Biological Science (7th Edition)
7th Edition
ISBN: 9780134678320
Author: Scott Freeman, Kim Quillin, Lizabeth Allison, Michael Black, Greg Podgorski, Emily Taylor, Jeff Carmichael
Publisher: PEARSON
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Chapter 17, Problem 13PIAT
Summary Introduction
Introduction:
In the experiment, the different RNA (ribonucleic acid) polymerases (I, II, and III) purified from the rat liver were incubated with the α-(alpha) amanitin. The results of the experiment are shown in the figure below:
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The structure of a prodrug used for treating people with HIV virus (human immunodeficiency virus) and AIDS is shown below. This molecule is a precursor of a protease inhibitor that competitively inhibits HIV protease due to its resemblance to the proteolytic site of the enzyme. Which process in the viral life cycle does this inhibitor target directly ?
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Chapter 17 Solutions
Biological Science (7th Edition)
Ch. 17 - Prob. 1TYKCh. 17 - Prob. 3TYKCh. 17 - Prob. 4TYKCh. 17 - 5. RNases and proteases are enzymes that destroy...Ch. 17 - Prob. 6TYUCh. 17 - The nucleotide shown below is called cordycepin...Ch. 17 - Prob. 10TYPSSCh. 17 - What better not be for dinner? Eating even a...Ch. 17 - 12. α-Amanitin inhibits transcription by binding...Ch. 17 - Prob. 13PIAT
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- 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_forwardStudy the depiction of the lac operon in Figure 11.2. Normally, the genes are turned off when lactose is not present. Lactose activates the genes, which code for enzymes that enable the cell to use lactose. Mutations can alter the function of this operon. Predict how the following mutations would affect the function of the operon in the presence and absence of lactose: a. mutation of regulatory gene; repressor will not bind to lactose b. mutation of operator; repressor will not bind to operator c. mutation of regulatory gene; repressor will not bind to operator d. mutation of promoter; RNA polymerase will not attach to promoterarrow_forwardIn 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.arrow_forward
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