Essentials of Genetics (9th Edition) - Standalone book
9th Edition
ISBN: 9780134047799
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Chapter 8, Problem 1PDQ
HOW DO WE KNOW?
In this chapter, we have focused on genetic systems present in bacteria and the viruses that use bacteria as hosts (bacteriophages). In particular, we discussed mechanisms by which bacteria and their phages undergo genetic recombination, the basis of chromosome mapping. Based on your knowledge of these topics, answer several fundamental questions:
(a) How do we know that bacteria undergo genetic recombination, allowing the transfer of genes from one organism to another?
(b) How do we know that conjugation leading to genetic recombination between bacteria involves cell contact, which precedes the transfer of genes from one bacteriumto another?
(c) How do we know that during transduction bacterial cell-to- cell contact is not essential?
(d) How do we know that intergenic exchange occurs in bacteriophages?
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With respect to F+ and F- bacterial matings, answer the following questions: (a) How was it established that physical contact between cells was necessary? (b) How was it established that chromosome transfer was unidirectional? (c) What is the genetic basis for a bacterium’s being F+?
A 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…
With respect to F+ and F- bacterial matings, (a) How was it established that physical contact was necessary? (b) How was it established that chromosome transfer was unidirectional? (c) What is the genetic basis of a bacterium being F+?
Chapter 8 Solutions
Essentials of Genetics (9th Edition) - Standalone book
Ch. 8 -
CASE STUDY | To treat or not to treat
A...Ch. 8 - CASE STUDY | To treat or not to treat A...Ch. 8 - Prob. 3CSCh. 8 - Prob. 4CSCh. 8 - HOW DO WE KNOW? In this chapter, we have focused...Ch. 8 -
CONCEPT QUESTION
2. Review the Chapter Concepts...Ch. 8 -
3. Distinguish among the three modes of...Ch. 8 - With respect to F+ and F- bacterial matings, (a)...Ch. 8 - List all of the differences between F+ × F– and...Ch. 8 - Describe the basis for chromosome mapping in the...
Ch. 8 - Why are the recombinants produced from an Hfr × F–...Ch. 8 - Describe the origin of F' bacteria and...Ch. 8 -
9. Describe the mechanism of transformation.
Ch. 8 - .
10. The bacteriophage genome consists primarily...Ch. 8 - Prob. 11PDQCh. 8 - In the plaque assay, what is the precise origin of...Ch. 8 -
13. In the plaque assay, exactly what makes up a...Ch. 8 - A plaque assay is performed beginning with 1.0 mL...Ch. 8 -
15. Describe the difference between the lytic...Ch. 8 - Prob. 16PDQCh. 8 -
17. Explain the observations that led Zinder and...Ch. 8 -
18. Describe the execution of and rationale...Ch. 8 - If a single bacteriophage infects one E. coli cell...Ch. 8 - A phage-infected bacterial culture was subjected...
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- Consider the following experiment. First, large populations of two mutant strains of Escherichia coli are mixed, each requiring a different, single amino acid. After plating them onto a minimal medium, 45 colonies grew. Which of the following may explain this result? A) The colonies may be due to back mutation (reversion). B) The colonies may be due to recombination. C) Either A or B is possible. D) Neither A nor B is possible.arrow_forwardBacterial conjugation has medical consequences. For example, certain plasmids contain genes that code for toxins. The causative agent of a deadly form of food poisoning, E. coli 0157, synthesizes a toxin that causes massive bloody diarrhea and kidney failure. This toxin is now believed to have originated in Shigella, another bacterium that causes dysentery. Similarly, the growing problem of antibiotic resistance is partly attributable to the spread of antibioticresistant genes among bacterial populations. Antibiotic resistance develops because antibiotics are overused in medical practice and in livestock feeds. Suggest a mechanism by which this extensive use promotes antibiotic resistance.arrow_forwardA geneticist isolates two mutations in a bacteriophage. One mutation causes clear plaques (c), and the other produces minute plaques (m). Previous mapping experiments have established that the genes responsible for these two mutations are 8 m.u. apart. The geneticist mixes phages with genotype c+ m+ and genotype c− m− and uses the mixture to infect bacterial cells. She collects the progeny phages and cultures a sample of them on plated bacteria. A total of 1000 plaques are observed. What numbers of the different types of plaques (c+ m+, c− m−, c+ m−, c− m+) should she expect to see?arrow_forward
- In E. coli, the genes for histidine (his), arginine (arg), tyrosine (tyr), and valine (val) synthesis are closely linked. A wild-type strain is used as a donor for transducing viruses. Various recipient strains were treated with these viruses. Wild-type recombinant colonies were then counted, with the results shown in the following table. Genotype of recipient Number of wild-type colonies Genotype of recipient Number of wild-type colonies his- arg+ val+ tyr+ 254 his+ arg- val- tyr+ 132 his+ arg+ val- tyr- 36 his+ arg- val+ tyr+ 240 his- arg- val+ tyr+ 28 his- arg+ val- tyr+ 102 his- arg+ val+ tyr- 100 his+ arg- val+ tyr- 2 What is the cotransduction coefficient of histidine and valine? For consistency, standardize on histidine. 0.38 (wrong) 0.537 0.402 None of thesearrow_forwardE. coli cells are simultaneously infected with two strains of phage λ. One strain has a mutant host range, is temperature sensitive, and produces clear plaques (genotype h st c); another strain carries the wildtype alleles (genotype h+ st+ c+). Progeny phages are collected from the lysed cells and are plated on bacteria. The following numbers of different progeny phages are obtained: Progeny phage genotype Number of plaques h+ c+ st+ 321 h c st 338 h+ c st 26 h c+ st+ 30 h+ c st+ 106 h c+ st 110 h+ c+ st 5 h c st+ 6 a. Determine the order of the three genes on the phage chromosome. b. Determine the map distances between the genes. c. Determine the coefficient of coincidence and the interferencearrow_forwardWe have two specific strains of E. coli that have shown horizontal gene transfer (HGT) when mixed. To experimentally determine the method of HGT that is happening, the following conditions are set up in different tubes of culture media: A) Donor and recipient strain mixed together (control - no treatment). B) Donor and recipient strains mixed together, DNase added (can digest DNA in solution, not within cells).C) Special tube containing a membrane filter (with pores that allow DNA and viruses to pass through, but not bacterial cells) that separates two compartments. Donor strain is added on one side, the recipient strain on the other (they are separated by the filter).D) Donor and recipient strains mixed together, with chemical that inactivates viruses (chemical affects bacteriophages in solution so they are unable to attach to cells). The results: Tubes A, B, and D: HGT was observed. Tube C: HGT was NOT observed. Based on this, which type of HGT was occurring? Conjugation,…arrow_forward
- Two mutations that affect plaque morphology in phages (a and b ) have been isolated. Phages carrying both mutations (a b) are mixed with wild-type phages (a* b*) and added to a culture of bacterial cells. Once the phages have infected and lysed the bacteria, samples of the phage lysate are collected and cultured on plated bacteria. The following numbers of plaques are observed: Plaque phenotype Number a* b* 2043 a* b- 320 a b* 357 2134 What is the frequency of recombination between the a and b genes?arrow_forwardBy conducting conjugation experiments between Hfr and recipientstrains, Wollman and Jacob mapped the order of many bacterialgenes. Throughout the course of their studies, they identified severaldifferent Hfr strains in which the F-factor DNA had been integratedat different places along the bacterial chromosome. A sample of theirexperimental results is shown in the following table:What information do you know based on the question and your understanding of the topic?arrow_forwardConsider three genes in E. coli: thr+, ara+, and leu+ (which give the cell the ability to synthesize threonine, arabinose, and leucine, respectively). All three of these genes are close together on the E. coli chromosome. Phages are grown in a thr+ ara+ leu+ strain of bacteria (the donor strain). The phage lysate is collected and used to infect a strain of bacteria that is thr− ara− leu −. The recipient bacteria are then tested on selective medium lacking leucine. Bacteria that grow and form colonies on this medium (leu+ transductants) are then replica-plated on medium lacking threonine and on medium lacking arabinose to see which are thr+ and which are ara+. Another group of the recipient bacteria are tested on medium lackingthreonine. Bacteria that grow and form colonies on this medium (thr+ transductants) are then replica-plated on medium lacking leucine and onto medium lacking arabinose to see which are ara+ and which are leu+. Results from these experiments are as follows:…arrow_forward
- Nine rII− mutants of bacteriophage T4 were used inpairwise infections of E. coli K(λ) hosts. Six of themutations in these phages are point mutations; theother three are deletions. The ability of the doubly infected cells to produce progeny phages in large numbers is scored in the following chart.1 2 3 4 5 6 7 8 91 − − + + − − − + +2 − + + − − − + +3 − − + − + − −4 − + − + − −5 − − − + +6 − − − −7 − + +8 − −9 −The same nine mutants were then used in pairwise infections of E. coli B hosts. The production of progenyphages that can subsequently lyse E. coli K(λ) hosts isnow scored. In the table, 0 means the progeny do notproduce any plaques on E. coli K(λ) cells; − meansthat only a very few progeny phages produce plaques;and + means that many progeny produce plaques(more than 10 times as many as in the − cases).1 2 3 4 5 6 7 8 91 − + + + + − − + +2 − + + + + − + +3 0 − + 0 + + −4 − + − + + +5 − + − + +6 0 0 − +7 0 + +8 − +9 −a. Which of the mutants are the three deletions? Whatcriteria did…arrow_forwardBy conducting conjugation experiments between Hfr and recipientstrains, Wollman and Jacob mapped the order of many bacterialgenes. Throughout the course of their studies, they identified severaldifferent Hfr strains in which the F-factor DNA had been integratedat different places along the bacterial chromosome. A sample of theirexperimental results is shown in the table:Draw a map that shows the order of genes and the locations ofthe origins of transfer among these different Hfr strains?arrow_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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genetic recombination strategies of bacteria CONJUGATION, TRANSDUCTION AND TRANSFORMATION; Author: Scientist Cindy;https://www.youtube.com/watch?v=_Va8FZJEl9A;License: Standard youtube license