Brock Biology of Microorganisms 15th Edition By Michael T. Madigan -Kelly-S.-Bender – Test Bank

$20.00

Description

INSTANT DOWNLOAD COMPLETE TEST BANK WITH ANSWERS

 

Brock Biology of Microorganisms 15th Edition By Michael T. Madigan – Kelly-S.-Bender – Test Bank

 

Sample  Questions

 

Brock Biology of Microorganisms, 15e (Madigan et al.)

Chapter 5   Microbial Growth and Its Control

 

5.1   Multiple Choice Questions

 

1) Which of the following is/are transferred to daughter cells during bacterial cell division?

  1. A) only genomic DNA, the blueprint for everything the cell needs
  2. B) individual genes and enzymes to later be brought together to form the genome
  3. C) only the chromosome and enzymes to immediately begin cellular metabolism
  4. D) chromosome, proteins, and all other cellular constituents

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.1

 

2) Bacterial growth refers to

  1. A) an increase in the number of cells.
  2. B) the occurrence of binary fission.
  3. C) the time required for one cell to divide into two.
  4. D) the occurrence of mitosis.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.1

 

3) The time interval required for the formation of two cells from one is called the

  1. A) generation time.
  2. B) growth time.
  3. C) growth rate.
  4. D) division rate.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.1

 

4) Turbidity measurements are commonly utilized for monitoring

  1. A) planktonic cultures.
  2. B) biofilms.
  3. C) conies.
  4. D) sessile cultures.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.8

 

 

5) The time between inoculation and the beginning of growth is usually called the

  1. A) lag phase.
  2. B) log phase.
  3. C) dormant phase.
  4. D) death phase.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.3

6) Optical density and viable cell concentration are LEAST proportional to each other during the

  1. A) lag phase.
  2. B) exponential growth phase.
  3. C) stationary phase.
  4. D) death phase.

Answer:  D

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.8

 

7) When attempting to determine viable cell counts of a heat-sensitive bacterial isolate, which cell counting method should be avoided?

  1. A) spread-plate method
  2. B) pour-plate method
  3. C) live staining and direct counting with a microscope
  4. D) both spread- and pour-plate methods

Answer:  B

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.7

 

8) Bacteria that are able to grow in humans and cause disease would be classified as

  1. A) psychrophiles.
  2. B) mesophiles.
  3. C) thermophiles.
  4. D) hyperthermophiles.

Answer:  B

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.9

 

9) An agar plate for counting colonies and maximizing statistical validity should ideally contain

  1. A) 1 to 100 colonies.
  2. B) 50 to 100 colonies.
  3. C) 30 to 300 colonies.
  4. D) 100 to 1000 colonies.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.7

 

 

10) Relative to enzymes in mesophilic microorganisms, which of the following is NOT characteristic of enzymes in psychrophiles?

  1. A) decreased alpha helix content
  2. B) decreased beta sheet content
  3. C) fewer hydrogen bonds
  4. D) fewer ionic bonds

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.10

11) The number of colonies obtained in a plate count does NOT depend on the

  1. A) inoculum volume.
  2. B) size of the colonies.
  3. C) type of culture medium.
  4. D) incubation time.

Answer:  B

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.7

 

12) Cell density in a chemostat is controlled by

  1. A) the concentration of the limiting nutrient.
  2. B) the dilution factor.
  3. C) the size of the initial inoculum.
  4. D) the concentration of the limiting nutrient, the dilution factor, and the size of the initial inoculum.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.4

 

13) What temperature is most commonly used in autoclaves to sterilize growth media and other devices prior to experimentation?

  1. A) 95°C
  2. B) 101°C
  3. C) 121°C
  4. D) 140°C

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.15

 

14) Most mesophilic organisms can grow in a temperature range of

  1. A) 0-15°C.
  2. B) 10-20°C.
  3. C) 20-45°C.
  4. D) 50-65°C.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.9

 

15) Obligate anaerobes which are sensitive to O2 would be found growing

  1. A) throughout a tube of thioglycolate broth.
  2. B) only at the very top of a tube of thioglycolate broth.
  3. C) only at the bottom of a tube of thioglycolate broth.
  4. D) approximately one-third of the way down the thioglycolate broth.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.16

16) A microbe growing in a refrigerator is likely

  1. A) psychrophilic.
  2. B) mesophilic.
  3. C) psychrotolerant or psychrophilic.
  4. D) hyperthermophilic.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.10

 

17) Which of these statements is/are TRUE?

  1. A) In general, species that can grow at higher temperatures are prokaryotic.
  2. B) The most thermophilic prokaryotes are species of Archaea.
  3. C) Chemoorganotrophic organisms are able to grow at higher temperatures than phototrophic organisms.
  4. D) All of the statements are true.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.11

 

18) A chemical that denatures proteins is most likely to be classified as a(n) ________ agent.

  1. A) antiseptic
  2. B) bacteriostatic
  3. C) bacteriocidal
  4. D) detergent

Answer:  C

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.17

 

19) The ratio of the vapor pressure of the air in equilibrium with a substance to the vapor pressure of pure water is known as

  1. A) water activity.
  2. B) vapor activity.
  3. C) positive water balance.
  4. D) osmosis.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.13

 

20) Organisms able to live in environments with high sugar concentrations are

  1. A) halotolerant.
  2. B) osmophiles.
  3. C) xerophiles.
  4. D) anaerobic fermenting bacteria.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.13

21) When water activity is low, an organism must

  1. A) increase its internal solute concentration.
  2. B) increase its external solute concentration.
  3. C) decrease its internal solute concentration.
  4. D) decrease its external solute concentration.

Answer:  A

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.13

 

22) A bacterium possessing ________ will produce O2 bubbles when placed in the presence of hydrogen peroxide.

  1. A) catalase
  2. B) superoxide dismutase
  3. C) superoxide reductase
  4. D) peroxidase

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.14

 

23) A halotolerant facultative anaerobe would grow BEST in which of the following environments?

  1. A) oxygen depleted saline
  2. B) oxygenated saline
  3. C) oxygen depleted non-saline
  4. D) oxygenated non-saline

Answer:  D

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.13

 

24) Which of the following forms of oxygen is/are generally toxic to living organisms?

  1. A) superoxide anion
  2. B) hydrogen peroxide
  3. C) hydroxyl radical
  4. D) Superoxide, hydrogen peroxide, and hydroxyl radicals are all toxic.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.14

 

25) Superoxide dismutase and catalase work together to convert superoxide into

  1. A) peroxide.
  2. B) oxygen.
  3. C) water
  4. D) water and oxygen.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.14

26) Which of the following methods used to enumerate cells often requires specialized staining to observe non-pigmented bacteria?

  1. A) spectrophotometry/turbidity
  2. B) spread-plating
  3. C) microscopy
  4. D) spread-plating, turbidity, and microscopy

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.6

 

27) Which term is most relevant in describing the efficacy of an antimicrobial for use in a clinical setting?

  1. A) effective dose
  2. B) sterilization coefficient
  3. C) lethal dosage
  4. D) minimum inhibitory concentration (MIC)

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.17

 

28) Which method would be LEAST effective at sterilizing a glass hockey stick to use in the spread-plate method?

  1. A) autoclaving
  2. B) gamma radiation
  3. C) ethanol soaking
  4. D) ultrahigh-temperature pasteurization

Answer:  D

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.15

 

 

29) When counting colonies on an agar plate

  1. A) it is assumed that each colony arose from division of one or a few cells.
  2. B) only viable colonies are counted.
  3. C) the medium must be suitable for colony growth.
  4. D) All of these answer choices are true.

Answer:  A

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.5

 

30) By controlling the concentration of nutrients continuously added to a chemostat, cells can constantly be maintained at the

  1. A) exponential growth phase.
  2. B) stationary growth phase.
  3. C) lag growth phase.
  4. D) death growth phase.

Answer:  A

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.4

31) Serial dilutions are generally made

  1. A) when the appropriate viable count is unknown.
  2. B) based on powers of 10, although other serial dilution factors are possible.
  3. C) to reach a suitable dilution while minimizing error.
  4. D) All of these answer choices apply.

Answer:  D

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.7

 

32) Which of the following statements is FALSE?

  1. A) In the death phase, bacterial growth may cease as a result of oxygen and nutrient depletion.
  2. B) The generation time of bacteria may vary from species to species.
  3. C) Bacteria typically divide by binary fission, producing two daughter cells.
  4. D) In the lag phase, cell death exceeds cell division.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.3

 

33) The optimal growth temperature of a bacterium is most closely related to the optimal temperature for

  1. A) transcription of DNA.
  2. B) DNA replication.
  3. C) enzyme function.
  4. D) mRNA attachment to ribosomes.

Answer:  C

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.9

 

34) Why is the enzyme catalase important to the survival of bacteria?

  1. A) Catalase protects the cell from desiccation.
  2. B) Catalase breaks down toxic hydrogen peroxide.
  3. C) Catalase allows for growth in high salt concentrations.
  4. D) Catalase aids in the transport of oxygen into the cell.

Answer:  B

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.14

 

35) A medium containing known substances in precise amounts is known as a ________ media.

  1. A) selective
  2. B) complex
  3. C) natural
  4. D) defined

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.5

36) Selective medium differs from differential medium because

  1. A) selective medium permits growth of more organisms than differential medium.
  2. B) selective medium permits growth of a particular microbial type while differential medium is used to distinguish between types of organisms.
  3. C) differential medium differentiates pathogens from nonpathogens while selective medium grows only pathogens.
  4. D) differential medium contains growth inhibitors while selective medium does not.

Answer:  B

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  5.5

 

37) The cell membranes of thermophiles and hyperthermophilic bacteria typically have

  1. A) more saturated fatty acids.
  2. B) more long chain fatty acids.
  3. C) fewer unsaturated fatty acids.
  4. D) All of these answer choices are true of thermophiles and hyperthermophilic bacteria.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.11

 

38) The use of chemical agents directly on exposed body surfaces to destroy or inhibit vegetative pathogens is

  1. A) disinfection.
  2. B) sterilization.
  3. C) antisepsis.
  4. D) sanitization.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.17

 

39) Which of the following statements is TRUE of disinfectants?

  1. A) They are always effective in destroying endospores.
  2. B) They are used on living tissue.
  3. C) They are used for sterilization.
  4. D) They are used on inanimate surfaces.

Answer:  D

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.17

 

40) A ________ agent is a chemical that inhibits bacteria from reproducing, but does NOT necessarily kill them.

  1. A) bacteriostatic
  2. B) bacteriolytic
  3. C) bacteriocidal
  4. D) xerophilic

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.17

41) Microbial contamination is prevented by the

  1. A) use of the quadrant streak method.
  2. B) use of aseptic technique.
  3. C) use of cultures containing special nutrients.
  4. D) spread plate method.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.5

 

42) A common antimicrobial chemical (or chemicals) used as both an antiseptic and a general disinfectant is

  1. A) iodophors.
  2. B) 70% alcohol.
  3. C) hydrogen peroxide.
  4. D) Each of these chemicals at appropriate concentrations can be used as antiseptics or disinfectants.

Answer:  D

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.17

 

 

43) Cultures of a bacterial species were incubated on the shelf of a refrigerator (5°C), on a lab benchtop (25°C), on the shelf of a 37°C incubator, and on the shelf of a 50°C incubator. After incubation, there was no growth at 37°C or 50°C, slight growth on the benchtop, and abundant growth at refrigeration. What term could be used for this species?

  1. A) halophile
  2. B) mesophile
  3. C) anaerobe
  4. D) psychrophile

Answer:  D

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.9

 

44) A microbiology student in a research lab repeated an experiment described by others using the identical bacterial isolate, the same growth medium, and the same growth conditions. However, the student was unable to achieve the same O.D. at 600 nm. What is the LEAST likely cause for this discrepancy in measured turbidity?

  1. A) One experiment used 16 mm wide test tubes while the other used 18 mm wide test tubes.
  2. B) One experiment subtracted the yellow color of the growth medium away from the final turbidity reported whereas the other used colorless water.
  3. C) One experiment vigorously dispersed the bacterial pellicle with vortexing while the other did not.
  4. D) The two experiments were carried out at vastly different elevations and did not consider the influence of pressure.

Answer:  D

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  5.8

5.2   True/False Questions

 

1) Microbial growth is generally described as an increase in cell number rather than the expansion in size of an individual microbial cell.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.1

 

2) The duration of exponential growth would increase if bacterial cells divided into three equal daughter cells rather than two.

Answer:  FALSE

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  5.3

 

3) A selective medium is used to suppress the growth of certain bacteria in order to facilitate the growth of other bacteria.

Answer:  FALSE

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  5.5

 

4) The lag phase does NOT occur if an exponentially growing culture is transferred into a nutrient rich medium with the same components and growth conditions as the previous medium.

Answer:  TRUE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.3

 

5) The rate of exponential growth varies greatly according to bacterial species and bioavailable nutrients.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.3

 

6) A bacterium whose optimal growth temperature is 35˚C would be classified as a thermophile.

Answer:  FALSE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.3

 

7) A complex medium is one that has been completely made in the laboratory and the concentration of all the nutrients is known.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.5

 

8) Many fungi and bacteria grow best at pH 5 or below.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.12

 

9) In both the lag and stationary phase, there is no net increase or decrease in viable cells.

Answer:  TRUE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.3

10) The death phase applies to individual cells rather than populations.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.3

 

11) Direct microscopic counting of cells is an accurate method for estimating the number of viable cells in a sample.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.7

 

 

12) The optimal pH for growth of an organism refers to its intracellular environment.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.12

 

13) When viable cell concentrations are too high to count on an agar plate, it is common to use larger sized plates to increase the surface area for counting the colonies.

Answer:  FALSE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.7

 

14) Refrigeration controls microbial growth in food as a result of irreversible cell damage caused by low temperature.

Answer:  FALSE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.9

 

15) The optimum growth temperature for an organism is typically closer to its minimum temperature rather than the maximum temperature.

Answer:  FALSE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.9

 

16) A bacterium such as a snow alga that is able to grow in cold temperatures is called a psychrophile.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.9

 

17) Some microbes can grow in boiling water.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.11

 

18) Organisms living in boiling hot springs often grow rapidly and have short doubling times.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.11

19) Taq polymerase was isolated from a thermophile and is used in the polymerase chain reaction (PCR) technique because it does not become inactivated at high temperatures.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.11

 

 

20) Mesophiles typically have longer fatty acid tails and more saturated carbon-carbon bonds in their cytoplasmic tails compared with thermophiles.

Answer:  FALSE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.11

 

21) Knowledge of microbial growth patterns is useful in working with the control of microbial growth.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.15

 

22) Irreversible cell damage is more likely to occur at low rather than high temperatures.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.9

 

23) Based on studies with obligate acidophiles, high concentrations of hydrogen ions are required for their membrane stability.

Answer:  TRUE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.12

 

24) A bacterial cell in nutrient broth with a generation time of 15 minutes will produce 16 cells in 2 hours.

Answer:  FALSE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.2

 

25) Planktonic cells remain in suspension while sessile cells adhere to a surface.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.1

 

26) The thermal death time is the time needed to kill all the bacteria in a particular culture at a certain temperature.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.15

 

27) Autoclaving and pasteurization are two processes that are both very effective for sterilization.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.15

 

28) Most microorganisms are particularly susceptible to antimicrobial agents during the logarithmic growth phase.

Answer:  TRUE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.17

 

29) Sterility of a laminar flow hood is accomplished by filter-sterilized air passed through the hood quickly enough that non-sterile air does not flow into the work area.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.16

 

5.3   Essay Questions

 

1) Use the formulas N = N02n and g = t/n.

N = final cell number, N0 = initial cell number,

n = number of generations, t = time, g = generation time.

Find the generation time if N = 2 × 108, N0 = 3 × 106 and t = 3 hours.

Answer:  The number of generations (n) must first be determined through the rearranged formula: n = [(log N)-(log N0)]/(log 2) to determine n = 6.059 generations. Finally, the generation time (g) = t/n, which equates to nearly 0.5 hours per generation or almost 30 minutes per generation.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.2

 

2) Explain the difference between the division rate and the generation time.

Answer:  Generation time (g) and division rate (v) are reciprocal values of each other; both are determined by monitoring exponential growth over time. Division rate is reported as the number of generations per time and is calculated by taking 1/g. The units for g are in time per generation, which is the doubling time during exponential growth.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.2

 

3) Knowing the concentration of microorganisms in a sample is often an important consideration for environmental, industrial, and medical microbiologists, yet a microscope is sometimes not used to accomplish this task. Explain five major limitations to using a light microscope to directly count microorganisms.

Answer:  All of the following eight are potential issues associated with the direct counting method: 1) live and dead cells are indistinguishable without special staining, 2) low cell concentrations are challenging to enumerate, 3) motile cells must be immobilized, 4) precision can be quite poor with manually counting hundreds or thousands of cells, 5) small cells are difficult to observe and count even at 1,000X total magnification, 6) small particulates can be mistaken for microbial cells, 7) unstained cells require more sophisticated microscopy, and 8) it is very time-consuming.

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.6

4) Compare and contrast the spread plate method and the pour plate method of doing plate counts. Also describe which group of organisms would not be quantifiable in the pour plate method but would still be observed in the spread plate method.

Answer:  Answers will vary, but one benefit to the pour plate method is that larger volumes of cells can be dispensed, whereas large liquid volumes dispensed on the surface of agar plates is often impractical. Many cells, especially psychrophiles and some mesophiles, cannot withstand the warm temperature of molten agar (~50°C) and therefore must be spread on top of an agar plate. Colonies are easier to enumerate on a two- rather than three-dimensional surface that favors the spread plate technique over the pour plate method.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.7

 

5) Enumerating the viable and total cell concentration of a population of a microbial isolate can be a laborious task for a microbiologist, especially when studying the same isolate for several years. It is often more practical to determine the relationship between optical density (OD) and cell concentration. Once this relationship (determined by a standard curve) is determined, the OD of an isolate in a broth can be used to determine the populationʹs concentration. Why must a standard curve be prepared for each isolate when using OD measurements to determine cell concentration? Also describe an experiment that would generate this type of standard curve.

Answer:  Answers will vary, but it should be noted that a particular OD value will correspond to a specific cell concentration due to different shapes and sizes of microorganisms. One such example could be a culture of Citricella sp. SE45 which was determined to have an OD540 of 0.485. This OD fit into the already-constructed standard curve to indicate that 5 × 105 viable cells are in each milliliter of broth, which can be helpful in experiments where the final cell concentration or number is critical. To initially create the standard curve, a traditional growth curve experiment could be performed over time where both OD and cell concentration (viable with viability plating or total with direct counts) would be measured. The values of OD and cell concentration for each time point would then be used to create a standard curve, and a linear regression curve would show the relationship of OD to cell concentration such that subsequent experiments OD could be measured as a proxy for cell concentration.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  5.8

 

6) Explain why a hyperthermophile is unlikely to be a human pathogen.

Answer:  The human bodyʹs temperature is approximately 37°C, which is not a favorable environment for a hyperthermophile, which has an optimal growth temperature of 80°C or higher.

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.11

 

 

7) Explain the present understanding of molecular adaptations to the cytoplasmic membrane as found in psychrophiles.

Answer:  Non-psychrophiles do not thrive in the cold environment, in part due to predominantly having large or saturated fatty acid chains in their cytoplasmic membrane, which become increasingly rigid and wax-like with colder temperatures. The psychrophiles have higher relative concentrations of short and unsaturated fatty acids in their membrane and therefore maintain a semi-permeable membrane when exposed to less heat.

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.10

8) Explain the present understanding of molecular adaptations to the cytoplasmic membrane that are present in thermophiles.

Answer:  In thermophiles and most hyperthermophilic Bacteria, the cytoplasmic membrane has a higher content of long-chain and saturated fatty acids and a lower content of unsaturated fatty acids than are found in the cytoplasmic membranes of mesophiles. Saturated fatty acids form a stronger hydrophobic environment than do unsaturated fatty acids, and longer-chain fatty acids have a higher melting point than shorter-chain fatty acids; collectively these properties increase membrane stability.

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.11

 

9) Would you expect a xerophilic organism to be halotolerant? Why or why not?

Answer:  Yes, xerophiles are typically halotolerant. For example, the xerophilic halotolerant fungus Xeromyces bisporus inhabits cereal, candy, and dried fruit, all of which are very dry environments that also contain salt. With all else being equal, a decrease in solvent (such as water) translates to an increase in solute (salt) concentration, and therefore a microbe capable of tolerating a low water activity likely can also tolerate higher salt concentrations.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.13

 

10) Describe the autoclave, and explain the roles of time, temperature, and pressure in the functioning of the autoclave.

Answer:  High pressure is used in an autoclave so that liquids do not boil at temperatures beyond 100°C. The heat (not pressure) is used to kill living cells and is usually used for 15 minutes at 121°C. The time must be increased for bulky samples to ensure the objects or liquids are heated to 121°C for at least 15 minutes.

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  5.15

 

 

11) Discuss the pros and cons of using radiation to sterilize items such as foods, drugs, and surgical supplies.

Answer:  UV radiation sterilizes only the surfaces of objects which can be problematic for foods that are not fully cooked on the inside. Ionizing radiation can penetrate so they are ideal for foods such as grain cereals and ground beef. The high costs associated with the specialized equipment make it feasible only for large industrial applications. Due to the high energy of some of these radiation procedures, there are many concerns of potential radioactive contamination, production of carcinogenic or toxic products, and nutritional value alteration.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.16

12) Why is 100% ethanol less effective as an antimicrobial compared to a 70-95% concentration of ethanol?

Answer:  70% percent alcohol is ideal as opposed to a stronger solution. Pure alcohol coagulates protein on contact. If 100% alcohol is poured over a single-celled organism, the alcohol goes through the cell wall of the organism in all direction and coagulates the proteins found in the cell membrane. The coagulated membrane proteins would then stop the alcohol from penetrating further into the cell, preventing denaturation of cytoplasmic proteins. If this happened, the cell would become inactive but not dead. Under favorable conditions, the cell would then begin to function. If 70 percent alcohol is poured over a single celled organism, the diluted alcohol also coagulates the protein, but at a slower rate, so that it is able to penetrate all the way through the cell before protein coagulation can block it. Thus, the protein found in the entire cell is coagulated and the organism dies.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  5.17

 

13) Why are microbial doubling times in nature typically longer than those obtained in the laboratory?

Answer:  The general theme should be on how lab conditions are designed to optimally grow a microorganism (e.g., pH, substrate(s), temperature, vitamins) and minimize pressure. In the environment this microbe often is in conditions that are less than ideal, where they likely will encounter competition for growth substrates and other essential chemicals.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  5.2

Brock Biology of Microorganisms, 15e (Madigan et al.)

Chapter 7   Molecular Biology of Microbial Growth

 

7.1   Multiple Choice Questions

 

1) Fluorescence microscopy allows visualization of all of the following EXCEPT

  1. A) cytoskeletal proteins.
  2. B) nuclear proteins.
  3. C) interactions between cytoplasmic proteins.
  4. D) All of these answer choices can be visualized.

Answer:  C

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.1

 

2) Once replication of the chromosome has begun, DnaA is inactivated

  1. A) as a result of being blocked from oriC
  2. B) by increased expression of FtsZ.
  3. C) as a result of increased binding to ATP.
  4. D) by increased binding of DnaA-ATP to oriC.

Answer:  A

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.2

 

3) ________ prohibits re-initiation of chromosome replication.

  1. A) DnaA-ATP
  2. B) SeqA
  3. C) oriC
  4. D) FtsZ

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.2

 

4) Prior to DNA replication, both strands of the chromosome are methylated on the ________ residue of the sequence-GATC-.

  1. A) A
  2. B) G
  3. C) C
  4. D) T

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.2

 

 

5) Hemimethylated DNA is found on

  1. A) the newly synthesized strand of DNA immediately after replication.
  2. B) the parental strand of DNA immediately after replication.
  3. C) both strands of dsDNA immediately after replication.
  4. D) neither strand of DNA immediately after replication.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.2

6) The typical time required for chromosome replication in Escherichia coli is

  1. A) 10 minutes.
  2. B) 20 minutes.
  3. C) 30 minutes.
  4. D) 40 minutes.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.2

 

7) The typical generation time of Escherichia coli is approximately ________ under optimal environmental conditions.

  1. A) 10 minutes
  2. B) 20 minutes
  3. C) 30 minutes
  4. D) 40 minutes

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.2

 

8) The Par system is necessary for

  1. A) formation of a replication fork in Escherichia coli.
  2. B) distribution of genetic material in replicating Caulobacter.
  3. C) septum formation.
  4. D) elongation of bacillus prior to cell division.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.2

 

9) Which of the following proteins is most active in divisome complexes?

  1. A) FtsZ
  2. B) crescentin
  3. C) MinCD
  4. D) MreB

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.3

 

 

10) A bacterium that lacks the mreB gene will have a ________ shape.

  1. A) bacillus
  2. B) coccoid
  3. C) short bacillus
  4. D) vibrio

Answer:  B

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.4

11) What are the primary regulator units that control endospore formation?

  1. A) allosteric proteins
  2. B) antisense RNAs
  3. C) riboswitches
  4. D) sigma factors

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.6

 

12) All of the following are TRUE statements concerning binary fission of microbial cells EXCEPT

  1. A) the chromosome of the cell is replicated.
  2. B) elongation of the cell occurs and the chromosomes are pushed apart.
  3. C) a septum is formed across the midline of the cell.
  4. D) daughter cells produced can be of different sizes.

Answer:  D

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.6

 

13) There are four basic stages to biofilm formation. Which of the following is the correct order of these stages?

  1. A) attachment, colonization, development, dispersal
  2. B) colonization, attachment, development, dispersal
  3. C) development, colonization, attachment, dispersal
  4. D) attachment, development, colonization, dispersal

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.9

 

14) Pseudomonas aeruginosa is a(n) ________ pathogen.

  1. A) opportunistic
  2. B) idiopathic
  3. C) pessimistic
  4. D) iatrogenic

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.9

 

15) Which of the following activities is a result of cyclic di-guanosine monophosphate synthesis?

  1. A) transition from planktonic to sessile growth during biofilm formation
  2. B) reduction in activity of flagellar motor
  3. C) biosynthesis of extracellular matrix during biofilm formation
  4. D) All of these answer choices activities are a result of cyclic di-guanosine monophosphate synthesis.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.9

16) The major symptoms of the genetic disorder cystic fibrosis are caused due to biofilm formation by

  1. A) Serratia marcesans.
  2. B) Pseudomonas aeruginosa.
  3. C) Vibrio cholera.
  4. D) Bacillus subtilis.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.9

 

17) All of the following are targets of currently used antibiotics EXCEPT

  1. A) DNA synthesis.
  2. B) biofilm formation.
  3. C) transcription.
  4. D) translation.

Answer:  B

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.10

 

18) When a Bacillus anthracis population suddenly must form spores to survive a harsh nutrient poor environment, how do the cells obtain energy?

  1. A) Cells in a growth phase that have not used up all of their energy will be the only ones to make endospores, which is why relatively few endospores are often made from a large population.
  2. B) Intracellular energy reserves are quickly made available to produce endospores.
  3. C) Slow responding cells are cannibalized by others that already began spore formation.
  4. D) Global regulation is initiated to minimize energy waste in biosynthetic pathways and catabolic pathways are increased to consume remaining usable substrates to fuel spore formation.

Answer:  C

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.6

 

 

19) Bacteria from the genus Caulobacter are used to model cellular differentiation in eukaryotes. The abundance of CtrA, DnaA, and GcrA separately control activity of other genes necessary for differentiation in Caulobacter. Thus, these three proteins can be classified as

  1. A) activating sensors.
  2. B) heterologous regulators.
  3. C) differentiating regulons.
  4. D) transcriptional regulators.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.7

20) A bacterium that either partially or fully catabolizes an acyl-homoserine lactone will likely disrupt

  1. A) attenuation.
  2. B) chemotaxis.
  3. C) endospore formation.
  4. D) quorum sensing.

Answer:  D

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.9

 

21) Quorum sensing generally follows the mechanism of which type of regulation?

  1. A) feedback inhibition
  2. B) negative transcriptional regulation
  3. C) positive transcriptional regulation
  4. D) a two-component regulation system

Answer:  D

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.9

 

22) Penicillins work by

  1. A) activating autolytic enzymes.
  2. B) blocking the transpeptidation reaction that forms peptidoglycan cross-links.
  3. C) stimulating autolysins to form holes in the plasma membrane.
  4. D) All of these answer choices are true.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.10

 

23) Differential gene expression occurs in each of the following bacterium EXCEPT

  1. A)
  2. B)
  3. C)
  4. D)

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.6

 

24) σF functions as a(n) ________ during endospore formation.

  1. A) transcription factor
  2. B) anti-sigma factor
  3. C) protease
  4. D) RNase

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.6

25) When Spo0A is highly phosphorylated

  1. A) heterocyst formation is initiated.
  2. B) stalk formation is initiated.
  3. C) endospore formation is initiated.
  4. D) photosynthesis is initiated.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.6

 

26) What is the function of heterocysts in cyanobacteria such as Anabaena?

  1. A) nitrogen fixation
  2. B) sexual reproduction
  3. C) asexual reproduction
  4. D) photosynthesis

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.6

 

27) The major shape-determining factor in Bacteria is the protein

  1. A) FtsZ.
  2. B) crescentin.
  3. C) MreB.
  4. D) MinD.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.3

 

28) In the process of binary fission in Bacteria, which action occurs first?

  1. A) formation of the divisome
  2. B) DNA replication
  3. C) cell elongation
  4. D) cytokinesis

Answer:  B

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.3

 

 

29) FtsZ is related to ________, an important protein involved in cell division.

  1. A) actin
  2. B) myosin
  3. C) tubulin
  4. D) collagen

Answer:  C

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.3

30) FtsA is related to ________, an important cytoskeletal protein in eukaryotes.

  1. A) actin
  2. B) myosin
  3. C) tubulin
  4. D) collagen

Answer:  A

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.3

 

31) Endospore formation is stimulated when

  1. A) bacterial growth ceases due to limitation of an essential nutrient.
  2. B) the bacterium is undergoing binary fission.
  3. C) bacteria are dividing exponentially.
  4. D) nutrient levels rise.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.6

 

32) Proteins MinC, MinD, and MinE interact to

  1. A) help guide FtsZ to the cell midpoint.
  2. B) initiate peptidoglycan synthesis.
  3. C) assist chromosome segregation.
  4. D) determine cell shape.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.4

 

33) The curved-rod shape typical of Caulobacter is due to the protein(s)

  1. A) crescentin.
  2. B) MreB.
  3. C) crescentin and MreB.
  4. D) neither crescentin nor MreB.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.4

 

 

34) Which of the following traits is NOT characteristic of the Caulobacter life cycle?

  1. A) Cell division only occurs in the swarmer stage.
  2. B) Cell division only occurs in the stalked cell stage.
  3. C) Chromosome replication only occurs in the swarmer stage.
  4. D) Chromosome replication and cell division only occurs in the stalked cell stage.

Answer:  D

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.7

35) Biofilm formation in Pseudomonas aeruginosa is triggered by ________ cell densities and repressed by ________ cell densities.

  1. A) high; high
  2. B) low; low
  3. C) high; low
  4. D) low; high

Answer:  C

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.9

 

36) Biofilm formation in Vibrio cholera is triggered by ________ cell densities and repressed by ________ cell densities.

  1. A) high; high
  2. B) low; low
  3. C) high; low
  4. D) low; high

Answer:  D

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.7

 

37) What is the function of bactoprenol?

  1. A) It is a hydrophobic alcohol that transports peptidoglycan precursors across the cytoplasmic membrane.
  2. B) It is responsible for forming the peptide cross-links between muramic acid residues in adjacent glycan chains.
  3. C) It triggers the recruitment of FtsZ and the initiation of the divisome.
  4. D) It supplies the energy necessary for transpeptidation to occur.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.5

 

 

38) Shortage of key nutrients such as ________ results in transcription and secretion of a toxic protein that leads to cannibalization of neighboring cells of the same species.

  1. A) magnesium
  2. B) phosphate
  3. C) calcium
  4. D) iron

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.6

 

39) Heterocysts are specialized cells that undergo

  1. A) oxidative phosphorylation.
  2. B) photophosphorylation.
  3. C) nitrogen fixation.
  4. D) lysine decarboxylation.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.8

40) Which of the following is NOT a function of DnaA in Caulobacter?

  1. A) initiation of DNA replication
  2. B) initiation of swarming
  3. C) transcriptional regulation
  4. D) All of these answer choices are functions of DnaA.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.7

 

41) Why is the presence of a cell wall significant from a clinical standpoint?

  1. A) All types of cells have a cell wall, and it makes identification of the causative agent of disease difficult.
  2. B) The cell wall protects microorganisms from destruction by the immune system.
  3. C) Animal cells do not have cell walls, so antibiotics that target cell walls can selectively destroy invading microorganisms.
  4. D) Only gram-negative Bacteria have cell walls.

Answer:  C

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.5

 

 

42) Which of the following molecules functions as an intracellular signaling molecule during quorum sensing?

  1. A) acylated homoserine lactones
  2. B) hydrophilic lipids
  3. C) quinones
  4. D) proteorhodopsins

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.9

 

43) Dormancy is a result of all of the following EXCEPT

  1. A) phenotypic heterogeneity.
  2. B) ineffective efflux pumps.
  3. C) TA modules.
  4. D) the stringent response.

Answer:  B

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.11

 

7.2   True/False Questions

 

1) The purpose of hemimethylated DNA is to block chromosome replication.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.2

 

2) Plasmids are replicated in Escherichia coli using the Par system.

Answer:  FALSE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.2

3) PopZ plays a critical role in localizing proteins to the cell poles.

Answer:  TRUE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.2

 

4) The Par system is analogous to spindle fibers in eukaryotic cells.

Answer:  TRUE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.2

 

5) In the Caulobacter life cycle, the role of swarmer cells is strictly for reproduction.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.7

 

 

6) Biofilm formation is more likely to occur when Vibrio cholerae is found in low densities such as sea water.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.9

 

7) Multiple sigma factors are essential to induce the biosynthesis of endospores, consequently a complex regulatory mechanism such as this has a higher chance of mutation leading to incorrect functioning compared to a simple repression mechanism.

Answer:  TRUE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.6

 

8) Generation of new bacillus-shaped cells starts with one cell elongating and terminates when it splits into two separate bacilli.

Answer:  TRUE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.4

 

9) The activity of MinC and MinD direct whether a bacterial cell will be coccoid or bacillus shaped.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.4

 

10) In fast-growing Escherichia coli cells, multiple replication forks of genomic DNA allows binary fission to occur before the genome has been fully duplicated.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.2

 

11) Multiple sigma factors are essential to induce the formation of endospores.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.6

12) During elongation of a cell during binary fission, small gaps caused by transglycosylases are formed before cell membrane precursors can be inserted.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.3

 

13) Antibiotic resistance can develop in a bacterium as a result of spontaneous mutation.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.10

 

14) Natural antibiotics are produced by all bacterial genera.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.10

 

15) All members of the genus Pseudomonas respond similarly to c-di-GMP.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.9

 

16) The spherical shape of Staphylococcus aureus is a result of MreB function.

Answer:  FALSE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.4

 

17) Both crescentin and MreB play a role in formation of the vibrio-shaped bacterium Caulobacter.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.4

 

18) FtsZ is analogous to the protein tubulin in eukaryotic cells.

Answer:  FALSE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.4

 

19) In both cocci as well as rod-shaped cells, new peptidoglycan synthesis occurs in both directions from the central location of the FtsZ ring.

Answer:  FALSE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.5

 

20) Oxygenic phototrophs require the uptake of oxygen for the reactions of photosynthesis.

Answer:  FALSE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.8

 

21) Mobile resistance genes encode enzymes that inactivate an antibiotic by altering its structure.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.10

22) Unlike penicillin, methicillin targets an alternative penicillin binding protein called MecA in nonresistant Staphylococcus aureus.

Answer:  FALSE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.10

 

23) MRSA strains synthesize MecA only in the presence of methicillin or other β-lactam antibiotics.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.10

 

24) Antibiotic-sensitive bacteria sometimes produce cells that are transiently resistant to multiple antibiotics.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.10

 

25) When Mycobacterium tuberculosis is in the dormant state it is still susceptible to antibiotics, but the rate of death is slower.

Answer:  FALSE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.11

 

26) Persistence is a heritable trait.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.11

 

27) The toxin component of TA modules helps ensure survival during stressful growth conditions.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.11

 

28) Free HipA toxin inhibits transcription.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.11

 

7.3   Essay Questions

 

1)  Describe the functions of the FtsZ protein and the Z-ring in bacterial cells.

Answer:  FtsZ is an essential cell division protein that forms a contractile ring structure (Z ring) at the future cell division site. The regulation of the ring assembly controls the timing and the location of cell division. One of the functions of the FtsZ ring is to recruit other cell division proteins to the septum to produce a new cell wall between the dividing cells.

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.3

 

2) What is the purpose of the MreB protein in bacteria?

Answer:  The bacterial actin homologue MreB is required for the maintenance of a rod-shaped cell and has been shown to form spirals that traverse along the longitudinal axis of Bacillus subtilis and Escherichia coli cells.

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.4

 

3) What role does the Par system play in bacterial cell division?

Answer:  The Par (partitioning) system can be found in many bacteria including Caulobacter which undergoes budding. Similar to mitotic spindle fibers which separate replicated chromosomes in eukaryotic cells, the Par system distributes chromosomes and plasmids equally to progeny cells during growth. This system is composed of the ParAATPase, the ParB chromosome-binding protein, and the PopZ complex, as well as a centromere-like parS sequence located near oriC.

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.2

 

4) A researcher unexpectedly identified mutant bacteria deficient in the FtsZ protein. Their growth appeared filamentous, and displayed incomplete cell division. Explain the role of the FtsZ protein and how a deficiency would account for this altered growth.

Answer:  FtsZ is an essential cell division protein that forms a contractile ring structure (Z ring) at the future cell division site. The regulation of the ring assembly controls the timing and the location of cell division. One of the functions of the FtsZ ring is to recruit other cell division proteins to the septum to produce a new cell wall between the dividing cells. Consequently, a mutation in this protein would result in incomplete cell division and a potentially filamentous appearance.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.3

 

5) How might a genetically modified strain of E.coli with a nonfunctional mreB gene compare to E. coli possessing a functional mreB gene?

Answer:  MreB forms a simple cytoskeleton in Bacteria by forming patchlike filaments around the inside of the cell just below the cytoplasmic membrane. The MreB cytoskeleton is thought to define cell shape by recruiting other proteins that function in cell wall growth to group in a specific pattern. Inactivation of the gene encoding MreB in rod-shaped bacteria causes the cells to become coccus-shaped.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  7.3

 

 

6) Explain how FtsZ-type proteins being found in mitochondria and chloroplasts strengthens or weakens the endosymbiosis theory.

Answer:  The protein FtsZ, encoded by the gene ftsZ, is specifically involved in prokaryotic cell division. FtsZ-like proteins identified in mitochondria and chloroplasts suggest the FtsZ-encoding gene was first present in prokaryotes and that these two organelles were subsequently developed and maintained in Eukarya. Ultimately this corroborates the ideas of the endosymbiotic theory and seemingly links chloroplast and mitochondrial origins to prokaryotes.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  7.3

7) Explain the role of CtrA in the formation of swarmer cells in Caulobacter.

Answer:  CtrA is activated by phosphorylation in emerging swarmer cells in response to external signals. Once phosphorylated, CtrA-P activates genes that encode synthesis of the flagellum and other functions specific to swarmer cells. Conversely, CtrA-P represses the synthesis of the protein CcrA and so inhibits the initiation of DNA replication in swarmer cells by binding to and blocking the origin of replication.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.7

 

8) Explain how green fluorescent protein (GFP) could be utilized to determine the role of FtsZ in binary fission in Escherichia coli.

Answer:  For visualizing molecular events, reporter genes that encode fluorescent products such as green fluorescent protein (GFP) are routinely used. In this case, the gene for the production of GFP would be incorporated into the genome of Escherichia coli in the region of the DNA that codes for the FtsZ gene so that both genes are controlled by the same regulatory sequence; that is, the gene’s regulatory sequence now controls the production of GFP, in addition to FtsZ. In cells where the gene is expressed, and the tagged FtsZ proteins are produced, GFP is produced at the same time. Thus, only those cells in which the tagged gene is expressed, or the target proteins are produced, will fluoresce when observed under fluorescence microscopy.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  7.3

 

9) Explain the impact spontaneous mutation of the gene minE in an Escherichia coli bacterium would have on future cell divisions.

Answer:  The MinE protein is one of three proteins of the Min system encoded by the minB operon required to generate pole to pole oscillations prior to bacterial cell division as a means of specifying the midzone of the cell. If a spontaneous mutation were to occur resulting in inactivation of this protein, it would affect cell division in the next generation since the FtsZ ring would not form and cell division would not occur.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  7.3

 

10) Compare the process of new peptidoglycan formation in cocci versus rod-shaped cells.

Answer:  In cocci, new cell wall material grows out in opposite directions from the FtsZ ring. In contrast, new cell wall grows at several locations along the length of a rod-shaped cell.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.5

 

11) The chemical composition of the peptidoglycan carrier bactoprenol is highly hydrophobic. Explain why this is necessary for bactroprenol to carry on its function.

Answer:  Bactoprenol is a hydrophobic C55 alcohol that bonds to peptidoglycan precursors. Bactoprenol transports peptidoglycan precursors across the cytoplasmic membrane by rendering them sufficiently hydrophobic to pass through the fatty acid tails of the membrane interior. Once in the periplasm, bactoprenol interacts with enzymes called transglycosylases that insert cell wall precursors into the growing point of the cell wall and catalyze glycosidic bond formation.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.5

12) How does the antibiotic penicillin lead to the death of pathogens?

Answer:  Several penicillin binding proteins have been identified in bacteria, including FtsI. When penicillin is bound to penicillin-binding proteins, the proteins lose their catalytic activity. In the absence of transpeptidation, the continued activity of autolysins so weakens the cell wall that the cell eventually bursts.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.5

 

13)  Describe the steps of binary fission.

Answer:  1) Replication of the circular prokaryotic chromosome begins at the origin of replication and continues in both directions at once. 2) The cell begins to elongate. FtsZ proteins migrate toward the midpoint of the cell. 3) The duplicated chromosomes separate and continue to move away from each other toward opposite ends of the cell. FtsZ proteins form a ring around the periphery of the midpoint between the chromosomes. 4) The FtsZ ring directs the formation of a septum that divides the cell. Plasma membrane and cell wall materials accumulate. 5) After the septum is complete, the cell pinches in two, forming two daughter cells. FtsZ is dispersed throughout the cytoplasm of the new cells.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  7.3

 

14) Explain why the antibiotic penicillin is effective at destroying bacteria but does not have a similar effect in eukaryotic cells.

Answer:  Animal cells do not possess a cell wall and therefore lack peptidoglycan. Consequently the drug can be administered in high doses and is typically nontoxic. Most pathogenic bacteria do have cell walls composed of peptidoglycan and are thus potential targets of the drug.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.5

 

 

15) The major trigger of endospore formation in Bacillus is nutrient deprivation. Without the appropriate nutrients, how can endospore formation occur?

Answer:  Sporulating cells cannibalize cells of their own species to obtain the appropriate nutrients. Those Bacilli in which Spo0A has become activated secrete a toxic protein that lyses nearby Bacillus cells whose Spo0A has not yet become activated. The lytic protein is produced in addition to another protein that functions to delay sporulation of neighboring cells. Cells committed to sporulation also make an antitoxin protein to protect themselves against the effects of their own toxic protein. Once lysed, the sacrificed cells are used as a source of nutrients.

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.6

 

16) Nitrogen fixation utilizes the enzyme nitrogenase, which is extremely sensitive to oxygen. How do cyanobacteria carry on both oxygenic photosynthesis and nitrogen fixation at the same time?

Answer:  Some filamentous cyanobacteria such as Anabaena and Nostoc form specialized cells called heterocysts that are exclusively dedicated to nitrogen fixation. These cells lack photosystem II, the complex that produces O2 during oxygenic photosynthesis.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  7.8

17) Your microbiology professor is interested in isolating a particular endospore-forming bacterium from its natural environment and asks you to devise a protocol to do this. Explain how you could go about separating the endospores from the other microbes in the population.

Answer:  The most direct method for isolating the endospores would be to expose the mixed population to high heat for a brief period to induce sporulation. This would also serve the purpose of killing any non-endospore forming bacteria.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  7.6

 

18) Various species of bacteria naturally produce antibiotics. What resistance mechanisms are utilized by bacteria to survive antibiotics produced by other microbes or from any antibiotics they may produce themselves?

Answer:  Resistance mechanisms are genetically encoded and fall into four classes: (1) modification of the drug target; (2) enzymatic inactivation; (3) removal from the cell via efflux pumps, and (4) metabolic bypasses.

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.10

 

 

19) Biofilm formation is known to increase the resistance of member bacteria to antibiotics. How does this happen?

Answer:  There are several causes of increased antibiotic resistance in biofilms. These include decreased permeability of antibiotics due to the exo-polysaccharide matrix produced by the biofilm members. Efflux pumps also play a major role. In E. coli, genes encoding the AcrAB-To1C efflux pump are upregulated when cells enter biofilm growth. Pseudomonas aeruginosa encodes several multidrug efflux pumps that are more active when cells grow in an attached state.

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  7.10

Brock Biology of Microorganisms, 15e (Madigan et al.)

Chapter 11   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)

  1. A) autotroph.
  2. B) auxotroph.
  3. C) heterotroph.
  4. D) organotroph.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  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?

  1. A) silent mutation
  2. B) nonsense mutation
  3. C) missense mutation
  4. D) frameshift mutation

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.2

 

3) A mutation that readily reverses to restore the original parental type would most likely be due to a(n)

  1. A) deletion.
  2. B) insertion.
  3. C) point mutation.
  4. D) frameshift mutation.

Answer:  C

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  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?

  1. A) conjugation of an F+ plasmid
  2. B) gene transfer through a gene transfer agent
  3. C) transduction by a dsDNA phage Mu
  4. D) transformation of a linear piece of DNA

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.5, 11.7

 

 

5) The mutagens 2-aminopurine and 5-bromouracil are examples of

  1. A) alkylating agents.
  2. B) nucleotide base analogs.
  3. C) chemicals reacting with DNA.
  4. D) None of the answers are correct.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.4

6) The killing of cells by UV irradiation involves

  1. A) absorption at 260 nm by proteins only.
  2. B) absorption at 260 nm by RNA only.
  3. C) formation of pyrimidine dimers.
  4. D) formation of purine dimers.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.4

 

7) Ionizing radiation does NOT include

  1. A) gamma rays.
  2. B) UV rays.
  3. C) X-rays.
  4. D) cosmic rays.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.4

 

8) Which of the following methods may introduce foreign DNA into a recipient?

  1. A) transformation
  2. B) transduction
  3. C) conjugation
  4. D) transformation, transduction, and conjugation

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  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 ________.

  1. A) transformation / conjugation
  2. B) transduction / conjugation
  3. C) conjugation / transformation
  4. D) transformation / transduction

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.8

 

 

10) Which of the following is NOT required for homologous recombination?

  1. A) an Hfr chromosome
  2. B) RecA
  3. C) proteins having helicase activity
  4. D) endonuclease

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  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?

  1. A) The colonies may be due to back mutation (reversion).
  2. B) The colonies may be due to recombination.
  3. C) Either A or B is possible.
  4. D) Neither A nor B is possible.

Answer:  C

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  11.3, 11.5

 

12) You have performed the following mating experiment using Hfr and F-strains of Escherichia coli:

Hfr (thr+ leu+ gal+ strs) × F- (thr- leu- gal- strr). Which of the following selective media would you use to score recombinant colonies?

  1. A) minimal medium
  2. B) minimal medium + streptomycin
  3. C) minimal medium + threonine
  4. D) minimal medium + streptomycin + threonine

Answer:  B

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  11.9

 

13) Horizontal gene transfer in Archaea

  1. A) is infrequent in nature and therefore difficult to use for genetic studies in the laboratory.
  2. B) has not been documented, thus all genetic studies of archaea are done via genomic

sequencing.

  1. C) frequently occurs in nature and has been used to perform genetic studies in the

laboratory as well.

  1. D) frequently occurs in nature, but there are very few laboratory studies because archaea do

not cause human disease.

Answer:  C

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  11.10

 

 

14) The minimal amount of genetic information required for specialized transduction would include

  1. A) the att
  2. B) the cos
  3. C) a helper phage.
  4. D) the att region, cos site, and a helper phage.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.7

15) Lysogeny probably carries a strong selective advantage for the host cell because it

  1. A) prevents cell lysis.
  2. B) confers resistance to infection by viruses of the same type.
  3. C) confers resistance to infection by viruses of a different type (or strain).
  4. D) confers resistance to infection by many virus types and prevent cell lysis.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.7

 

16) A plasmid may

  1. A) replicate independently of the chromosome.
  2. B) be transferred cell-to-cell during conjugation.
  3. C) be integrated into the chromosome.
  4. D) replicate independently of the chromosome, integrate into the chromosome, or be transferred cell-to-cell during conjugation.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.9

 

17) Plasmids that govern their own transfer are known as

  1. A) transformable.
  2. B) transmutable.
  3. C) conjugative.
  4. D) transfective.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.8

 

18) Homologous recombination has been observed in

  1. A) Archaea.
  2. B) Bacteria.
  3. C) Eukarya.
  4. D) All three domains (Archaea, Bacteria, and Eukarya).

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.5

 

19) Hfr strains of Escherichia coli

  1. A) do not possess an F factor.
  2. B) have the F factor as a plasmid.
  3. C) have an integrated F factor.
  4. D) transfer the complete F factor to recipient cells at a high frequency.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.9

20) F+ strains of Escherichia coli

  1. A) do not have an F factor.
  2. B) have the F factor as a plasmid.
  3. C) have an integrated F factor.
  4. D) transfer the F factor to recipient cells at a high frequency.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.8

 

21) F-strains of Escherichia coli

  1. A) do not have an F factor.
  2. B) have the F factor as a plasmid.
  3. C) have an integrated F factor.
  4. D) transfer the F factor to other strains at a high frequency.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.8

 

22) All Hfr strains integrate into the chromosome at

  1. A) the same locus.
  2. B) several specific sites.
  3. C) the same locus most of the time, although there may be some variation.
  4. D) loci that cannot be accurately determined.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.9

 

23) Transposition is a(n)

  1. A) homologous recombination event.
  2. B) analogous recombination event.
  3. C) site-specific recombination event.
  4. D) general recombination event.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.11

 

 

24) The enzyme transposase may be coded for by insertion sequences on a

  1. A) chromosome.
  2. B) phage.
  3. C) plasmid.
  4. D) chromosome, phage, or plasmid.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.11

25) A ʺpoint mutationʺ refers to mutations involving

  1. A) a base-pair substitution.
  2. B) the gain of a base pair (microinsertion).
  3. C) the deletion of a base pair (microdeletion).
  4. D) a substitution, deletion, or addition of one base-pair.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.2

 

26) A deleterious mutation in recA results in

  1. A) a decrease in specific recombination.
  2. B) a decrease in homologous recombination.
  3. C) an increase in homologous recombination.
  4. D) no change in either general or specific recombination.

Answer:  B

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  11.5

 

27) The production of a functional gene product by transforming bacteria that lack a lacZ gene with a plasmid containing a lacZ gene is known as

  1. A) complementation.
  2. B) mitosis.
  3. C) transfection.
  4. D) reversion.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.5

 

28) Consider conjugation in Escherichia coli. In which of the following matings would chromosomal genes be transferred most frequently?

  1. A) F+ × F-
  2. B) F- × F-
  3. C) Hfr × F-
  4. D) F+ x F+

Answer:  C

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  11.9

 

29) Which of the following features are common to transformation, transduction, and conjugation?

(1)  unidirectional transfer of genes

(2)  incomplete gene transfer

(3)  homologous recombination

(4)  meiosis occurring in the recipient

  1. A) 1, 2, 3
  2. B) 1, 2
  3. C) 3, 4
  4. D) 1, 2, 4

Answer:  A

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  11.5

30) Which of the following is most similar to lysogeny?

  1. A) Hfr state
  2. B) F+ state
  3. C) F- state
  4. D) Fʹ state

Answer:  A

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  11.9

 

31) In the bacterial world, a gene located on which of the following would be the LEAST likely to be transferred?

  1. A) a resistance plasmid
  2. B) An F plasmid
  3. C) the phage Mu
  4. D) the chromosome

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.9

 

32) Genetic recombination involving insertion sequences typically results in what type of mutation?

  1. A) base-pair substitution mutation
  2. B) silent mutation
  3. C) frameshift mutation
  4. D) base-pair deletion mutation

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.11

 

 

33) The SOS regulatory system is activated by

  1. A) the activity of DNA polymerase IV.
  2. B) DNA damage.
  3. C) transcription of LexA.
  4. D) repression of RecA.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.4

 

34) Which of the following factors has delayed the development of laboratory-based genetic systems in Archaea?

  1. A) There are no documented systems of conjugation in Archaea.
  2. B) Homologous recombination does NOT occur in Archaea.
  3. C) Archaea do NOT host viruses or plasmids.
  4. D) Many archaea grow in extreme or unusual conditions that make the use of agar and traditional mutant screening techniques problematic.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.10

35) Transformation and homologous recombination allow for the formation of heteroduplex DNA. Which of the following would occur during DNA replication of this molecule?

  1. A) One daughter strand is complementary to the recombinant DNA molecule, while the other daughter strand is complementary to the parent DNA molecule.
  2. B) Both daughter strands are complementary to the recombinant DNA molecule.
  3. C) Both daughter strands are complementary to the parent DNA molecule.
  4. D) None of the answers are correct.

Answer:  A

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  11.6

 

36) You work for a biotechnology company that uses Streptomyces strains to produce pharmaceutical products. A phage has infected and killed some of your Streptomyces strains during production, resulting in dramatically decreased yields. To protect the strains from infection you propose to

  1. A) introduce gene transfer agents into the Streptomyces cultures to transfer antibiotic resistance genes into your Streptomyces
  2. B) design and insert CRISPR spacer sequences into the genomes of your strains that are complementary to the genomes of the phages that are infecting the cultures.
  3. C) infect the Streptomyces strains with a helper phage that will help the strains resist infection.
  4. D) transform the Streptomyces strains with plasmids encoding antibody proteins that will protect them from phage infection.

Answer:  B

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  11.12

 

 

37) Hfr means high frequency of ________, and these cells are capable of transferring genes from their ________ to other cells.

  1. A) transformation / chromosome
  2. B) transduction / plasmids
  3. C) recombination / chromosome
  4. D) transduction / chromosome

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.9

 

38) Mutation rates are similar in Bacteria and Archaea, yet certain stressful conditions mutation rates increase. Why is the mutation rate not constant and close to zero all of the time?

  1. A) Increased mutation rates can be advantageous in rapidly changing environments because some random mutations may be useful for survival.
  2. B) Microorganisms carefully control the mutation rate of their DNA to match the environmental conditions and maximize evolution.
  3. C) The increased mutation rate under stressful conditions is an indication that the microorganisms can no longer replicate their DNA properly and are about to die.
  4. D) Constant mutation rates would halt evolution completely.

Answer:  A

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  11.3

39) Chemical mutagens, UV radiation, and ionizing radiation all increase mutation rates, but they have different mechanisms. Which type of mutagen would be best suited for creating large deletions and rearrangements within a genome?

  1. A) chemical mutagens
  2. B) UV radiation
  3. C) ionizing radiation
  4. D) Chemical, UV, and ionizing radiation would create large deletions and rearrangements if used in a very high dose.

Answer:  C

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  11.4

 

40) The SOS system repairs DNA that has gaps, breaks, and other lesions by

  1. A) cutting DNA from other parts of the genome and pasting it into the gaps or damaged areas.
  2. B) stabilizing single-stranded DNA until the next round of normal replication.
  3. C) using specialized DNA polymerases that will synthesize a new DNA strand even if there

is not a normal complementary DNA strand to act as a template.

  1. D) using available mRNA and a special RNA-dependent DNA polymerase to fill in the gaps and replace damaged DNA.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.4

 

 

41) Microinsertions and microdeletions often result in ________ mutations.

  1. A) auxotrophic
  2. B) advantageous
  3. C) silent
  4. D) frameshift

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.2

 

42) When damaged or single-stranded DNA activates the RecA protein, the RecA protein stimulates the cleavage of LexA. This results in

  1. A) repression of polymerase V and activation of endonuclease.
  2. B) activation of the Hfr system.
  3. C) derepression of the SOS system.
  4. D) increased transduction and recombination.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.4

 

43) The F (fertility) plasmid contains a set of genes that encode for the ________ proteins that are essential in conjugative transfer of DNA.

  1. A) pili
  2. B) SOS repair
  3. C) transduction
  4. D) transformation

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.8

44) The designations Phe-, Leu-, and Ser+ refer to an organismʹs

  1. A) plasmid type.
  2. B) genotype.
  3. C) phenotype.
  4. D) mutation type.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.1

 

45) If a bacterium carrying a plasmid that confers resistance to ampicillin is placed into medium without ampicillin, it may

  1. A) gain resistance to other antibiotics.
  2. B) transfer resistance to other cultures in the laboratory.
  3. C) undergo a reversion mutation.
  4. D) lose the plasmid because there is no selection for ampicillin resistance.

Answer:  D

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  11.3

 

46) When DNA is transferred into a prokaryotic cell it may

  1. A) be degraded by enzymes.
  2. B) replicate independent of the host chromosome.
  3. C) recombine with the host chromosome.
  4. D) be degraded by enzymes, replicate independent of the host chromosome, or recombine with the host chromosome.

Answer:  D

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  11.12

 

47) The process in which related DNA sequences from two different sources are exchanged is called

  1. A) transduction.
  2. B) phage conversion.
  3. C) reversion.
  4. D) homologous recombination.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.5

 

48) Integration of linear transforming DNA into the chromosome

  1. A) is not required for the expression for the transformed genes.
  2. B) is catalyzed by the RecA gene.
  3. C) almost never occurs because restriction endonuclease will degrade the DNA before it is integrated.
  4. D) only occurs in laboratory-based systems in artificial competent cells.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.6

49) High-efficiency, natural transformation

  1. A) is common in Bacteria and
  2. B) requires specialized DNA uptake, DNA binding, and integration proteins.
  3. C) is only common in Archaea.
  4. D) usually involves plasmids.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.6

 

 

50) The CRISPR system

  1. A) facilitates homologous recombination through a complex system of proteins and clustered repeats.
  2. B) recognizes foreign DNA sequences that have previously entered the cell and directs the Cas proteins to destroy them.
  3. C) repairs DNA and increases DNA damage tolerance during times of stress.
  4. D) synthesizes gene transfer agents during stationary phase.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.12

 

51) Which of the following is the correct abbreviation for a mutation in a gene that synthesizes one of the enzymes involved in tryptophan production?

  1. A) TrpC
  2. B) trp
  3. C) Trp-
  4. D) trpC1

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.1

 

52) Which of the following is TRUE of competence?

  1. A) It is not required for transformation.
  2. B) It commonly occurs with high efficiency in nature.
  3. C) It cannot occur naturally in bacteria.
  4. D) It requires special proteins such as a cell wall autolysin.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.6

 

53) What is one of the changes that occur when a cell contains an F plasmid that is NOT integrated into the chromosome?

  1. A) It is no longer able to produce a pilus.
  2. B) Cell surface receptors change, preventing the uptake of more plasmids through conjugation.
  3. C) Mutation rates are decreased.
  4. D) The cell is considered an Hfr cell.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.9

11.2   True/False Questions

 

1) UV radiation is a useful tool in producing mutants of microbial cultures.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.4

 

2) When UV radiation damage occurs, DNA repair occurs only in the absence of template instruction.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.4

 

3) Following uptake, transforming DNA becomes attached to a competence-specific protein that prevents it from nuclease attack until it reaches the chromosome.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.6

 

4) The evolutionary significance of phage conversion likely stems from an effective alteration of host cells.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.7

 

5) Bacterial mating (or conjugation) is a bidirectional process where nucleic acids (DNA or RNA) are transferred between two cells.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.8

 

6) Laboratory-based genetic systems have been difficult to develop for Archaea, because they do NOT naturally undergo conjugation or transduction.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.10

 

7) Almost all plasmids are double-stranded DNA.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.8

 

8) Most plasmids are circular rather than linear.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.8

 

9) When the F factor is integrated, all Hfr strains have the origin of replication functioning in the same direction.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.9

 

10) In a prokaryotic genome, either insertion elements OR transposons are present, but both are never present at the same time.

Answer:  FALSE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  11.11

 

11) The use of transposons to generate mutations is a convenient way to create bacterial mutations in the laboratory.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.11

 

12) Transposons can be found on many genetic elements, including plasmids, chromosomes, and viral genomes.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.11

 

13) In transformation experiments using a variety of Bacteria, it has been noted that essentially all of the cells in a population can become competent.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.6

 

14) In specialized transduction, virtually any genetic marker can be transferred from donor to recipient.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.7

 

15) In specialized transduction, as exemplified by lambda phage in E. coli, transduction occurs at high efficiency for only a restricted group of genes near the insertion site of lambda.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.7

 

16) Many Bacteria isolated from nature are natural lysogens.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.7

 

17) Lysogeny is essential for the virulence of many pathogenic bacterial strains.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.7

 

 

18) Proteins and nucleic acids absorb light maximally at 260 nm; hence, proteins protect cells from UV effects.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.4

19) Insertion sequences are found on both ends of transposons and encode for transposase.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  10.11

 

20) Toxigenicity in Corynebacterium diphtheriae is due to phage conversion.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.7

 

21) All Hfr strains possess an F factor integrated into the host chromosome.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.9

 

22) Intercalating agents, like acridine orange and ethidium bromide, lead to mutagenesis by pushing DNA base pairs apart, which can lead to insertions or deletions.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.4

 

23) A typical mutation rate for a bacterium is in the range of 10-6 to 10-7 per kbp.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  10.3

 

24) Archaea have a unique form of conjugation involving cytoplasmic bridges for bidirectional transfer of DNA.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.11

 

25) A phage can be infectious even if all of its DNA has been replaced by bacterial DNA.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.7

 

11.3   Essay Questions

 

1) Design an experiment (including controls) to determine if a new kitchen chemical cleaner is mutagenic. Include in your answer how you would interpret the results.

Answer:  The Ames test could be used to determine if the cleaner solution causes mutations in bacteria. An auxotrophic mutant strain (e.g., Salmonella enterica His-) containing a point mutation is most helpful for this test, so it can measurably revert in small populations of cells. A His-cell population exposed to various concentrations of the cleaner solution would then be put into agar plates lacking histidine, and the number of colonies that grew would represent cells that mutated back to wildtype (His +). An important control for the test is to plate an equal number of cells not exposed to the cleaner solution on to the same medium to calculate a baseline mutation rate of the cells when not exposed to the chemical cleaner. If there are significantly more His+ revertants that grew after exposure to the cleaner solution, then the chemical is mutagenic.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  11.2

 

2) Explain how electroporation works and the significance of the procedure for genetic experiments.

Answer:  Electroporation involves pulsing cells with a high voltage. The brief electrical charge makes cells temporarily permeable such that foreign DNA (e.g., plasmids) can be taken up. Electroporation can be used to introduce circular pieces of dsDNA into cells that are not naturally competent, such as gram-negative bacteria and some species of Archaea. This process yields artificially constructed transformants and is a key step in genetic manipulation.

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  11.6

 

3) Compare and contrast generalized transduction with specialized transduction.

Answer:  Transduction as a whole is the transfer of genetic information from one cell to another mediated by a virus. Generalized transduction allows for any portion of a cellʹs genome to be transferred, but this process occurs at a very low frequency (roughly 10-6 to 10-8). Specialized transduction transfers only genetic material adjacent to the integrated phage (prophage), however its transfer efficiency is much higher than in generalized transduction.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  11.7

 

4) Consider the map shown below of a portion of a prokaryotic genome. How might the elements indicated affect the stability and structure of the genome? Assume that IS2 encodes for a transposase that catalyzes conservative transposition. Re-draw the genome map to illustrate your prediction if necessary.

 

| IS2 |   gene1   gene2   gene3   | IS2 |   geneA  geneB  geneC  | IS2 |   geneD  | IS2 |   geneX

Answer:  Answers should indicate that having multiple IS2 elements in the genome could result in the deletion and/or rearrangement of gene order by transposition events. Composite transposons can form if the genes between to insertion sequences move, or transpose, with the insertion sequences. For instance, geneD could be removed from this part of the genome and inserted elsewhere. OR geneABCD could all move together to another part of the genome through transposition. Similarly gene123 could move together, or even gene123 and geneABCD (but not geneX). If conservative transposition is the only possibility then the genes will move from one spot to another in the genome, but if replicative transposition occurred the genes could be copied to another part of the genome.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  11.11

 

5) Discuss the importance of homologous recombination and transposition in natural horizontal gene transfer and evolution. Be sure to define all of the terms you use and connect them together logically in your answer.

Answer:  Natural horizontal gene transfer is the movement of genes between cells that are not direct descendants of each other. Three mechanisms (transduction, transformation, and conjugation) can transfer genetic material between cells. Once the genetic material is in a cell, it is not necessarily passed down to daughter cells unless it can replicate on its own or is integrated into the genome. Homologous recombination combines similar DNA molecules from two different sources, thus homologous recombination aids in the integration of transferred genetic material into the recipient cell. Transformation also can move segments of DNA between two genetic elements, thus transformation also aids in the movement of genes from temporary or separate genetic elements into the genome of a recipient. Without homologous recombination and transposition, horizontal gene transfer would result in fewer permanent or heritable genetic changes. Thus without homologous recombination and transposition, horizontal gene transfer would have a lesser effect on evolution.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  11.11

 

6) Why is the outcome of a base-pair substitution mutation variable? List and describe the possible outcomes of a base-pair substitution mutation in your answer.

Answer:  The outcome of a base-pair substitution mutation is variable because multiple codons can code for the same amino acid. For example, ACC, ACG, and ACU all encode for alanine (codon examples do not necessarily have to be correct, but should convey the concept). Often the third position of the codon can change without changing the amino acid. In addition, some amino acids have similar properties (e.g., leucine and isoleucine), thus changing particular amino acids may not have much of a phenotypic affect. There are three possible outcomes of a point mutation. Silent mutations can have slightly modified protein structures or unmodified ones, as long as the nucleotide change does not result in any change of the phenotype. Missense mutations are more restricted in that a nucleotide change results in a particular amino acid (codon) change. Although a missense mutation will change the amino acid sequence, this again may or may not change the structure of the protein and the phenotype. A nonsense mutation is the most specific of all where the nucleotide sequence now contains an inserted non-native stop site. The premature halt of protein synthesis will result in a shortened protein structure and will likely change the phenotype of the organism.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  11.2

 

7) How does the SOS system simultaneously fix damage DNA and increase the mutation rate? How does the SOS system increase the survival of microorganisms?

Answer:  The SOS system causes the expression of a number of proteins, and specifically DNA polymerases, that fix lesions in DNA molecules but are prone to errors or do not use templates at all to synthesize new DNA. This allows the organism to survive DNA damage by fixing large lesions or breaks in the DNA but also results in mutations. Even under some conditions that do not result in a lot of DNA damage, the SOS system proteins can increase mutation rate. A slight increase in mutation rate may increase the probability of an advantageous mutation in one cell (even if many other cells are disadvantaged or killed by mutations). The cell with the advantageous mutation would be better able to survive the stress and could go on to replicate and take over the population, thus increasing the survival of the microorganism.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  11.4

 

8) A mutant you made in the laboratory with 5-bromouracil suddenly regains the wild-type phenotype. You discuss this phenomenon with your advisor and colleagues, and they suggest you try transposon mutagenesis to avoid this problem and create stable mutants. Why would mutants created with 5-bromouracil be more likely to regain the wild-type phenotype than mutants created via transposon mutagenesis?

Answer:  Because a point mutation involves only a single base pair change, reversion is quite common relative to fixing an insertion of several nucleotides. Also common in Bacteria are suppressor mutations, which suppress the initial mutation and restore the wild-type phenotype (but not genotype) of a point mutation. However no analogous mechanism works to fix large-scale insertions or deletions.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  11.11

 

9) The CRISPR system is said to protect or preserve genome integrity in prokaryotic cells by acting as a type of ʺimmune system.ʺ Explain how the CRISPR system works and its limitations.

Answer:  The CRISPR (clustered regularly interspaced short palindromic repeats) system is based on a region of the genome that contains DNA sequences from previously encountered foreign DNA separated by short palindromic repeats. The CRISPR region is translated into a long RNA molecule and then cleaved at each palindromic repeat to create short RNA molecules (crRNAs) that are complementary to foreign nucleic acids. If the crRNA binds to a piece of nucleic acid, associated proteins called Cas proteins will destroy the foreign nucleic acid and prevent it from recombining with the genome or replicating. The CRISPR system only works if the foreign DNA has homology with the DNA sequences in the CRISPR region of the genome. This means that novel phages or other DNA sequences would not be recognized or destroyed. Another limitation is that the CRISPR system might destroy DNA that contains useful genes or sequences if it were allowed to recombine with the genome. This could slow down evolution and adaptation.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  11.12

 

10) How might mutations in the dnaQ sequence, whose enzyme is involved in DNA proofreading, be beneficial for a microbe in a highly competitive and dynamic community?

Answer:  Mutations in dnaQ would result in an increase in mutation rate; however unrepaired mutations are not always lethal. In this situation, mutagenesis might be thought of as creating diversity of a microbeʹs genome. Genetic diversity might create a genotype different enough that a new phenotype arises. If for example the new phenotype is an ability to catabolize a particular compound, this could provide the cell with the ability to survive using a different substrate for energy when others are limited. Also, if a certain cell surface component is the target for phage infection and lysis, a different gene sequence could modify the surface component to perhaps enhance survivability in the environment.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  11.4

Brock Biology of Microorganisms, 15e (Madigan et al.)

Chapter 13   Microbial Evolution and Systematics

 

13.1   Multiple Choice Questions

 

1) The earliest stromatolites were probably formed by

  1. A) anoxygenic phototrophs.
  2. B) anoxygenic lithotrophs.
  3. C) oxygenic phototrophs.
  4. D) oxygenic lithotrophs.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.2

 

2) What two gases were most abundant on early Earth?

  1. A) O2 and CO2
  2. B) N2 and H2
  3. C) CO2 and H2
  4. D) CO2 and N2

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.1

 

3) Compared with today, the temperature on Earth during its first half-billion years was probably

  1. A) considerably warmer.
  2. B) considerably colder.
  3. C) about the same as today.
  4. D) about the same as today on average, but the diurnal fluctuations were much greater.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.1

 

4) The earliest RNA probably functioned in

  1. A) catalysis.
  2. B) self-replication.
  3. C) both catalysis and self-replication.
  4. D) neither catalysis nor self-replication.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.1

 

 

5) Chemical reactions involving ________ have been proposed as energy-yielding reactions for the earliest organisms on Earth.

  1. A) Fe, O2, and H2
  2. B) S, O2, and H2
  3. C) S, and O2
  4. D) S, and H2

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.1

6) Evolution is driven by

  1. A) random mutation.
  2. B) novel metabolic pathways.
  3. C) selection pressure.
  4. D) selection pressure applied to random mutation.

Answer:  D

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  13.5

 

7) Microorganisms were probably restricted to the oceans and subsurface environments until

  1. A) aquatic life brought them onto land.
  2. B) chemoorganotrophy developed.
  3. C) phototrophy evolved.
  4. D) the ozone layer was made.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.2

 

8) Which of the following is an assumption used in the molecular clock approach?

  1. A) Nucleotide changes accumulate in a sequence in proportion to time.
  2. B) Nucleotide changes are generally NOT transferred to progeny.
  3. C) Nucleotide changes are NOT random.
  4. D) Nucleotide changes are subject to intense selective pressure.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.5

 

9) An important molecule in sequence-based evolutionary analyses of microbes are ________ genes.

  1. A) ATPase
  2. B) electron transport
  3. C) SSU rRNA
  4. D) tRNA

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.7

 

10) A monophyletic group is a group that

  1. A) descended from one ancestor.
  2. B) has the same fitness level.
  3. C) possesses one taxonomic trait that is the same.
  4. D) shares one phylogenetic marker.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.8

11) A gene for a specific trait may have more than one form, allowing the trait to vary. These sequence variants of a gene are called a(n)

  1. A) alleles.
  2. B) MLST.
  3. C) horizontal gene transfers.
  4. D) ribotypes.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.5

 

12) A key concept in evolution is that all mutations are

  1. A) random.
  2. B) neutral.
  3. C) deleterious.
  4. D) either deleterious or beneficial.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.5

 

13) LUCA is

  1. A) the last universal common ancestor.
  2. B) the individual ancestor of each of the three domains.
  3. C) actually somewhat of a misnomer because it is now believed that each of the domains arose independently.
  4. D) All of the answers are correct.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.3

 

 

14) Horizontal gene transfer

  1. A) is so rare over evolutionary history that it is not considered when examining microbial evolution.
  2. B) occurs within bacterial species.
  3. C) complicates the construction of phylogenetic trees and the interpretation of specific traits in relation to evolution.
  4. D) only affects the evolution of plasmids.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.3

 

15) Molecular sequencing suggests that mitochondria arose from a group of prokaryotic organisms within the

  1. A) Crenarchaeota.
  2. B) cyanobacteria.
  3. C) iron-oxidizing bacteria.
  4. D) Alphaproteobacteria.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.4

16) Which statement most closely expresses our present understanding?

  1. A) The chloroplast is an ancestor of the cyanobacteria.
  2. B) The cyanobacteria are descendents of the chloroplast.
  3. C) The chloroplast arose from the incorporation of a cyanobacterial-like organism.
  4. D) The chloroplast and the cyanobacteria are not closely (or specifically) related.

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.4

 

17) Two tubes are inoculated from the same tube containing a bacterial culture. The cultures are then transferred every day for 2 months. All of the media and growth conditions are the same in every tube. After 2 months of cultivation, the fitness and genotype frequencies of the populations in the two tubes are compared. The fitness of the two cultures is the same, but the genotype frequencies are very different in the two cultures. How is this possible?

  1. A) Two months is not long enough for different fitness levels to evolve even if the genotype frequencies change.
  2. B) This result is not possible because different genotype frequencies would result in different fitness levels under the same growth conditions.
  3. C) Natural selection caused the evolution of different genotype frequencies within the separate test tubes.
  4. D) Genetic drift within the small populations in the test tubes resulted in different genotype frequencies.

Answer:  D

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  13.5

 

18) If you allowed 10 identical parallel Salmonella cultures to evolve for 10,000 generations under new growth conditions with very little nitrogen, the parallel cultures would

  1. A) evolve identical adaptations to use the nitrogen source provided in the media.
  2. B) not change or adapt very much over this small number of generations.
  3. C) each accumulate different random mutations resulting in different adaptations to use the nitrogen in the media.
  4. D) direct mutations to occur in nitrogen utilization and uptake genes in order to adapt rapidly to the culture conditions.

Answer:  C

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  13.5

 

19) Both deletions and insertions occur during the evolution of microbial genomes. Insertions bring in new genes that may be useful for the cell. Deletions

  1. A) may increase fitness of a microorganism by eliminating unneeded genes.
  2. B) always result in a severe loss of fitness for the microorganism.
  3. C) are uncommon because they are usually lethal.
  4. D) always result in a severe loss of fitness but keep microbial genomes compact.

Answer:  A

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  13.6

20) Polyphasic taxonomy uses methods that include

  1. A) phenotypic methods.
  2. B) genotypic methods.
  3. C) phylogenetic methods.
  4. D) phenotypic, genotypic, and phylogenetic methods.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.6

 

21) Ribotyping

  1. A) bypasses sequencing and sequence alignments.
  2. B) exploits unique DNA restriction patterns.
  3. C) allows discrimination between species and different strains of a species.
  4. D) bypasses sequencing and sequence alignments, exploits unique DNA restriction patterns, and allows discrimination between species and different strains of a species.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.9

 

 

22) Which is NOT a characteristic of a ʺprimitiveʺ state of microbial evolution?

  1. A) hyperthermophilic
  2. B) aerobic metabolism
  3. C) small genome
  4. D) branching near the root of the evolutionary tree of life

Answer:  B

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  13.1

 

23) Because DNA-DNA hybridization reveals subtle differences in genes, it is useful for differentiating organisms from different

  1. A) domains.
  2. B) families.
  3. C) orders.
  4. D) strains or species.

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.8

 

24) Which taxonomic tool would scientists use if they wanted to determine if an outbreak of food poisoning was caused by a particular strain of a pathogen?

  1. A) fluorescence in situ hybridization
  2. B) DNA-DNA hybridization
  3. C) multilocus sequence typing
  4. D) fatty acid methyl ester analysis

Answer:  C

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  13.9

25) It is believed that phototrophy arose approximately 3.3 billion years ago in

  1. A) Bacteria.
  2. B) Archaea.
  3. C) Eukarya.
  4. D) LUCA.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.2

 

26) The oxic atmosphere created conditions that led to the evolution of various new metabolic pathways, such as

  1. A) sulfide oxidation.
  2. B) nitrification.
  3. C) iron oxidation.
  4. D) sulfide oxidation, nitrification, and iron oxidation.

Answer:  D

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  13.2

 

27) You are studying 12 new isolates from the human skin. Their average nucleotide identity for shared orthologous genes is 97%.The isolates would most likely be

  1. A) classified as individual strains of the same species.
  2. B) classified as individual species of the same genus.
  3. C) split into different families.
  4. D) classified as the same species if they can mate via conjugation.

Answer:  A

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  13.8

 

28) The pan genome of a microbial species is constantly changing because of

  1. A) bottleneck events.
  2. B) horizontal gene transfer.
  3. C) substitutions.
  4. D) substitution and bottleneck events.

Answer:  B

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  13.6

 

29) Microbial ________ studies the diversity of microorganisms and links their phylogeny with ________.

  1. A) taxonomy / genotype
  2. B) phylogeny / phenotype
  3. C) taxonomy / phenotype
  4. D) systematics / taxonomy

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.7

30) The earliest photosynthetic microbes, before the cyanobacterial lineage developed, oxidized substances other than water. What was produced by these microbes instead of oxygen?

  1. A) nitrate
  2. B) elemental sulfur
  3. C) ferrous iron
  4. D) hydrogen sulfide

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.2

 

 

31) Based on the phylogenetic tree below, which of the following statements is FALSE?

 

  1. A) Species E is more closely related to species A than to species D.
  2. B) Species D is more closely related to species C than to species E.
  3. C) Species F is more closely related to species D than to species E.
  4. D) Species C is more closely related to species B than to species D.

Answer:  B

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  13.7

 

32) The early energy reactions used hydrogen, which is a powerful

  1. A) oxidant.
  2. B) electron donor.
  3. C) electron acceptor.
  4. D) stabilizer.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.1

33) Oxygen did not accumulate in the early atmosphere until it reacted with reduced materials, especially ________, in the oceans.

  1. A) hydrogen
  2. B) elemental sulfur
  3. C) ferrous iron
  4. D) nitrate

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.2

 

 

34) As oxygen appeared in the atmosphere, ________ also accumulated, which formed a protective barrier that protects the Earth from ________.

  1. A) ozone / UV radiation
  2. B) elemental sulfur / volcanism
  3. C) ferrous iron / hydroxylating radicals
  4. D) sulfate / hydrogen sulfide

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.2

 

35) Two eukaryotic organelles that are hypothesized to be the result of endosymbiosis are the ________ and the ________.

  1. A) nucleus / Golgi body
  2. B) mitochondrion / chloroplast
  3. C) endoplasmic reticulum / Golgi body
  4. D) hydrogenosome / chloroplast

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.4

 

36) Microbial species have a core genome and a pan genome. What is the difference between the two?

  1. A) The core genome consists of all the nucleic acid polymerases and translation enzymes, while the pan genome consists of all the biosynthetic pathways.
  2. B) The core genome is the set of genes within the mitochondria, while the pan genome is the set of genes in the nucleus of a species.
  3. C) The core genome is a set of genes shared by all members of a species, while the pan genome includes the core genes as well as genes that are not shared by all members.
  4. D) The core genome is the set of genes introduced by horizontal gene transfer, while the pan genome is the set of genes that is not transferred horizontally.

Answer:  C

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  13.6

37) What characteristics make a gene a good candidate for determining the evolutionary relationships between organisms?

  1. A) highly conserved
  2. B) universally distributed
  3. C) transferred horizontally between species
  4. D) highly conserved and universally distributed

Answer:  D

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.7

 

 

38) Which of the following statements is FALSE concerning the universal tree of life?

  1. A) Previous versions of the universal tree of life were based largely on fossils and comparative biology, which overlooked the diversity of and relationships between most prokaryotes.
  2. B) The current universal tree of life is supported by multiple genes including SSU rRNA sequences.
  3. C) The current universal tree of life depicts three domains of life, two of which are prokaryotic.
  4. D) The current universal tree of life is based on molecular sequences and completely changes our view of evolutionary relationships between eukaryotic species.

Answer:  D

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  13.3

 

39) ________ formed the semipermeable membrane-like surfaces for the earliest life forms.

  1. A) Proteins
  2. B) Hydrogen sulfide
  3. C) Lipid bilayers
  4. D) RNA

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.4

 

40) The first catalytic and self-replication biological molecule was most likely

  1. A) RNA.
  2. B) DNA.
  3. C) proteins.
  4. D) ATP.

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.1

 

41) The evolutionary history of a group of organisms is called its ________ and it is inferred from ________.

  1. A) taxonomy / phenotype
  2. B) phylogeny / nucleotide sequence data
  3. C) phylogeny / phenotype
  4. D) taxonomy / morphology

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.3

 

42) One widely used method of examining evolutionary relationships is ________, which is based on the assumption that evolution is most likely to have proceeded by the path requiring fewest changes.

  1. A) topology
  2. B) neighbor joining
  3. C) parsimony
  4. D) systematics

Answer:  C

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.7

 

43) Multilocus sequence typing (MLST) involves sequencing several different

  1. A) genomes.
  2. B) housekeeping genes.
  3. C) tRNA genes.
  4. D) rRNA genes.

Answer:  B

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.9

 

44) Systematic analysis now commonly includes ________ to identify, characterize, and determine relationships between new strains of bacterial species.

  1. A) whole genome analysis
  2. B) microscopy
  3. C) staining
  4. D) pigments

Answer:  A

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.9

 

13.2   True/False Questions

 

1) The atmosphere of primitive Earth is usually described as an oxidizing atmosphere.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.2

 

2) The earliest nucleic acid was probably a simple DNA molecule.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.1

 

3) The establishment of DNA as the genome of the cell may have resulted from the need to store genetic information in a more stable form than RNA.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.1

 

4) Eukaryotes originated after the rise in atmospheric oxygen.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.4

5) A type strain is the first strain of a new species to be identified and described.

Answer:  TRUE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  13.10

 

6) Sequencing technology and molecular phylogenetic analyses have had very little impact on our understanding of the evolution and diversity of life on Earth.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.9

 

7) Molecular phylogeny and rRNA analysis provided the evidence used to separate Bacteria and Archaea into distinct domains.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.3

 

8) Organisms with greater phylogenetic distance in their genomes have less gene exchange than those with less phylogenetic distance.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.6

 

9) Microbial species are difficult to define because they are seldom monophyletic.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.8

 

10) Organisms within a species should have strong phenotypic cohesiveness.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.8

 

11) At present there are four phyla in the domain Bacteria.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.3

 

 

12) The universal phylogenetic tree shows that the Bacteria diverged from the Archaea before the Eukarya diverged from the Archaea.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.3

 

13) The 16S rRNA gene sequence is an approximately 1,465 bp linear strand of single stranded RNA.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.7

14) DNA-DNA hybridization is a sensitive method for revealing subtle genetic differences because it measures the degree of sequence similarity between two genomes.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.8

 

15) One phenotypic trait used for species identification and description is the analysis of the types and proportions of the fatty acids present in the cytoplasmic and outer membranes. The methodology employed is often nicknamed FAME.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.9

 

16) In taxonomy, family is a more general term than order.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.10

 

17) Oxygen was a driving factor in the formation of eukaryotic cells.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.2

 

18) The primary domains were founded based on comparative ribosomal RNA gene sequencing.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.3

 

19) All unicellular organisms belong to the same domain of life.

Answer:  FALSE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.3

 

 

20) Approximately 3.7 billion years ago Archaea and Bacteria diverged as being distinct from each other.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.3

 

21) When exposed to UV light, oxygen gas produces ozone gas.

Answer:  TRUE

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.2

 

22) Horizontal gene transfer is one plausible explanation as to why organisms in Archaea, Bacteria, and

Eukarya still share so many genes among such distinct domains.

Answer:  TRUE

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  13.3

13.3   Essay Questions

 

1) Explain the endosymbiotic theory. What is the evolutionary value of endosymbiosis?

Answer:  The theory states that a chemoorganotrophic bacterium and a cyanobacterium were stably incorporated into a cell and gave rise to a mitochondrion and chloroplast (respectively), which formed a eukaryotic cell. Endosymbiosis provides a mutualistic relationship where the incorporated cells are protected from direct contact with the environment, and the cell that incorporated the cells gains metabolic diversity.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  13.4

 

2) Describe the hydrogen hypothesis and why it is favored over other hypotheses.

Answer:  The hydrogen hypothesis states that a hydrogen-consuming archaea served as a host for the stable incorporation of a hydrogen-producing facultative anaerobic bacterium. This hypothesis describes the first eukaryotic cells as ones without nuclei. It accounts for eukaryotes having bacteria-like ATP-producing pathways in hydrogenosomes and the presence of bacteria-like lipids in eukaryotes. Neither of these is accounted for in other hypotheses that state that eukaryotes first contained nuclei and lacked mitochondria.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  13.4

 

 

3) Carl Woese was an important figure in microbial classification and changed our understanding of the evolution and diversity of life. What breakthrough did he make that allowed us to infer evolutionary relationships between organisms? What assumptions was his breakthrough based on?

Answer:  During the 1970s, Dr. Carl Woese pioneered the usage of SSU rRNA genes for phylogenetic analysis. He realized that SSU rRNA genes could be used to infer evolutionary relationships because they were (1) universally distributed, (2) have the same function in every organism, (3) change slowly over evolutionary time (highly conserved), and (4) of adequate length to provide sufficient information about very deep, or old, evolutionary relationships. He assumed that changes in DNA sequence are reliably passed on to offspring (they are heritable) and that nucleotide changes between two organisms will accumulate over time since they last shared a common ancestor. Homologous sequence comparison thus allows us to infer evolutionary relationships between nucleotide sequences. Genes that are not often exchanged through horizontal gene transfer (such as SSU rRNA or other housekeeping genes) can be used to infer evolutionary relationships between organisms.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  13.3

4) Why is the biological species concept useful for Eukarya but not meaningful for Archaea and Bacteria? How are prokaryotic species defined?

Answer:  A species is an interbreeding population of organisms that is isolated from other interbreeding populations, and this definition generally works well in defining eukaryotic species that reproduce sexually. However, prokaryotes undergo asexual reproduction and are subject to horizontal gene transfer in the environment, which complicates the definition of a species in Archaea and Bacteria. Prokaryotic species are defined as a monophyletic group of strains that share important phenotypic traits and have relatively high genomic and SSU rRNA sequence similarity. (Answers could include the accepted values of greater than 70% DNA-DNA hybridization and 97% SSU rRNA sequence similarity but the concept of similarity of a specific level and a monophyletic origin is most important.)

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  13.8

 

5) You have isolated a microbe from an environmental sample. The microbe has the ability to perform a new metabolic reaction at a very low temperature, so you are excited that it could be a new species. What experiments should you perform to determine if your isolate is truly a new species? What process will you follow to officially name your isolate?

Answer:  Experiments to determine the isolateʹs characteristics and distinguishing traits must be performed which, among others, involve a series of biochemical and physiological tests. The morphology and cell wall structure should also be determined. To define it as a species and determine its relationship to other organisms the SSU rRNA and DNA. DNA hybridization values in comparison to closely related organisms must be determined. If the isolate is sufficiently different, viable cultures must be sent to at least two international culture collections. A manuscript can then be published describing the organism and a proposed genus and species name.

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.9

 

6) Rhodobacter cells perform photosynthesis in the presence of light and grow heterotrophically in the dark. After being cultured in total darkness for multiple generations, phenotypic and genotypic changes occur. What phenotypic changes occur and what evolutionary processes are driving them?

Answer:  When a population of phototrophs is grown in the dark for multiple generations, the population will tend to lose the ability to produce photopigments and may even lose the ability to perform photosynthesis altogether. This occurs because there is no selection for the production of the pigments and enzymes necessary for photosynthesis. The production of photosynthetic pigments and enzymes is metabolically expensive, thus producing them when they are not needed results in a competitive disadvantage in the dark cultures, thus random mutants that produce fewer pigments take over Rhodobacter cultures grown in the dark. The evolutionary processes behind this phenotypic change are random mutations to the genome, followed by the selection of mutants that grow better in the dark.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  13.5

7) How did cyanobacteria, oxygen, and ozone impact the evolution of eukaryotic cells?

Answer:  The first phototrophs on Earth were anoxygenic bacteria, meaning that oxygen was NOT produced during photosynthesis. Eventually the Cyanobacteria evolved and oxygenic photosynthesis began to produce oxygen—the first major source of oxygen on Earth. As the oxygen was produced, it first reacted with all of the reduced chemicals found on early Earth, but eventually it began to accumulate in the atmosphere. As the oxygen gas accumulated, UV light converted the oxygen into ozone. Ozone absorbs UV irradiation and the formation of the ozone layer protected the Earthʹs surface from strong UV irradiation. Because UV irradiation damages DNA and other biomolecules, before the ozone layer life could only exist in protected environments, such as deep in the ocean or in the subsurface. The production of oxygen by Cyanobacteria thus did two very important things that may have helped eukaryotic and multicellular organisms evolve. The oxygen was a powerful oxidant and excellent source of energy (excellent electron acceptor for respiration) and the large majority of eukaryotic cells use aerobic respiration to generate large amounts of energy for cellular growth. All of this energy may have fueled the growth and diversity of multicellular organisms. In addition, the protection of the ozone layer allowed life to expand into new environments.

Bloom’s Taxonomy:  5-6: Evaluating/Creating

Chapter Section:  13.2

 

 

8) Life is believed to have originated at hydrothermal springs on the ocean floor rather than on or near the surface of Earth. What conditions made the hydrothermal springs the likely place for the beginning of life? Relate the conditions of the early oceans to the likely metabolism of the first cells.

Answer:  Aside from the temperature, the environmental conditions of Earth would have been less hostile for life in hydrothermal springs than on land where exposure to UV light would have destroyed biological molecules. The springs also likely provided a steady supply of energy, such as H2 and H2S, and where the hot water interfaced with colder water likely served to form precipitates that eventually became life. The metabolism of the first cells did not depend on oxygen, as the atmosphere lacked molecular oxygen, but probably utilized electron transfer reactions between hydrogen, hydrogen sulfide, and reduced iron to produce energy.

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.1

 

9) How does the Black Queen Hypothesis explain the effect of gene deletions on bacterial evolution?

Answer:  The Black Queen Hypothesis states that some organisms can increase their fitness by selection loss (deletion) of specific genes. The loss of the genes decreases the genome and proteome size and thus the amount of energy it takes to reproduce, increasing the organismʹs fitness. However, it also makes the organism dependent on other members of the community to provide the functions or gene products that have been deleted. The dependent organisms that have lost genes will only be competitive as long as others in the community provide the functions/products that have been lost. Thus in environments with a stable community composition, organism may evolve smaller genomes through gene deletion. This has been observed in the open ocean and in some symbiotic organisms.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  13.6

10) Explain the strengths and limitations of using 16S rRNA for phylogenetic analyses.

Answer:  Due to function constraints, the 16S rRNA gene sequence has changed relatively little over time and is therefore an excellent individual nucleotide sequence useful in phylogenetic studies. It is not especially good at distinguishing among different species within a genus, because there is often not enough variation between the sequences. There are large and freely available databases dedicated to cataloging 16S rDNA sequences, and it remains the ʺgold standardʺ today for phylogenetic analyses.

Bloom’s Taxonomy:  1-2: Remembering/Understanding

Chapter Section:  13.7

 

 

11) Can gene frequencies change in the absence of selection? Why or why not?

Answer:  Gene frequencies do change in the absence of selection. Genetic drift is a random process through which gene frequencies change. By chance, some members of a population will have more offspring than others, while other members may be killed or inhibited by CHANCE, not selection. An example is the random selection of very few cells from one test tube culture to another test tube. Which cells are transferred will affect which genes are passed on to offspring, but the cells transferred were random and unrelated to the fitness of the cells. Thus gene frequencies can change (drift) randomly over time even when there is no selection on that particular gene. Although genetic drift is separate from selection, it is still an important part of evolution.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  13.5

 

12) What type of methods are used in microbial systematics? Why are multiple methods used to characterize and classify microbial species? Be sure to define the methods and describe the general strengths and weaknesses of each type.

Answer:  Microbial systematics is polyphasic and uses three types of methods to identify and describe microbial species: (1) phenotypic — the observed characteristics of an organism, (2) genotypic — the genetic features of an organism, and (3) phylogenetic — the evolutionary relationship of an organism to other organisms. Phenotypic methods are important because they describe the important traits of an organism that define the organismʹs function and impact on humans or the environment. However, many phenotypic traits are not unique to a species, so identifying the huge number of microbial species based only on phenotypic traits is not feasible. Genetic methods complement phenotypic methods and help us quickly characterize both gene content and genetic similarity to other organisms. Lastly, phylogenetic methods help to organize microbial species in a way that is congruent with our understanding of evolution, and thus give both structure and meaning to the taxonomy of microorganisms.

Bloom’s Taxonomy:  3-4: Applying/Analyzing

Chapter Section:  13.9