100 60 40 // 24 14 6 8 10 12 Time (hours) Key 1 = V. melanops, 2 = V. sativa, 3 = V. benghalensis, 4 = V. atropurpurea, 5 = V. faba, 6 = V. narbonensis, 7 = E. coli Percentage of labeled DNA not hybridized 20 2. [After W. V. Choi, 1971, Genetics 68:213-230.]
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In DNA-hybridization experiments on six species of plants in the genus Vicia, DNA was isolated from each of the six species, denatured by heating, and sheared into small fragments (W. Y. Chooi. 1971. Genetics 68:213–230). In one experiment, DNA from each species and from E. coli was allowed to renature. The graph shows the results of this renaturation experiment.
Q. Notice that, for the Vicia species, the rate of renaturation is much faster in the first hour and then slows down. What might cause this initial rapid renaturation and the subsequent slowdown?
![100
60
40
//
24
14
6 8
10
12
Time (hours)
Key
1 = V. melanops, 2 = V. sativa, 3 = V. benghalensis, 4 = V.
atropurpurea, 5 = V. faba, 6 = V. narbonensis, 7 = E. coli
Percentage of labeled DNA
not hybridized
20
2.
[After W. V. Choi, 1971, Genetics 68:213-230.]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa7d50630-534f-4ef5-b722-b17cf01b1378%2Fbad341a1-abd3-4c49-9130-4b5c05cce8fe%2Ff6nohmq.jpeg&w=3840&q=75)
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- What would be the expected distribution of centrifuged DNA (in the Meselson /Stahl experiment), using only the intact chromosomes of E. coli, by the second generation (the F2 generation)? 12.5% heavy-weight DNA, and 87.5% intermediate-weight DNA 25% heavy-weight DNA, and 75% light-weight DNA 87.5% light-weight DNA, and 12.5% intermediate-weight DNA 75% heavy-weight DNA, and 25% light-weight DNA 50% intermediate-weight DNA, and 50% light-weight DNA The above experiment would demonstrate which of the following DNA replication mechanisms? completely discontinuous DNA replication completely continuous DNA replication semi-conservative DNA replication completely dispersive DNA replication completely conservative DNA replicationA newly-identified relative of Staphylococcus is causing great concern at the Center for Disease Control. It has been observed that some strains of this new bacterial species are resistant to multiple antibiotics used in clinical settings. Thanks to your bacterial genetic prowess, you are hired as part of a research team whose mission is to characterize this new species. It appears that at least three different antibiotic resistance genes have integrated onto the bacterial chromosome: min+ = confers resistance to minocycline cep+ = confers resistance to cephalexin van+ = confers resistance to vancomycin The following table shows the results of your experiments. Transformation of the recipient cells was conducted, using the DNA of the donor cells. You next selected for recipient cells that had undergone transformation by plating the transformed cells on media containing minocycline. This enabled you to select for min+ cells. The min+ cells were then replica plated onto media…Explain how PCR/OLA (polymerase chain reaction/oligonucleotide ligation assay) can be used in the diagnosis of sickle cell disorder . Would you recommend this method for routine diagnosis of sickle cell disorder? Explain
- Sam is working in a genetics lab this summer to gain more hands-on experience before applying to graduate school. The professor with whom he is working is particularly interested in studying a new strain of E. coli and assigns Sam an interrupted mating experiment to perform. Sam is attempting to determine the order of 7 different genes on the bacterial chromosome: Ger, Fra, Lod, Moc, Bla, Jat, and Arp. Sam used four different Hfr strains (Hfr1, Hfr2, Hfr3, Hfr4) to help him determine the gene order. He observed that each Hfr strains transferred genes at the times shown in the table below. However, Sam is a little forgetful and neglected to record the times at which several of the genes were transferred. Afraid that the professor will become angry with him, Sam comes to you for assistance. Using the data from the experiment; fill in the missing data in the table attached and draw a single circular map of the seven genes as they would appear on the E. coli chromosome. Be sure to…Using bacteriophage P22, you performed a threefactor cross in Salmonella typhimurium. The crosswas between an Arg− Leu− His− recipient bacteriumand bacteriophage P22, which was grown on an Arg+Leu+ and His+ strain. You selected for 1000 Arg+transductants and tested them on several selective media by replica plating. You obtained the followingresults:Arg+ Leu− His− 585Arg+ Leu− His+ 300Arg+ Leu+ His+ 114Arg+ Leu+ His− 1a. What are the cotransduction frequencies of argwith either his or leu?b. Is arg closer to his than it is to leu, or is arg closerto leu than it is to his? What are the relative mapdistances?c. Why can you not obtain an accurate cotransductionfrequency for his and leu from the data provided?d. What is the order of the three markers?e. Draw the crossovers that would produce the classwith only one transductant.f. Estimate the proportion of the 4.9 Mb S. typhimuriumgenome that contains these three genes. Assume thatbacteriophage P22 can package the same amountof DNA as the…Robert Bost and Richard Cribbs studied a strain of E. coli (araB14)that possessed a nonsense mutation in the structural gene that encodes Lribulokinase,an enzyme that allows the bacteria to metabolize the sugararabinose (R. Bost and R. Cribbs. 1969. Genetics 62:1–8). From thearaB14 strain, they isolated some bacteria that possessed mutations thatcaused them to revert back to the wild type. Genetic analysis of theserevertants showed that they possessed two different suppressormutations. One suppressor mutation (R1) was linked to the originalmutation in L-ribulokinase and probably occurred at the same locus. Byitself, this mutation allowed the production of L-ribulokinase, but theenzyme produced was not as effective in metabolizing arabinose as theenzyme encoded by the wild-type allele. The second suppressormutation (SuB) was not linked to the original mutation. In conjunctionwith the R1 mutation, SuB allowed the production of L-ribulokinase, butSuB by itself was not able to suppress the…
- In 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.What is dideoxy sequencing? Explain it please.In a transformation experiment, donor DNA was obtained from aprototroph bacterial strain (a+b+c+), and the recipient was a tripleauxotroph (a-b-c-). What general conclusions can you draw aboutthe linkage relationships among the three genes from the followingtransformant classes that were recovered? a+ b- c- 180 a- b+ c- 150 a+ b+ c- 210 a- b- c+ 179 a+ b- c+ 2 a- b+ c+ 1 a+ b+ c+ 3
- Nucleic acid hybridization techniques have been essential tools of molecular diagnostics for plant, animal and human diseases. Describe how the following hybridization techniques are Nucleic Acid Hybridization Technique Northern Blot Western Blot Transblot Molecules involved (DNA, RNA or Protein) Steps involved Immobilization phase (Medium) Electrophoretic run (Yes or No) Confirmatory test for hybridization Detection method for hybridization Potential applicationsJerome Vinograd found that when circular DNA from apolyoma virus is subjected to cesium chloride gradientcentrifugation, it separates into two distinct bands, oneconsisting of supercoiled DNA and the other of relaxedDNA. Explain how you would identify each band.DNA 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?