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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Textbook Question
Chapter 6, Problem 26ESP
In a cotransformation experiment, using various combinations of genes two at a time, the following data were produced. Determine which genes are “linked” to which others.
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Following a dye terminator DNA sequencing reaction using 2',3'-dideoxynucleotide triphosphates
(ddNTP's), separation of the primer reaction products was achieved using capillary gel
electrophoresis. In the following example, ddATP was labeled with a 'green' fluorophore, ddTTP was
labeled with 'red', ddCTP was labeled with 'black', and ddGTP was labeled with 'blue.
From left to right (i.e., the shortest to longest retention time), the sequence was:
AACGGTTGTCTCTGATTTGTATTATGTT.
What is the sequence of the template DNA, from its 3' to 5' end?
о ТTIGCCAACAGAGACTAAACATААТАСАА
O TTGTATTATGTTTAGTCTCTGTTGGCAA
О ААСАТААТАСАААТСAGAGAСAАССGTT
O AACGGTTGTCTCTGATTTGTATTATGTT
Four Hfr strains are derived from an F+ strain of E. Coli to serve as donors for an interrupted-mating experiment. Use the time-table and partial map of the F+ strain (shown below) to determine the genes’ respective positions. Keep in mind that the map distances are NOT proportional, only the FIRST 5 markers are indicated per strain, and the entry times, recorded in minutes, are in parentheses. Transferred genes represent wild-type alleles. Based on the data, which gene can be located at position 5 on the map?
The answer bank only has one 9 minute option, therefore 9 minutes can only be used once. Is it possible that there is a mistake in the above solution?
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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- Consider the following experiment. First, large populations of two mutant strains of Escherichia coli are mixed, each requiring a different, single amino acid. After plating them onto a minimal medium, 45 colonies grew. Which of the following may explain this result? A) The colonies may be due to back mutation (reversion). B) The colonies may be due to recombination. C) Either A or B is possible. D) Neither A nor B is possible.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_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_forwardDescribe the outcome of a chain-terminator sequencing procedure in which (a) too little ddNTP is added or (b) too much ddNTP is added.arrow_forwardDNA from a strain of Bacillus subtilis with genotype a+ b+ c+ d+ e+ is used to transform a strain with genotype a− b− c− d− e−. Pairs of genes are checked for cotransformation, and the following results are obtained: Pair of genes Cotransformation Pair of genes Cotransformation a+ and b+ No b+ and d+ No a+ and c+ No b+ and e+ Yes a+ and d+ Yes c+ and d+ No a+ and e+ Yes c+ and e+ Yes b+ and c+ Yes d+ and e+ No On the basis of these results, what is the order of the genes on the bacterial chromosome?arrow_forward
- In a cotransformation experiment (see question 4 of More GeneticTIPS), DNA was isolated from a donor strain that was proA+ andstrC+ and sensitive to tetracycline. (The proA and strC genes conferthe ability to synthesize proline and confer streptomycin resistance,respectively.) A recipient strain is proA− and strC− and isresistant to tetracycline. After transformation, the bacteria werefirst streaked on a medium containing proline, streptomycin, andtetracycline. Colonies were then restreaked on a medium containingstreptomycin and tetracycline. (Note: Each type of medium hadcarbon and nitrogen sources for growth.) The following resultswere obtained:70 colonies grew on the medium containing proline, streptomycin,and tetracycline, but only 2 of these 70 colonies grew whenrestreaked on the medium containing streptomycin and tetracyclinebut lacking proline. If we assume the average size of the DNA fragments is 2 minutes,how far apart are these two genes?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_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_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_forwardGenomic DNA from a family where sickle-cell disease is known to be hereditary, is digested with the restriction enzyme MstII and run in a Southern Blot. The blot is hybridised with two different 0.6 kb probes, both probes (indicated in red in the diagram below) are specific for the β-globin gene (indicated as grey arrow on the diagram below). The normal wild-type βA allele contains an MstII restriction site indicated with the asterisk (*) in the diagram below; in the mutated sickle-cell βS allele this restriction site has been lost. What size bands would you expect to see on the Southern blots using probe 1 and probe 2 for an individual with sickle cell disease (have 2 βS alleles)? Probe 1 Probe 2 (a) 0.6kb 0.6kb and 1.2kb (b) 0.6kb and 1.8kb 0.6kb, 1.2kb and 1.8kb (c) 1.2kb 0.6kb (d) 1.8kb 1.8kb a. (a) b. (b) c. (c) d. (d)arrow_forwardWhen demonstrating horizontal gene transfer, why is it important to use strains that each require at least two different amino acids?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