielded the following results:arrow_forward
- In 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_forwardE. 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_forwardAfter isolating six Neurospora mutations, you mate A and a representatives of each in all pairwise combinations and determine which mutations complement. Table 1gives the results, with a (+) indicating complementation, and a (-) representing lack of complementation. With the exception of mutant 1, how many different genes are represented in this experiment, or can you tell from these experiments?arrow_forward
- A fluctuation test was carried out to determine the rate of mutation to Azetidine resistance (a toxic proline analog) in S. typhimurium. Twenty tubes of rich medium were each inoculated with a few wild-type cells and the cultures were grown to 10° cells / ml. A 0.1 ml sample of each culture was then plated on each plate (a total of 20 plates) to detect AztR mutants. The results are shown in the following table. Calculate the mutation rate of S. typhimurium to AztR. Culture # # AztR mutants Culture # # AztR mutants 11 12 3 4. 13 14 15 4. 30 303 97 69 14 16 17 18 19 20 10 19arrow_forwardIn 1944, Avery, Macleod, and McCarty provided strong evidence that DNA is the hereditary material in Streptococcus pneumoniae by Group of answer choices showing that avirulent cells could become virulent by the process of transduction none of these is true. showing that virulent cells could become avirulent if the DNA was destroyed after transformation showing that avirulent cells could not gain the ability to become virulent cells if conjugation was interrupted. showing that avirulent cells could not gain the ability to become virulent if DNA was destroyed after transformation.arrow_forwardantibodies were produced in a rabbit against a 25 KDa human soluble protien. when the rabbit antiserum was used in a western blot of human soluble protiens, it detected 25-KDa protien,but also bound to protiens of 65KDa and 39 KDa.assuming that a pure protien was used to make the antibodies,how do yuo explain the results of the western blot?arrow_forward
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