8) The diagram to the right shows some of the DNA in a bacterium intowhich a human gene, X, has been successfully inserted. The bacteria thatresults from reproduction of this cell will most likely have the ability toDNA ofbacteriumCell wall ofbacteriumreplicate all of the genetic instructions found in humansproduce vaccines to be used to immunize humansproduce a human blood cells according to instruction in gene XO produce the human protein coded for by gene X

The process shown in the above diagram is known as Recombinant DNA technology. This recombinant DNA…

DNA of bacterium Cell wall of bacterium replicate all of the genetic instructions found in humans O produce vaccines to be used to immunize humans O produce a human blood cells according to instruction in gene X produce the human protein coded for by gene X

DNA of bacterium Cell wall of bacterium replicate all of the genetic instructions found in humans O produce vaccines to be used to immunize humans O produce a human blood cells according to instruction in gene X produce the human protein coded for by gene X

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter9: Gene Expression And Gene Regulation

Section: Chapter Questions

Problem 1CS: There have been recurring cases of mad-cow disease in the United Kingdom since the mid-1990s....

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Row C D. B B. one: A C O phage with radiolabelled protein coat phage with radiolabelled DNA 100 10. phage infects The experiment shown above was designed by bacterium phage infects bacterium EXPERIMENT 1 phage shell is removed EXPERIMENT 2 요 Hershey & Hershey & Chase Meselson & Stahl Meselson & Stahl 8 phage shell is removed 28 no radioactivity in cells ii 48 radioactivity in cells LL and proved that ii DNA replication is semiconservative DNA is the hereditary material DNA replication is semiconservative DNA is the hereditary material (select the row that correctly completes the statement)
There have been recurring cases of mad-cow disease in the United Kingdom since the mid-1990s. Mad-cow disease is caused by a prion, an infectious particle that consists only of protein. In 1986, the media began reporting that cows all over England were dying from a mysterious disease. Initially, there was little interest in determining whether humans could be affected. For 10 years, the British government maintained that this unusual disease could not be transmitted to humans. However, in March 1996, the government did an about-face and announced that bovine spongiform encephalopathy (BSE), commonly known as mad-cow disease, can be transmitted to humans, where it is known as variant Creutzfeldt-Jakob disease (VCJD). As in cows, this disease eats away at the nervous system, destroying the brain and essentially turning it into a spongelike structure filled with holes. Victims experience dementia; confusion; loss of speech, sight, and hearing; convulsions; coma; and finally death. Prion diseases are always fatal, and there is no treatment. Precautionary measures taken in Britain to prevent this disease in humans may have begun too late. Many of the victims contracted it over a decade earlier, when the BSE epidemic began, and the incubation period is long (VCJD has an incubation period of 10 to 40 years). A recent study concluded that 1 in 2,000 people in Great Britain carry the abnormally folded protein that causes VCJD. In spite of these numbers, the death rate from VCJD remains low. It is not clear whether this means that the incubation period for the disease is much longer than previously thought, or whether they may never develop the disease. How can a prion replicate itself without genetic material?
Before development of a vaccine against this microbe, thedisease it caused accounted for two-thirds of bacterial meningi-tis cases during the first year of life but is still the number oneleading cause of mental retardation in patients who survive seri-ous disease due to permanent central nervous system disorders.What is the microorganism?(a) Haemophilus influenzae type B(b) Haemophilus influenzae type A(c) Neisseria meningitidis(d) Streptococcus pneumoniae(e) Listeria monocytogenes
If a single bacteriophage infects one E. coli cell present in a cultureof bacteria and, upon lysis, yields 200 viable viruses, howmany phages will exist in a single plaque if three more lytic cyclesoccur?
Describe and explain the difference in results if the serial dilution step of the T4 bacteriophage suspension inour experiment was skipped (i.e., all of Step #3 of the procedures was skipped)? _________________________Current EventsThe Johnson and Johnson COVID-19 vaccine consists of inactivated adenovirus that has been altered to containgenetic information encoding SARS-CoV-2 surface spike proteins. The virus targets and delivers this geneticinformation to the cells of our respiratory tract, which produce benign SARS-CoV-2 surface spike proteins butare effective for training our immune system to defend us against SARS-CoV-2.Based on this information and your knowledge of viruses in general, could T4 bacteriophage be used in place of
Katelyn had been working for Dr. Johnson for a month, and while she had become quite good at measuring inhibitionzones, she didn’t know why she was doing all this work. She had gotten very curious after she began doing all themeasurements on a new set of antibiotics. # is experiment involved infecting mice with MRSA and tracking how theMRSA grew over time.Data were collected by counting the cells of MRSA taken from $ uid samples from the mice. # e cells were measuredby taking one gram of the $ uid and spreading it over plates, but now Katelyn counted the colonies that grew on theplate after 24 hours. Because there were so many, she actually measured the colonies as “log CFU/g.” A CFU is acolony forming unit, or essentially a cell that will divide into a colony that can be seen. Because there can be so many,Katelyn measured them on a logarithmic (log) scale. # e raw data in her lab notebook looked like the following:Table 1. E% ect of treatment on MRSA in mice after 24 hours of drug…
Please answer with yes or no with correcting the falte statement. Felline to cottart the will cost the whole statements mark, 56) Prionsare small circular RNA Viruses, those viruses don't have capsid and they can't replicate in their own. They need & helper virus like hepatitis If virus to replicate 57) Car receptor can act as a receptor for multiple viruses, like adenoviruses and Herpesviruses. 58) For infiuenza viruses they enter the host cell via Clathrien-coated pits endocytosis; in which the cell receptors will be digested by the lysozyme enzymes 59) Uncosting is the removal of the outer capsid in order to release the genome to start the viral replication, In case of adenovirus the uncoalo appen.. membrane
Each species of bacteria has its own distinctive cell surface. Thecharacteristics of the cell surface play an important role in processessuch as conjugation and transduction. For example, certainstrains of E. coli have pili on their cell surface. These pili enableE. coli to conjugate with other E. coli and also enable certain bacteriophages (such as M13) to bind to the surface of the E. coli andgain entry into the cytoplasm. With these ideas in mind, explainwhich forms of genetic transfer (i.e., conjugation, transduction,and transformation) are more likely to occur between different speciesof bacteria. Discuss some of the potential consequences ofinterspecies genetic transfer.
As described in Figure, host DNA is hydrolyzed into smallpieces, which are occasionally assembled with phage proteins, creatinga phage with bacterial chromosomal DNA. If the breakage ofthe chromosomal DNA is not random (i.e., it is more likely tobreak at certain spots as opposed to other spots), how might nonrandombreakage affect cotransduction frequency?
With regard to the experiment described in Figure shown, Explain why an antibody was used to remove the bacteria thatwere not transformed. What would the results look like, in allfive cases, if the antibody/centrifugation step had not beenincluded in the experimental procedure?
20. P2 and P4 are bacteriophages of E. coli. They have the following properties: (1) When P2 phage infects a bacterium, the bacterium eventually bursts, yielding about 100 P2 progeny. (2) When P4 infects a bacterium, the bacterium survives because P4 is a defective phage. (3) When P2 phage and P4 phage coinfect the same bacterium, lysis of the bacterium yields 100 P4 progeny and no P2 progeny, because the presence of P4 inhibits the growth of P2. Suppose 3 x 108 P2 and 2 x 108 P4 are added to 108 bacterial cells. How many bacterial cells will produce P2 progeny? A. 1.28 x 107 В. 6.75 х 105 С. 5.00 х 106 D. 2.56 x 107 E. 3.23 x 104
Indicate whether each statement applies to the lytic or lysogenic cycle of phage A, or to both. 1. The host cell is lysed. (Click to select) v 2. The genetic material of the phage is copied. (Click to select) v 3. The genetic material of the phage is integrated into the bacterium's chromosome. (Click to select) v 4. Two ends of the phage DNA become covalently attached to one another. (Click to select) v 5. The integrase gene is turned on.: (Click to select) v (Click to select) lytic lysogenic both