Biology: The Dynamic Science (MindTap Course List)
4th Edition
ISBN: 9781305389892
Author: Peter J. Russell, Paul E. Hertz, Beverly McMillan
Publisher: Cengage Learning
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Chapter 14, Problem 2TYK
Summary Introduction
Introduction:
Hershey and Chase carried a series of experiments to show that DNA (deoxyribonucleic acid) is the genetic material. Earlier biologists believed that proteins are the genetic material. In this experiment, a bacteriophage T2 phage was used to infect a bacterium. The phage DNA was labeled with radioactive phosphorus-32 (32P), and the phage protein was labeled with radioactive sulfur-35 (35S). The amount of radioactive isotopes was observed in the infected bacterium.
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If a researcher accidentally transformed the bacterial two hybrid plasmids ( pKT25-X and pUT18-Y) into a normal strain of E coli that produces adenylate cyclase, his/her experiment will:
a. show a high lacZ activity in all tubes (control and experiment) because Proteins X and Y do not need to be interacting in order to activate lacZ gene prescription.
b. not work because cells lack cAMP.
c. need replicates in order to show true interaction between X and Y.
In Hershey-Chase experiment, bacteriophages protein coats were tagged with radioactive isotope S-32. These phages were used to infect E. coli cells and the cells were further centrifuged to form pellets.
Why was the radioactivity level of S-32 found greater outside the cells compared to the E. coli cell pellets? Explain briefly.
If the experiment is repeated in the same manner but this time the phage protein coats are labelled with isotope X and the phage DNA with isotope Y, which isotope’s radioactivity will be found in greater amounts in the E. coli cell pellets after centrifugation? Explain briefly.
What observation led Hershey and Chase to the conclusion that DNA, not the protein coat, carried the genetic information needed to make new virus or phage particles?
Select one:
a. The cell membrane of the bacteria infected with a virus tagged with radioactive phosphorus exhibited radioactivity after it’s initial cell division.
b. Bacterial cells infected with viruses tagged with radioactive phosphorus were radioactive, indicating viral DNA entered the host cell.
c. The cell was vigorously shaken, causing the protein coats to detach from the cell membrane of the bacteria. The protein coats contain genetic information.
d. Bacterial cells infected with viruses tagged with radioactive sulfur were radioactive, indicating viral protein entered the host cell.
Chapter 14 Solutions
Biology: The Dynamic Science (MindTap Course List)
Ch. 14.1 - Prob. 1SBCh. 14.2 - Prob. 1SBCh. 14.2 - Prob. 2SBCh. 14.2 - Prob. 3SBCh. 14.2 - Prob. 4SBCh. 14.3 - What is the importance of complementary base...Ch. 14.3 - Why is a primer needed for DNA replication? How is...Ch. 14.3 - DNA polymerase III and DNA polymerase I are used...Ch. 14.3 - Prob. 4SBCh. 14.4 - Why is a proofreading mechanism important for DNA...
Ch. 14 - Working on the Amazon River, a biologist isolated...Ch. 14 - Prob. 2TYKCh. 14 - Pyrimidines built from a single carbon ring are:...Ch. 14 - Which of the following statements about DNA...Ch. 14 - Which of the following statements about DNA is...Ch. 14 - Prob. 6TYKCh. 14 - Prob. 7TYKCh. 14 - Prob. 8TYKCh. 14 - Prob. 9TYKCh. 14 - Prob. 10TYKCh. 14 - Discuss Concepts Eukaryotic chromosomes can be...Ch. 14 - Prob. 12TYKCh. 14 - Prob. 13TYKCh. 14 - Discuss Concepts During replication, an error...Ch. 14 - Design an Experiment Design an experiment using...Ch. 14 - Prob. 16TYKCh. 14 - Prob. 1ITDCh. 14 - Prob. 2ITD
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- In the Hershey and Chase experiment involving T2 phage,a. most of the 32P entered the bacterial cells whereas most of the 35S did not.b. most of the 35S entered the bacterial cells whereas most of the 32P did not.c. equal amounts of 32P and 35S entered the bacterial cells.d. none of the above was observed.arrow_forwardIn a genetics lab, Kim and Maria infected a samplefrom an E. coli culture with a particular virulent bacteriophage. They noticed that most of the cells werelysed, but a few survived. The survival rate in theirsample was about 1 × 10−4. Kim was sure the bacteriophage induced the resistance in the cells, whileMaria thought that resistant mutants probably alreadyexisted in the sample of cells they used. Earlier, for adifferent experiment, they had spread a dilute suspension of E. coli onto solid medium in a large petri dish,and, after seeing that about 105colonies were growingup, they had replica-plated that plate onto three otherplates. Kim and Maria decide to use these plates totest their theories. They pipette a suspension of thebacteriophage onto each of the three replica plates.What should they see if Kim is right? What shouldthey see if Maria is right?arrow_forwardAlfred Hershey and Martha Chase performed experiments to show that A: the genetic material was DNA contained in the bacteriophage, which entered bacteria during phage infection. B:the genetic material was RNA contained in the bacteriophage, which entered bacteria during phage infection C:the genetic material was protein contained in the bacteriophage, which entered bacteria during phage infection D:none of these choices are correctHomework question i am lost onarrow_forward
- The figure above shows a schematic of genes and transcription control elements from phage λ. Use this figure as an aid to help you describe the molecular events involved in: a) The establishment of lysogeny b) The establishment of a lytic life cyclearrow_forwardBacteria exposed to viruses incorporate sections of the virus’s DNA into the CRISPR array sequences in their genome. This mechanism allows bacteria to fight off the viruses, like an immune response: the information in CRISPR spacers served as “coordinates” for recognizing and cutting up invading DNA sequences. Describe what might happen under the conditions described after a bacteriophage infects a bacterial cell and releases its DNA into the bacterial cell. Explain why: 1. The invading phage DNA is recognized by the Cas proteins but not inserted into the CRISPR array region of the bacterial genome: The bacteria will be unable to elicit an immune response and will succumb to the phase infection 2. The cas genes on the bacterial genome contains a missense mutation that increases its cleavage/cut activityThe bacteria will elicit an immune response that will successfully fight the phage infectionarrow_forwardPhage Mu may be integrated into E. coli genomic DNA via Select one: O a. recombination catalyzed by homologous recombination enzymes. O b. conservative transposition O c. integration catalyzed by integrase. d. replicative transpositionarrow_forward
- A man was brought to the hospital showing "pneumonia"-like symptoms. They took the viral load and determined this to be the new RNA retrovirus infecting lung cells. Since antibodies were unavailable, doctors decided to treat him with drugs. Which of the following would most benefit the patient? A. Drug A is an inhibitor of the binding of 4OS with 60S ribosome of the lung cells. B. Drug B is a structural analogue of the viral peptide that binds to the receptor of the host cell enabling viral entry into the cell. C. Drug C is a polar molecule that inhibits the lytic enzymes activated upon viral infection. D. Drug D is an inducer of a proteolytic enzyme. E. Drug E is a steroid that binds at the regulatory site of a gene causing the synthesis of a repressor that binds at the gene site for the receptor.arrow_forwardWrite: A – if the item is associated with Transcription B – if the item is associated with Translation a.) TATA box b.) Shine-Dalgarno sequence c.) DNA template strand d.) Ribosome binding site e.) Single strand binding proteinsarrow_forwardThe following DNA sequence is from a bacteriophage that infects a pathogenic bacterium and scientists want to know if this bacteriophage could prove to be a potential treatment against it. But first scientists need to discover if different strains of this pathogen have restriction endonucleases that it may use for its own protection. They try 3 different RE’s:a) EcoR1 b) HaeIII c) BamH1 Look up the recognition sequences for the 3 Res. Enzymes above and check whether the phage genome (a snippet of which is shown below) will or will not be ‘cut’. Tell me how their experiment worked out and what their conclusion was.G A A A A G G C C A C A A G G C C G T C G A C T T T T A A A A G G C C A C A T G C G G C T T T T C C G G T G T T C C G G C AG C T GA A A AT T T T C C G G T G T A C G CCarrow_forward
- In the transformation experiment by Griffith, two strains of an organism was injected into a mouse model. One, he labelled S was a heat-killed but pathogenic strain while the other, R was a non- virulent or non-pathogenic strain. Within 48 hours, the mouse died and only the S strain was obtained from the dead mouse. What conclusions can you derived from this experiment? A. Some of the pathogenic S strains survived, proliferated and subsequently overpowered the R strain. B. The R strains mutated and were converted to a pathogenic S strain. C. The R strains assimilated the DNA from the S strain and were transformed into the S strain. D. The R strain DNA hybridized with the S strain DNA such that the daughter DNA of the new generation of R all contained the S DNA. E. All of these conclusions are possible.arrow_forwardRetrotransposons _______. A) are found only in animal cells B) use an RNA molecule as an intermediate in transposition C) generally move by a cut-and-paste mechanismarrow_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_forward
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