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In a genetics lab, Kim and Maria infected a sample from an E. coli culture with a particular virulent bacteriophage. They noticed that most of the cells were lysed, but a few survived. The survival rate in their sample was about 1 × 10-4. Kim was sure the bacteriophage induced the resistance in the cells, while Maria thought that resistant mutants probably already existed in the sample of cells they used. Earlier, for a different
experiment, they had spread a dilute suspension of E. coli onto solid medium in a large petri dish, and, after seeing that about 105 colonies were growing up, they had replica-plated that plate onto three other plates. Kim and Maria decide to use these plates to test their theories. They pipette a suspension of the bacteriophage onto each of the three replica plates. What should they see if Kim is right? What should they see if Maria is right?
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- The Pfizer Covid-19 vaccine is one of the vaccines currently being rolled out for mass vaccination in South Africa to protect the population against SARS CoV-2, and the active compound in this vaccine is an mRNA molecule. a) The starting material for productio of this vaccine is a copy of the coronavirus spike protein gene cloned into a ppasmid. List the main experimental steps in the vaccine production process, to go from a plamid to an mRNA molecule. b) Which molecular processes in the cell will allow an injected strand of mRNA to produce anti-Covid immunity in the vaccine recipient?arrow_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_forwardWhen various strains of λ phage are seeded on a lawn of E. coli, they can form clear or turbid plaques. (b) For mutant λ phages that can only form clear plaques, give two different types of mutation in the phage that can explain the clear plaque phenotype.arrow_forward
- To understand the genes responsible for growth and infectivity in a disease-causing bacterial strain, you perform chemical mutagenesis on a culture of these bacteria. In the course of your investigation into the properties of the resulting mutants, you identify a set of mutant bacteria that is still viable but their virulence is significantly impaired. How might these mutants be useful for vaccine development?arrow_forwardOne of the reasons why phage therapy has not been applied widely is that bacteria can become resistant to bacteriophages as well, through mutations in genes encoding for specific proteins. What would be a protein in the bacterial cell that, if mutated, would make that cell resistant to phage infection?arrow_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_forward
- Austin 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_forwardEight mutant bacteriophage strains cannot lyse a certain type of bacteria that can be lysed by wild-type bacteriophages. The mutant strains were allowed to infect the bacteria in a complementation test. A "+" indicates that lysis occurred with coinfection. A "-" indicates that lysis did not occur. GKWTMAQC G- ++++- K W T M A Q C - + +++ - ++++ - + + - + - + - - + + + A cistron is defined by no complementation in the How many genes are controlling lysis in this bacteriophage? (Use a number not a word in the space) configuration.arrow_forwardAustin 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_forward
- In order to determine the genetic material of a T2 phage, Alfred Hershey and Martha Chase conducted experiments using T2 phages that infected bacteria. In one treatment, they grew phages with radioactive sulfur. In another treatment, they grew phages with radioactive phosphorous. They allowed both types of phages to infect bacterial cells. After infection, they found that only bacteria infected with phages grown with radioactive phosphorous showed any radioactivity. Why did they use radioactive sulfur and phosphorous for this experiment? * O Sulfur is part of the DNA molecule but not part of a protein molecule. Sulfur and phosphorous are some of the most reactive molecules and are easily traced. Sulfur and phosphorous are able to survive the centrifuge, a crucial component of the experiment. O Phosphorous is part of the DNA molecule but not part of a protein molecule.arrow_forwardBacteriophage 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…arrow_forwardInterrupted conjugation was used to map three genes in E. coli. The donor genes first appeared in the recipient cells at the following times: gal, 10 minutes; his, 8 minutes; pro, 15 minutes. Which gene is in the middle?arrow_forward
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