Brock Biology Of Microorganis 15th Edition By Madigan, Kelly S. – Test Bank
Brock Biology of Microorganisms, 15e (Madigan et al.)
Chapter 11Â Â Genetics of Bacteria and Archaea
 11.1  Multiple Choice Questions
1) A mutant that has a nutritional requirement for growth is an example of a(n)
- A) autotroph.
- B) auxotroph.
- C) heterotroph.
- D) organotroph.
Answer:Â B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section:Â 11.1
2) Consider a mutation in which the change is from UAC to UAU. Both codons specify the amino acid tyrosine. Which type of point mutation is this?
- A) silent mutation
- B) nonsense mutation
- C) missense mutation
- D) frameshift mutation
Answer:Â A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section:Â 11.2
3) A mutation that readily reverses to restore the original parental type would most likely be due to a(n)
- A) deletion.
- B) insertion.
- C) point mutation.
- D) frameshift mutation.
Answer:Â C
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section:Â 11.2
4) Which process listed below allows genetic material to be transferred from a virus-like particle that lacks genes for its own replication?
- A) conjugation of an F+ plasmid
- B) gene transfer through a gene transfer agent
- C) transduction by a dsDNA phage Mu
- D) transformation of a linear piece of DNA
Answer:Â B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section:Â 11.5, 11.7
5) The mutagens 2-aminopurine and 5-bromouracil are examples of
- A) alkylating agents.
- B) nucleotide base analogs.
- C) chemicals reacting with DNA.
- D) None of the answers are correct.
Answer:Â B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section:Â 11.4
6) The killing of cells by UV irradiation involves
- A) absorption at 260 nm by proteins only.
- B) absorption at 260 nm by RNA only.
- C) formation of pyrimidine dimers.
- D) formation of purine dimers.
Answer:Â C
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section:Â 11.4
7) Ionizing radiation does NOT include
- A) gamma rays.
- B) UV rays.
- C) X-rays.
- D) cosmic rays.
Answer:Â B
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section:Â 11.4
8) Which of the following methods may introduce foreign DNA into a recipient?
- A) transformation
- B) transduction
- C) conjugation
- D) transformation, transduction, and conjugation
Answer:Â D
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section:Â 11.8
9) The uptake of free DNA from the environment ________, while transfer of DNA with cell-to-cell contact would most likely result in ________.
- A) transformation / conjugation
- B) transduction / conjugation
- C) conjugation / transformation
- D) transformation / transduction
Answer:Â A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section:Â 11.8
10) Which of the following is NOT required for homologous recombination?
- A) an Hfr chromosome
- B) RecA
- C) proteins having helicase activity
- D) endonuclease
Answer:Â A
Bloom’s Taxonomy: 1-2: Remembering/Understanding
Chapter Section:Â 11.5
11) 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.
Answer:Â C
Bloom’s Taxonomy: 3-4: Applying/Analyzing
Chapter Section:Â 11.3, 11.5
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