Genetics: Analysis and Principles
Genetics: Analysis and Principles
6th Edition
ISBN: 9781259616020
Author: Robert J. Brooker Professor Dr.
Publisher: McGraw-Hill Education
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Chapter 7, Problem 9EQ

Acridine orange is a chemical that inhibits the replication of F-factor DNA but does not affect the replication of chromosomal DNA, even if the chromosomal DNA contains an Hfr. Let's suppose that you have an E. coli strain that is unable to metabolize lactose and has an F factor that carries a streptomycin-resistant gene. You also have an F strain of E. coli that is sensitive to streptomycin and has the genes that allow the bacterium to metabolize lactose. This second strain can grow on a lactose-containing medium. How would you generate an Hfr strain that is resistant to streptomycin and can metabolize lactose? (Hint: F factors occasionally integrate into the chromosome to become Hfr strains, and occasionally Hfr strains excise their DNA from the chromosome to become F + strains that carry an F factor.)

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In your laboratory, you have an F − strain of E. coli that is resistant to streptomycin and is unable to metabolize lactose, but it can metabolize glucose. Therefore, this strain can grow on media that contain glucose and streptomycin, but it cannot grow on media containing only lactose. A researcher has sent you two E. coli strains in two separate tubes. One strain, let’s call it strain A, has an F factor that carries the genes that are required for lactose metabolism. On its chromosome, it also has the genes that are required for glucose metabolism. However, it is sensitive to streptomycin. This strain can grow on media containing lactose or glucose, but it cannot grow if streptomycin is added to the media. The second strain, let’s call it strain B, is an F − strain. On its chromosome, it has thegenes that are required for lactose and glucose metabolism. StrainB is also sensitive to streptomycin. Unfortunately, when strains A and B were sent to you, the labels had fallen off the…
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.
In a process of production of a recombinant protein by E. coli cells, it was observed accumulation of acetate in the culture medium. In this situation, it can be said that: (a) certainly the process in question was being conducted in anaerobiosis (B).Acetate accumulation is advantageous for the process as the acetate formation reaction generates 1 molecule of ATP (c)Knowing that decreasing the temperature of the process causes a reduction in the rate of glycolysis, this could be a strategy to reduce the accumulation of acetate (d).the acetate formed can be re-assimilated by the cell if the glyoxylate pathway is activated at some point in the culture

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Genetics: Analysis and Principles

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