Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 6, Problem 23P
Synthesis of the amino acid histidine is a multistep anabolic pathway that uses the products of
a. Why is growth observed in experiment
b. What is the genotype of exconjugants in experiment 2?
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Chapter 6 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 6 - For bacteria that are F+, Hfr, F', and F-, perform...Ch. 6 - The flow diagram identifies relationships between...Ch. 6 - Conjugation between an Hfr cell and an F-cell does...Ch. 6 - Bacteria transfer genes by conjugation,...Ch. 6 - Explain the importance of the following features...Ch. 6 - Prob. 6PCh. 6 - Describe what is meant by the term site-specific...Ch. 6 - What is a prophage, and how is a prophage formed?Ch. 6 - How is the frequency of cotransduction related to...Ch. 6 - Describe the differences between genetic...
Ch. 6 - Among the mechanisms of gene transfer in bacteria,...Ch. 6 - What is lateral gene transfer? How might it take...Ch. 6 - Lateral gene transfer is thought to have played a...Ch. 6 - Prob. 14PCh. 6 - A 2013 CDC report identified the practice of...Ch. 6 - Hfr strains that differ in integrated F factor...Ch. 6 - Five Hfr strains from the same bacterial species...Ch. 6 - An interrupted mating study is carried out on Hfr...Ch. 6 - An Hfr strain with the genotype cys+leu+met+strS...Ch. 6 - A triple-auxotrophic strain of E. coli having the...Ch. 6 - Penicillin was first used in the 1940 s to treat...Ch. 6 - An attribute of growth behavior of eight...Ch. 6 - Synthesis of the amino acid histidine is a...Ch. 6 - The phage P1 is used as a generalized transducing...Ch. 6 - Prob. 25PCh. 6 - Prob. 26PCh. 6 - Look closely at the consolidated Hfr map and the...Ch. 6 - Fifty bacterial colonies are on a complete-medium...
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- Five individual His- mutants (hisA, his B, hisC, hisD and his E) were isolated. The genes are all involved in the histidine biosynthetic pathway. A crossfeeding experiment was carried out to determine the order of the enzymes encoded by these genes. The result is shown (A, B, C, D and E represent the mutants hisA, his B, hisC, hisD and his E, respectively. Dark area indicates bacterial growth). From the result shown, what might be the histidine biosynthetic pathway (the order of enzymes)?arrow_forwardIn a cotransformation experiment (see question 4 of More GeneticTIPS), DNA was isolated from a donor strain that was proA+ andstrC+ and sensitive to tetracycline. (The proA and strC genes conferthe ability to synthesize proline and confer streptomycin resistance,respectively.) A recipient strain is proA− and strC− and isresistant to tetracycline. After transformation, the bacteria werefirst streaked on a medium containing proline, streptomycin, andtetracycline. Colonies were then restreaked on a medium containingstreptomycin and tetracycline. (Note: Each type of medium hadcarbon and nitrogen sources for growth.) The following resultswere obtained:70 colonies grew on the medium containing proline, streptomycin,and tetracycline, but only 2 of these 70 colonies grew whenrestreaked on the medium containing streptomycin and tetracyclinebut lacking proline. If we assume the average size of the DNA fragments is 2 minutes,how far apart are these two genes?arrow_forwardAll are correct about DNA gyrase in E. coli EXCEPT: It works to remove positive supercoiling introduced by the DnaB protein (helicase). It is a topoisomerase that hydrolyzes ATP during its reaction mechanism. Its mechanism involves the breaking of a single phosphoester bond in one strand of dsDNA. It works to relieve supercoiling in DNA to overcome the torsion stress imposed upon unwinding.arrow_forward
- Telomerase activity has been found to be 10 to 20 times more active in cancer cells than in normal somatic cells. What is the significance of this circumstance?arrow_forwardConsider the structure of Cro repressor protein from bacteriophage lambda E. It is a DNA binding protein, and like many sequence- specific DNA binding proteins, it must function as a homodimer Ex. Notice the mutual docking of a phenylalanine residue from each subunit into a hydrophobic pocket of the partner subunit. These hydrophobic interactions are required for dimerization. The noncovalent interactions highlighted in yellow are also required for dimerization. These interactions represent examples of: Osecondary structure O tertiary structure O quaternary structure O secondary AND quaternary structure Ⓒ tertiary AND quaternary structurearrow_forwardA number of yeast-derived elements were added to thecircular bacterial plasmid pBR322. Yeast that requireuracil for growth (Ura− cells) were transformed withthese modified plasmids and Ura+ colonies were selected by growth in media lacking uracil. For plasmidscontaining each of the elements listed in parts (a) to(c), indicate whether you expect the plasmid to integrate into a chromosome by recombination, or insteadwhether it is maintained separately as a plasmid. If theplasmid is maintained autonomously, is it stably inherited by all of the daughter cells of subsequent generations when you no longer select for Ura+ cells (that is,when you grow the yeast in media containing uracil)?a. URA+ geneb. URA+ gene, ARS c. URA+ gene, ARS, CEN (centromere)d. What would need to be added in order for these sequences to be maintained stably in yeast cells as alinear artificial chromosome?arrow_forward
- Many of the gene products involved in DNA synthesis wereinitially defined by studying mutant E. coli strains that could notsynthesize DNA. Question: The dnaE gene encodes the a subunit of DNA polymeraseIII. What effect is expected from a mutation in this gene?How could the mutant strain be maintained?arrow_forwardThe following table lists 4 bacterial strains that are partial diploids for lac operon genes. Given the activity of beta-galactosidase measured for each strain in the absence (-lac) or presence (+lac) of lactose, complete the table by choosing the appropriate symbol (+, -, C, S) to indicate the allele of the gene or site missing from the table (blue numbers). strain A BC 5 C D 7 chromosome I O 1 2 4 1 [Select] 9 3 [Select] [Select] [Select] 9 [Select] + + Z + + 6 + I +5 + 10 plasmid O 3 + 7 C Z + 8 8 2 [Select] 4 [Select] 6 [Select] B-gal act. -lac +lac 0.002 0.003 0.002 0.058 0.063 0.121 0.059 0.062 Select] 1 ✔ [ Select] + is C Sarrow_forwardA pyrimidine dimer which is a bulky lesion has mutated an E. coli cell's DNA. Describe both the photoreactivation enzyme repair (PRE) and Nucleotide excision repair describe how the cell uses Uvr A, B, C, D gene products to effect repair.arrow_forward
- A plasmid vector pBS281 is cleaved by the enzymeBamHI (5′ G^GATCC 3′), which recognizes only onesite in the DNA molecule. Human DNA is digestedwith the enzyme MboI (5′ ^GATC 3′), which recognizes many sites in human DNA. These two digestedDNAs are now ligated together. Consider only thosemolecules in which the pBS281 DNA has been joinedwith a fragment of human DNA. Answer the following questions concerning the junction between thetwo different kinds of DNA. a. What proportion of the junctions between pBS281and all possible human DNA fragments can becleaved with MboI?b. What proportion of the junctions between pBS281and all possible human DNA fragments can becleaved with BamHI?c. What proportion of the junctions between pBS281and all possible human DNA fragments can becleaved with XorII (5′ C^GATCG 3′)?d. What proportion of the junctions between pBS281and all possible human DNA fragments can becleaved with BstYI (5′ R^GATCY 3′)? (R and Ystand for purine and pyrimidine, respectively.)e.…arrow_forwardF ′strains in E. coli are derived from Hfr strains. In some cases, these F ′strains show a high rate of integration back into the bacterial chromosome of a second strain. Furthermore, the site of integration is often the site occupied by the sex factor in the original Hfr strain (before production of the F ′strains). Explain these results.arrow_forwardBudding yeasts such as S. cerevisiae exhibit telomerase activity throughout their life cycles, whereas human somatic cells do not exhibit telomerase activity. Suggest why this is so.arrow_forward
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