Microbiology 13th Edition Tortora – Test Bank

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Microbiology: An Introduction, 13e (Tortora et al.)

Chapter 5   Microbial Metabolism

 

5.1   Multiple-Choice Questions

 

1) Which of the following compounds is NOT an enzyme?

  1. A) dehydrogenase
  2. B) cellulase
  3. C) coenzyme A
  4. D) β-galactosidase
  5. E) sucrase

Answer:  C

Section:  5.2

Bloom’s Taxonomy:  Understanding

Learning Outcome:  5.3

 

2) Figure 5.1

 

 

Which compound is being reduced in the reaction shown in Figure 5.1?

  1. A) isocitric acid and α-ketoglutaric acid
  2. B) α-ketoglutaric acid and NAD+
  3. C) NAD+
  4. D) NADH
  5. E) NADH and isocitric acid

Answer:  C

Section:  5.3

Bloom’s Taxonomy:  Understanding

Learning Outcome:  5.8

Global Outcome:  3

 

 

3) Which organism is NOT correctly matched to its energy source?

  1. A) photoheterotroph – light
  2. B) photoautotroph – CO2
  3. C) chemoautotroph – Fe2+
  4. D) chemoheterotroph – glucose
  5. E) chemoautotroph – NH3

Answer:  B

Section:  5.9

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  3.1

Learning Outcome:  5.23

Global Outcome:  2

 

4) Which of the following statements about anaerobic respiration is FALSE?

  1. A) It yields lower amounts of ATP when compared to aerobic respiration.
  2. B) The complete Krebs cycle is utilized.
  3. C) It involves the reduction of an organic final electron acceptor.
  4. D) It generates ATP.
  5. E) It requires cytochromes.

Answer:  C

Section:  5.4

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  3.1

Learning Outcome:  5.15

 

 

5) Figure 5.2

 

 

What type of reaction is in Figure 5.2?

  1. A) decarboxylation
  2. B) transamination
  3. C) dehydrogenation
  4. D) oxidation
  5. E) reduction

Answer:  B

Section:  5.10

Bloom’s Taxonomy:  Analyzing

Learning Outcome:  5.24

Global Outcome:  2

 

6) What is the fate of pyruvic acid in an organism that uses aerobic respiration?

  1. A) It is reduced to lactic acid.
  2. B) It reacts with oxaloacetate to form citrate.
  3. C) It is oxidized in the electron transport chain.
  4. D) It is catabolized in glycolysis.
  5. E) It is converted into acetyl CoA.

Answer:  E

Section:  5.4

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.15

 

 

7) Figure 5.3

 

 

How would a noncompetitive inhibitor interfere with a reaction involving the enzyme shown in Figure 5.3?

  1. A) It would bind to a.
  2. B) It would bind to b.
  3. C) It would bind to c.
  4. D) It would bind to d.
  5. E) The answer cannot be determined based on the information provided.

Answer:  B

Section:  5.2

Bloom’s Taxonomy:  Understanding

Learning Outcome:  5.6

 

8) Figure 5.4

 

 

How is ATP generated in the reaction shown in Figure 5.4?

  1. A) glycolysis
  2. B) fermentation
  3. C) photophosphorylation
  4. D) oxidative phosphorylation
  5. E) substrate-level phosphorylation

Answer:  E

Section:  5.3

Bloom’s Taxonomy:  Understanding

Learning Outcome:  5.9

Global Outcome:  2

 

9) Fatty acids are oxidized in

  1. A) the Krebs cycle.
  2. B) the electron transport chain.
  3. C) glycolysis.
  4. D) the pentose phosphate pathway.
  5. E) the Entner-Doudoroff pathway.

Answer:  A

Section:  5.5

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  3.1; 6.3

Learning Outcome:  5.17

 

10) Figure 5.5

 

 

Which of the graphs in Figure 5.5 best illustrates the activity of an enzyme that is saturated with substrate?

  1. A) a
  2. B) b
  3. C) c
  4. D) d
  5. E) e

Answer:  C

Section:  5.2

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  7.2

Learning Outcome:  5.5

Global Outcome:  3

 

11) Which of the following is the best definition of oxidative phosphorylation?

  1. A) Electrons are passed through a series of carriers to O2.
  2. B) A proton gradient allows hydrogen ions to flow back into the cells through transmembrane protein channels, releasing energy that is used to generate ATP.
  3. C) ATP is directly transferred from a substrate to ADP.
  4. D) Electrons are passed through a series of carriers to an organic compound.

Answer:  B

Section:  5.3

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.9

 

12) From the list below, which is NOT produced during the Krebs cycle?

  1. A) FADH2
  2. B) NADH
  3. C) ATP
  4. D) NADPH
  5. E) CO2

Answer:  D

Section:  5.4

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.13

 

13) Which of the following statements about photophosphorylation is FALSE?

  1. A) Light liberates an electron from chlorophyll.
  2. B) The oxidation of carrier molecules releases energy.
  3. C) Energy from oxidation reactions is used to generate ATP from ADP.
  4. D) It requires CO2.
  5. E) It occurs in photosynthesizing cells.

Answer:  D

Section:  5.3

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  3.1

Learning Outcome:  5.9

 

14) A bacterium that only possesses the ability to ferment obtains energy

  1. A) by glycolysis only.
  2. B) by aerobic respiration only.
  3. C) by fermentation or aerobic respiration.
  4. D) only in the absence of oxygen.
  5. E) only in the presence of oxygen.

Answer:  A

Section:  5.4

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  3.1

Learning Outcome:  5.16

 

15) The advantage of the pentose phosphate pathway is that it produces all of the following EXCEPT

  1. A) precursors for nucleic acids.
  2. B) precursors for the synthesis of glucose.
  3. C) three ATPs.
  4. D) NADPH.
  5. E) precursors for the synthesis of amino acids.

Answer:  C

Section:  5.4

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  3.1

Learning Outcome:  5.12

 

16) Which biochemical process is NOT used during glycolysis?

  1. A) substrate-level phosphorylation
  2. B) oxidation-reduction
  3. C) carbohydrate catabolism
  4. D) beta oxidation
  5. E) enzymatic reactions

Answer:  D

Section:  5.5

Bloom’s Taxonomy:  Understanding

Learning Outcome:  5.17

Global Outcome:  7

 

17) In noncyclic photophosphorylation, O2 is released from

  1. A) CO2.
  2. B) 2H2
  3. C) C6H12O6.
  4. D) sunlight.
  5. E) chlorophyll.

Answer:  B

Section:  5.7

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  3.1

Learning Outcome:  5.19

 

18) Which of the following is the best definition of fermentation?

  1. A) the partial reduction of glucose to pyruvic acid
  2. B) the partial oxidation of glucose with organic molecules serving as electron acceptors
  3. C) the complete catabolism of glucose to CO2and H2O
  4. D) the production of energy by oxidative-level phosphorylation
  5. E) the production of energy by both substrate and oxidative phosphorylation

Answer:  B

Section:  5.4

Bloom’s Taxonomy:  Understanding

Learning Outcome:  5.16

 

19) All of the following are required in the reactions of microbial respiration EXCEPT

  1. A) electron transport system.
  2. B) cytochromes.
  3. C) a source of electrons.
  4. D) oxygen.
  5. E) final electron acceptor.

Answer:  D

Section:  5.4

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  3.1

Learning Outcome:  5.15

 

20) Increasing the temperature of a reaction will do all of the following EXCEPT

  1. A) increase the reaction rate.
  2. B) increase the number of molecules attaining activation energy.
  3. C) increase the number of molecular collisions.
  4. D) increase the activation energy.
  5. E) increase kinetic energy of the molecules.

Answer:  D

Section:  5.2

Bloom’s Taxonomy:  Understanding

Learning Outcome:  5.5

 

21) In green and purple bacteria, electrons to reduce CO2 can come from

  1. A) CO2.
  2. B) H2
  3. C) C6H12O6.
  4. D) H2
  5. E) chlorophyll.

Answer:  D

Section:  5.9

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  3.1

Learning Outcome:  5.23

 

22) Assume you are growing bacteria on a lipid medium that started at pH 7. The action of bacterial lipases should cause the pH of the medium to

  1. A) increase (become more alkaline).
  2. B) decrease (become more acidic).
  3. C) stay the same.

Answer:  B

Section:  5.5

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  3.2

Learning Outcome:  5.17

 

23) Which of the following uses CO2 for carbon and H2 for energy?

  1. A) chemoautotroph
  2. B) chemoheterotroph
  3. C) photoautotroph
  4. D) photoheterotroph

Answer:  A

Section:  5.9

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  3.1

Learning Outcome:  5.23

 

24) Which of the following uses glucose for carbon and energy?

  1. A) chemoautotroph
  2. B) chemoheterotroph
  3. C) photoautotroph
  4. D) photoheterotroph

Answer:  B

Section:  5.9

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  3.1

Learning Outcome:  5.23

 

25) Which of the following has bacteriochlorophylls and uses alcohols for carbon?

  1. A) chemoautotroph
  2. B) chemoheterotroph
  3. C) photoautotroph
  4. D) photoheterotroph

Answer:  D

Section:  5.9

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  3.1

Learning Outcome:  5.23

 

26) Cyanobacteria are a type of

  1. A) chemoautotroph.
  2. B) chemoheterotroph.
  3. C) photoautotroph.
  4. D) photoheterotroph.

Answer:  C

Section:  5.9

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  3.1

Learning Outcome:  5.23

 

 

27) Which of the following statements are TRUE?

1-Electron carriers are located at ribosomes.

2-ATP is a common intermediate between catabolic and anabolic pathways.

3-ATP is used for the long-term storage of energy and so is often found in storage granules.

4-Anaerobic organisms are capable of generating ATP via respiration.

5-ATP can be generated by the flow of protons across protein channels.

  1. A) 2, 4, 5
  2. B) 1, 3, 4
  3. C) 2, 3, 5
  4. D) 1, 2, 3
  5. E) All of the statements are true.

Answer:  A

Section:  5.1

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.2

 

28) Microorganisms that catabolize sugars into ethanol and hydrogen gas would most likely be categorized as

  1. A) aerobic respirers.
  2. B) anaerobic respirers.
  3. C) heterolactic fermenters.
  4. D) homolactic fermenters.
  5. E) alcohol fermenters.

Answer:  C

Section:  5.4

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  3.1

Learning Outcome:  5.16

Global Outcome:  2

 

29) Which of the following statements regarding metabolism is FALSE?

  1. A) Heat may be released in both anabolic and catabolic reactions.
  2. B) ATP is formed in catabolic reactions.
  3. C) ADP is formed in anabolic reactions.
  4. D) Anabolic reactions are degradative.

Answer:  D

Section:  5.1

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.1

 

 

30) A bacterium such as Pseudomonas uses nitrate as a final electron acceptor in an electron transport system. All the below statements are true EXCEPT

  1. A) the process does not yield as much ATP.
  2. B) they can respire without O2.
  3. C) they require light.
  4. D) the process produces nitrite ion.
  5. E) the process requires an electron donor.

Answer:  C

Section:  5.4

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  3.1

Learning Outcome:  5.15

Global Outcome:  2

 

31) Which of the following statements regarding the Entner-Doudoroff pathway is TRUE?

  1. A) It involves glycolysis.
  2. B) It involves the pentose phosphate pathway.
  3. C) NADH is generated.
  4. D) ATP is generated.
  5. E) NADH and ATP are generated.

Answer:  E

Section:  5.4

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  3.1

Learning Outcome:  5.12

 

32) Assume you are working for a chemical company and are responsible for growing a yeast culture that produces ethanol. The yeasts are growing well on the maltose medium but are not producing alcohol. What is the most likely explanation?

  1. A) The maltose is toxic.
  2. B) O2is in the medium.
  3. C) Not enough protein is provided.
  4. D) The temperature is too low.
  5. E) The temperature is too high.

Answer:  B

Section:  5.4

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  3.2

Learning Outcome:  5.15

 

 

33) Figure 5.6

 

 

The rates of O2 and glucose consumption by a bacterial culture are shown in Figure 5.6. Assume a bacterial culture was grown in a glucose medium without O2. Then O2 was added at the time marked X. The data indicate that

  1. A) these bacteria don’t use O2.
  2. B) these bacteria get more energy anaerobically.
  3. C) aerobic metabolism is more efficient than fermentation.
  4. D) these bacteria cannot grow anaerobically.

Answer:  C

Section:  5.4

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  7.2a

Learning Outcome:  5.15

Global Outcome:  3

 

34) An enzyme, citrate synthase, in the Krebs cycle is inhibited by ATP. This is an example of all of the following EXCEPT

  1. A) allosteric inhibition.
  2. B) competitive inhibition.
  3. C) feedback inhibition.
  4. D) noncompetitive inhibition.

Answer:  B

Section:  5.2

Bloom’s Taxonomy:  Understanding

Learning Outcome:  5.6

 

 

35) If a cell is starved for ATP, which of the following pathways would most likely be shut down?

  1. A) Krebs cycle
  2. B) glycolysis
  3. C) pentose phosphate pathway
  4. D) Krebs cycle and glycolysis

Answer:  C

Section:  5.4

Bloom’s Taxonomy:  Understanding

Learning Outcome:  5.12

 

36) Which of the following statements regarding the glycolysis pathway is FALSE?

  1. A) Two pyruvate molecules are generated.
  2. B) Four ATP molecules are generated via substrate-level phosphorylation.
  3. C) Two NADH molecules are generated.
  4. D) One molecule of ATP is expended.
  5. E) Two molecules of water are generated.

Answer:  D

Section:  5.4

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.11

 

37) Figure 5.7

 

 

The graph at the left in Figure 5.7 shows the reaction rate for an enzyme at its optimum temperature. Which graph shows enzyme activity at a higher temperature?

  1. A) a
  2. B) b
  3. C) c
  4. D) d

Answer:  B

Section:  5.2

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  7.2a

Learning Outcome:  5.5

Global Outcome:  3

 

 

38) A bacterial culture grown in a glucose-peptide medium causes the pH to increase. The bacteria are most likely

  1. A) fermenting the glucose.
  2. B) oxidizing the glucose.
  3. C) using the peptides.
  4. D) not growing.

Answer:  C

Section:  5.5

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  3.2

Learning Outcome:  5.17

 

39) Gallionella bacteria can get energy from the reaction Fe2+ → Fe3+. This reaction is an example of

  1. A) oxidation.
  2. B) reduction.
  3. C) fermentation.
  4. D) photophosphorylation.
  5. E) the Calvin-Benson cycle.

Answer:  A

Section:  5.3

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  3.1

Learning Outcome:  5.8

 

 

Figure 5.8

 

 

40) In Figure 5.8, where is ATP produced?

  1. A) a
  2. B) b
  3. C) c
  4. D) d
  5. E) e

Answer:  E

Section:  5.4

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.14

 

41) Refer to Figure 5.8. In aerobic respiration, where is water formed?

  1. A) a
  2. B) b
  3. C) c
  4. D) d
  5. E) e

Answer:  D

Section:  5.4

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.14

 

 

42) In Figure 5.8, the structure labeled “1” is

  1. A) NAD+.
  2. B) ATP synthase.
  3. C) a plasma membrane.
  4. D) a cell wall.
  5. E) cytoplasm.

Answer:  C

Section:  5.4

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.14

 

43) In Figure 5.8, the path labeled “2” is the flow of

  1. A) electrons.
  2. B) protons.
  3. C) energy.
  4. D) water.
  5. E) glucose.

Answer:  B

Section:  5.4

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.14

 

44) What is the most acidic place in Figure 5.8?

  1. A) a
  2. B) b
  3. C) c
  4. D) d
  5. E) e

Answer:  A

Section:  5.4

Bloom’s Taxonomy:  Understanding

Learning Outcome:  5.14

Global Outcome:  7

 

45) A urease test is used to identify Mycobacterium tuberculosis because

  1. A) urease is a sign of tuberculosis.
  2. B) tuberculosis produces urease.
  3. C) urea accumulates during tuberculosis.
  4. D) some bacteria reduce nitrate ion.
  5. E) bovis can cause tuberculosis.

Answer:  B

Section:  5.9

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  8.3

Learning Outcome:  5.18

 

 

46) Researchers are developing a ribozyme that cleaves the HIV genome. This pharmaceutical agent could be described as

  1. A) an RNA molecule capable of catalysis.
  2. B) a hydrolase.
  3. C) a genetic transposable element.
  4. D) a protease inhibitor.
  5. E) a competitive inhibitor for reverse transcriptase.

Answer:  A

Section:  5.2

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  3.4

Learning Outcome:  5.7

Global Outcome:  7

 

47) Which statements correspond to amphibolic pathways?

  1. Anabolic and catabolic reactions are joined through common intermediate.
  2. They are shared metabolic pathways.
  3. Feedback inhibition can help regulate rates of reactions.
  4. Both types of reactions are necessary but do not occur simultaneously.
  5. A) 1 only
  6. B) 1, 2, 3
  7. C) 1, 2, 3, 4
  8. D) 2, 4
  9. E) 2, 3, 4

Answer:  B

Section:  5.11

Bloom’s Taxonomy:  Understanding

Learning Outcome:  5.25

 

48) Why do eukaryotes generate only about 36 ATP per glucose in aerobic respiration but prokaryotes may generate about 38 ATP?

  1. A) eukaryotes have a less efficient electron transport system.
  2. B) eukaryotes do not transport as much hydrogen across the mitochondrial membrane as prokaryotes do across the cytoplasmic membrane.
  3. C) eukaryotes must shuttle pyruvate across the mitochondrial membrane by active transport.
  4. D) eukaryotes do not completely oxidize glucose in their respiration reactions.
  5. E) prokaryotes possess an alternate to the Krebs cycle that generates more reduced electron.carriers.

Answer:  C

Section:  5.4

Bloom’s Taxonomy:  Understanding

Learning Outcome:  5.14

 

 

49) In non-cyclic photophosphorylation, excited electrons ultimately

  1. A) return to chlorophyll.
  2. B) are used to form water.
  3. C) combine with hydrogen ions and NADP+ to produce NADPH.
  4. D) flow through ATP synthase.
  5. E) generate light within the spectrum of green wavelengths.

Answer:  C

Section:  5.7

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  3.1

Learning Outcome:  5.19

 

50) Insert non-labeled graph from page 150- replica created and inserted below

 

 

The graph shows the normal reaction rate of an enzyme and the reaction rate when a competitive inhibitor is present. Which description below explains the appearance of the graph?

  1. A) As the substrate concentration increases, the activity of the enzyme decreases.
  2. B) As the substrate concentration increases, the effect of the inhibitor was overcome and enzyme activity was restored.
  3. C) As the enzyme concentration increased, the effect of the inhibitor was overcome and enzyme activity was restored.
  4. D) As the enzyme concentration increased, the effect of the inhibitor was more pronounced.
  5. E) As the competitive inhibitor concentration decreased, the reaction rate also decreased.

Answer:  B

Section:  5.2

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  7.2a

Learning Outcome:  5.5

Global Outcome:  3

 

5.2   True/False Questions

 

1) Catabolic reactions are generally degradative and hydrolytic.

Answer:  TRUE

Section:  5.1

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.1

 

2) The pentose phosphate pathway can be characterized as an anabolic pathway.

Answer:  FALSE

Section:  5.4

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.12

 

3) In general, ATP is generated in catabolic pathways and expended in anabolic pathways.

Answer:  TRUE

Section:  5.1

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.2

 

4) An apoenzyme that loses its coenzyme subunit will be non-functional.

Answer:  TRUE

Section:  5.2

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.3

 

5) The use of enzymes is necessary to increase the activation energy requirements of a chemical reaction.

Answer:  FALSE

Section:  5.2

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.4

 

6) Glycolysis is utilized by cells in both respiration and fermentation.

Answer:  TRUE

Section:  5.4

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.15

 

7) Carbon fixation occurs during the light-independent phase of photosynthesis.

Answer:  TRUE

Section:  5.7

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.20

 

8) Both respiration and photosynthesis require the use of an electron transport chain.

Answer:  TRUE

Section:  5.4; 5.7

Bloom’s Taxonomy:  Understanding

Learning Outcome:  5.21

 

 

9) Both respiration and photosynthesis use water molecules for the donation of hydrogen ions.

Answer:  FALSE

Section:  5.4; 5.7

Bloom’s Taxonomy:  Understanding

Learning Outcome:  5.21

 

10) Once an enzyme has converted substrates into products, the active site reverts back to its original form.

Answer:  TRUE

Section:  5.2

Bloom’s Taxonomy:  Remembering

Learning Outcome:  5.4

 

5.3   Essay Questions

 

1) Compare and contrast photophosphorylation and oxidative phosphorylation.

Section:  5.4; 5.7

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  7.3a

Learning Outcome:  5.21

Global Outcome:  8

 

2) Rhodopseudomonas is an anaerobic photoautotroph that uses organic compounds as an electron donor. It is also capable of chemoheterotrophic metabolism. Diagram the metabolic pathways of this bacterium.

Section:  5.9

Bloom’s Taxonomy:  Applying

ASMcue Outcome:  3.1

Learning Outcome:  5.23

Global Outcome:  2

 

3) Differentiate the following two laboratory tests: starch hydrolysis and starch fermentation.

Section:  5.6

Bloom’s Taxonomy:  Applying

ASMcue Outcome:  7.3a; 8.3

Learning Outcome:  5.18

Global Outcome:  8

 

4) Streptococcus lacks an electron transport chain. How does this bacterium reoxidize NADH? Where is the NADH formed?

Section:  5.4

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  7.3a

Learning Outcome:  5.16

Global Outcome:  8

 

 

5) You look in the refrigerator and find some orange drink you had forgotten was there. The drink now has an “off” taste and bubbles. What is the most likely explanation for the changes in the drink?

Section:  5.4

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  7.3a

Learning Outcome:  5.16

Global Outcome:  8

 

6) Explain the overall purpose of metabolic pathways.

Section:  5.3

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  7.3a

Learning Outcome:  5.10

Global Outcome:  8

 

7) Summarize energy production in cells in a single sentence.

Section:  5.8

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  7.3a

Learning Outcome:  5.22

Global Outcome:  8

 

 

 

Microbiology: An Introduction, 13e (Tortora et al.)

Chapter 18   Practical Applications of Immunology

 

18.1   Multiple-Choice Questions

 

1) All of the following are generally used in vaccines EXCEPT

  1. A) toxoids.
  2. B) parts of bacterial cells.
  3. C) live, attenuated viruses.
  4. D) inactivated viruses.
  5. E) antibodies.

Answer:  E

Section:  18.1

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  6.3

Learning Outcome:  18.3

 

2) Patient’s serum, influenza virus, and red blood cells are mixed in a tube. What happens if the patient has antibodies against influenza virus?

  1. A) agglutination
  2. B) hemagglutination
  3. C) complement fixation
  4. D) hemolysis
  5. E) hemagglutination-inhibition

Answer:  E

Section:  18.2

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  6.3

Learning Outcome:  18.14

Global Outcome:  2

 

3) A patient shows the presence of antibodies against diphtheria toxin. Which of the following statements is FALSE?

  1. A) The patient may have the disease.
  2. B) The patient may have had the disease and has recovered.
  3. C) The patient may have been vaccinated.
  4. D) A recent transfusion may have passively introduced the antibodies.
  5. E) The patient was near someone who had the disease.

Answer:  E

Section:  18.1

Bloom’s Taxonomy:  Understanding

Learning Outcome:  18.5

Global Outcome:  2

 

 

4) In an agglutination test, eight serial dilutions to determine antibody titer were set up. Tube #1 contained a 1:2 dilution; tube #2, a 1:4, etc. If tube #6 is the last tube showing agglutination, what is the antibody titer?

  1. A) 6
  2. B) 1:6
  3. C) 64
  4. D) 1:32
  5. E) 32

Answer:  C

Section:  18.2

Bloom’s Taxonomy:  Understanding

Learning Outcome:  18.11

Global Outcome:  4

 

5) An ELISA for Hepatitis C has 95 percent sensitivity and 90 percent specificity. This means that the test

  1. A) detects 95 percent of the true positive samples and has 10 percent false positive results.
  2. B) detects 5 percent of the true positive samples and has 90 percent false positive results.
  3. C) detects 90 percent of the true positive samples and has 5 percent false positive results.
  4. D) detects 95 percent of the true positive samples and has 90 percent false positive results.
  5. E) detects 5 percent of the true positive samples and has 10 percent false positive results.

Answer:  A

Section:  18.2

Bloom’s Taxonomy:  Analyzing

Learning Outcome:  18.8

Global Outcome:  2

 

6) Which of the following are sources of antibodies for serological testing?

  1. A) vaccinated animals
  2. B) cells producing monoclonal antibodies
  3. C) viral cultures
  4. D) vaccinated animals and cells producing monoclonal antibodies
  5. E) vaccinated animals, cells producing monoclonal antibodies, and viral cultures

Answer:  D

Section:  18.2

Bloom’s Taxonomy:  Analyzing

Learning Outcome:  18.9

 

 

7) A reaction between an antibody and soluble antigen, forming larger, interlocking molecular lattices, is called a(n)

  1. A) agglutination reaction.
  2. B) complement fixation.
  3. C) immunofluorescence.
  4. D) neutralization reaction.
  5. E) precipitation reaction.

Answer:  E

Section:  18.2

Bloom’s Taxonomy:  Remembering

Learning Outcome:  18.10

 

8) A reaction between antibody and particulate antigen is called a(n)

  1. A) agglutination reaction.
  2. B) complement fixation.
  3. C) immunofluorescence.
  4. D) neutralization reaction.
  5. E) precipitation reaction.

Answer:  A

Section:  18.2

Bloom’s Taxonomy:  Remembering

Learning Outcome:  18.12

 

9) A reaction that uses the absence of hemolysis of red blood cells to indicate an antigen–antibody reaction is called a(n)

  1. A) agglutination reaction.
  2. B) complement fixation.
  3. C) immunofluorescence.
  4. D) neutralization reaction.
  5. E) precipitation reaction.

Answer:  B

Section:  18.2

Bloom’s Taxonomy:  Remembering

Learning Outcome:  18.16

 

10) A DNA plasmid encoding a protein antigen from West Nile virus is injected into muscle cells of a horse. This is an example of a(n)

  1. A) subunit vaccine.
  2. B) conjugated vaccine.
  3. C) nucleic acid vaccine.
  4. D) attenuated whole-agent vaccine.
  5. E) live whole-agent vaccine.

Answer:  C

Section:  18.1

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  4.4

Learning Outcome:  18.4

 

11) Toxoid vaccines, such as the vaccines against diphtheria and tetanus, elicit a(n)

  1. A) TCcell response.
  2. B) immune complex.
  3. C) dendritic cell proliferation.
  4. D) antibody response against these bacterial toxins.
  5. E) antibody response against gram-positive bacteria.

Answer:  D

Section:  18.1

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  3.4

Learning Outcome:  18.3

 

12) The clumping of test red blood cells indicates a negative test result (no antibodies against the virus in the patient’s serum) in the

  1. A) direct agglutination test.
  2. B) indirect agglutination test.
  3. C) complement-fixation test.
  4. D) precipitation test.
  5. E) neutralization test.

Answer:  E

Section:  18.2

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  8.3

Learning Outcome:  18.14

 

13) What type of vaccine involves host synthesis of viral antigens?

  1. A) conjugated vaccine
  2. B) subunit vaccine
  3. C) nucleic acid vaccine
  4. D) attenuated whole-agent vaccine
  5. E) toxoid vaccine

Answer:  C

Section:  18.1

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  4.4

Learning Outcome:  18.4

 

 

14) Purified protein from Bordetella pertussis is used in a(n)

  1. A) conjugated vaccine.
  2. B) subunit vaccine.
  3. C) nucleic acid vaccine.
  4. D) attenuated whole-agent vaccine.
  5. E) toxoid vaccine.

Answer:  B

Section:  18.1

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  6.3

Learning Outcome:  18.3

Global Outcome:  5

 

15) What type of vaccine is the live, weakened measles virus?

  1. A) conjugated vaccine
  2. B) subunit vaccine
  3. C) nucleic acid vaccine
  4. D) attenuated whole-agent vaccine
  5. E) toxoid vaccine

Answer:  D

Section:  18.1

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  6.3

Learning Outcome:  18.3

Global Outcome:  5

 

16) A test used to identify antibodies against Treponema pallidum in a patient’s serum is the

  1. A) direct fluorescent-antibody test.
  2. B) indirect fluorescent-antibody test.
  3. C) direct agglutination test.
  4. D) direct ELISA test.
  5. E) neutralization test.

Answer:  B

Section:  18.2

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  8.3

Learning Outcome:  18.17

 

 

17) A test used to identify Streptococcus pyogenes in a patient’s throat swab is the

  1. A) direct fluorescent-antibody test.
  2. B) indirect fluorescent-antibody test.
  3. C) hemagglutination test.
  4. D) neutralization test.
  5. E) indirect ELISA test.

Answer:  A

Section:  18.2

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  8.3

Learning Outcome:  18.17

 

18) A test used to detect anti-Rickettsia antibodies in a patient’s serum is the

  1. A) direct fluorescent-antibody test.
  2. B) indirect fluorescent-antibody test.
  3. C) direct ELISA test.
  4. D) direct agglutination test.

Answer:  B

Section:  18.2

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  8.3

Learning Outcome:  18.17

Global Outcome:  2

 

19) Which of the following is a pregnancy test used to find the fetal hormone HCG in a woman’s urine using anti-HCG and latex spheres?

  1. A) direct agglutination reaction
  2. B) indirect agglutination reaction
  3. C) immunofluorescence
  4. D) neutralization reaction
  5. E) precipitation reaction

Answer:  B

Section:  18.2

Bloom’s Taxonomy:  Analyzing

Learning Outcome:  18.11

Global Outcome:  2

 

 

20) Which of the following is a test to determine a patient’s blood type by mixing the patient’s red blood cells with antisera?

  1. A) direct agglutination reaction
  2. B) passive agglutination reaction
  3. C) immunofluorescence
  4. D) neutralization reaction
  5. E) precipitation reaction

Answer:  A

Section:  18.2

Bloom’s Taxonomy:  Analyzing

Learning Outcome:  18.11

Global Outcome:  2

 

21) Which of the following is a test to determine the presence of soluble antigens in a patient’s saliva?

  1. A) direct agglutination reaction
  2. B) passive agglutination reaction
  3. C) immunofluorescence
  4. D) neutralization reaction
  5. E) precipitation reaction

Answer:  E

Section:  18.2

Bloom’s Taxonomy:  Analyzing

Learning Outcome:  18.10

Global Outcome:  2

 

22) A patient’s serum, Mycobacterium, guinea pig complement, sheep red blood cells, and anti-sheep red blood cell antibodies are mixed in a test tube. What happens if the patient has antibodies to Mycobacterium?

  1. A) Bacteria fluoresce.
  2. B) Hemagglutination occurs.
  3. C) Hemagglutination-inhibition occurs.
  4. D) Hemolysis occurs.
  5. E) No hemolysis occurs.

Answer:  E

Section:  18.2

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  8.3

Learning Outcome:  18.16

Global Outcome:  2

 

 

23) A vaccine against HIV proteins made by a genetically-engineered vaccinia virus that has infected a eukaryotic cell line is a(n)

  1. A) conjugated vaccine.
  2. B) subunit vaccine.
  3. C) nucleic acid vaccine.
  4. D) inactivated whole-agent vaccine.
  5. E) toxoid vaccine.

Answer:  B

Section:  18.1

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  4.4

Learning Outcome:  18.3

Global Outcome:  2

 

24) Inactivated tetanus toxin is a(n)

  1. A) conjugated vaccine.
  2. B) subunit vaccine.
  3. C) nucleic acid vaccine.
  4. D) inactivated whole-agent vaccine.
  5. E) toxoid vaccine.

Answer:  E

Section:  18.1

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  6.3

Learning Outcome:  18.3

 

25) A hybridoma results from the fusion of a(n)

  1. A) B cell with a T cell.
  2. B) B cell with a myeloma cell.
  3. C) antigen with an antibody.
  4. D) antigen with a B cell.
  5. E) myeloma cell with a virus.

Answer:  B

Section:  18.2

Bloom’s Taxonomy:  Remembering

Learning Outcome:  18.9

 

 

Table 18.1

 

Antibody Titer

 

  Day 1 Day 7 Day 14 Day 21
Patient A 0 0 256 512
Patient B 128 256 512 1024
Patient C 0 0 0 0
Patient D 128 128 128 128

 

26) In Table 18.1, who probably has the disease?

  1. A) Patients A and B
  2. B) Patients B and C
  3. C) Patients A and C
  4. D) Patients C and D
  5. E) Patients A and D

Answer:  A

Section:  18.2

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  7.1b

Learning Outcome:  18.11

Global Outcome:  3

 

27) In Table 18.1, who is most likely protected from the disease, as observed by the test results over time?

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

Answer:  D

Section:  18.2

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  7.1b

Learning Outcome:  18.11

Global Outcome:  3

 

28) In Table 18.1, who showed seroconversion during these observations?

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

Answer:  A

Section:  18.2

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  7.1b

Learning Outcome:  18.11

Global Outcome:  3

 

29) In Table 18.1, what is the most likely and reasonable explanation for the results observed in Patient C?

  1. A) The patient has advanced HIV/AIDS, with no ability to make an immune response.
  2. B) The patient was not exposed to the pathogen and has not become infected, so no response is being produced over time.
  3. C) The patient has significantly stronger innate immunity than adaptive immunity mechanisms.
  4. D) The patient is producing a protective adaptive response that doesn’t involve antibodies.

Answer:  B

Section:  18.2

Bloom’s Taxonomy:  Evaluating

ASMcue Outcome:  7.1b

Learning Outcome:  18.11

 

30) If the table continued on to another week of testing, what is the one outcome below that you might NOT expect to see for Patient B at Day 28?

  1. A) a drop to zero antibody titer
  2. B) antibody titer continuing to rise
  3. C) antibody titer leveling off
  4. D) antibody titer starting to decrease slightly

Answer:  A

Section:  18.2

Bloom’s Taxonomy:  Evaluating

ASMcue Outcome:  7.1b

Learning Outcome:  18.11

 

 

Figure 18.1

 

 

31) Which component in Figure 18.1 came from the patient in this indirect ELISA test?

  1. A) a
  2. B) b
  3. C) c
  4. D) d
  5. E) e

Answer:  B

Section:  18.2

Bloom’s Taxonomy:  Applying

Learning Outcome:  18.18

Global Outcome:  3

 

32) Figure 18.1 is an illustration of a(n)

  1. A) negative indirect ELISA test.
  2. B) positive indirect ELISA test.
  3. C) complement fixation test.
  4. D) hemagglutination test.
  5. E) precipitation test.

Answer:  B

Section:  18.2

Bloom’s Taxonomy:  Analyzing

Learning Outcome:  18.18

 

33) In Figure 18.1, which component represents the substrate for the enzyme in the assay?

  1. A) a
  2. B) b
  3. C) c
  4. D) d
  5. E) e

Answer:  C

Section:  18.2

Bloom’s Taxonomy:  Applying

Learning Outcome:  18.18

 

34) In Figure 18.1, items c and e are different from each other. How are they different?

  1. A) Item c is the enzyme that modifies item e, the substrate.
  2. B) Item e is the enzyme that modifies the substrate, item c.
  3. C) Item c is the substrate acted on by the enzyme in the assay, and item e is the colored end product of that enzymatic reaction.
  4. D) Item e is the substrate acted on by the enzyme in the assay, and item c is the colored end product of that enzymatic reaction.

Answer:  C

Section:  18.2

Bloom’s Taxonomy:  Applying

Learning Outcome:  18.18

 

35) Monoclonal antibodies are used in diagnostic tests and disease treatments because they

  1. A) are highly specific.
  2. B) can be produced in large quantities.
  3. C) contain a mixture of antibodies.
  4. D) are highly specific and they can be produced in large quantities.
  5. E) are highly specific, they can be produced in large quantities, and they contain a mixture of antibodies.

Answer:  D

Section:  18.2

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  8.3

Learning Outcome:  18.20

 

36) The following steps are used to produce monoclonal antibodies. What is the fourth step?

  1. A) A B cell is activated to produce antibodies.
  2. B) Culture the hybridoma in a selective medium.
  3. C) Fuse a B cell to a myeloma cell.
  4. D) Isolate antibody-producing B cells.
  5. E) Vaccinate a mouse.

Answer:  C

Section:  18.2

Bloom’s Taxonomy:  Analyzing

Learning Outcome:  18.20

 

 

37) Palivizumab is used to treat respiratory syncytial virus disease. This antiviral drug is a(n)

  1. A) toxoid.
  2. B) monoclonal antibody.
  3. C) vaccine.
  4. D) immunosuppressive.
  5. E) nucleoside analog.

Answer:  B

Section:  18.2

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  3.4

Learning Outcome:  18.20

Global Outcome:  5

 

38) Live weakened polio virus can be used directly in a(n)

  1. A) inactivated whole-agent vaccine.
  2. B) attenuated whole-agent vaccine.
  3. C) conjugated vaccine.
  4. D) subunit vaccine.
  5. E) toxoid vaccine.

Answer:  B

Section:  18.1

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  6.3

Learning Outcome:  18.3

Global Outcome:  5

 

39) Haemophilus capsule polysaccharide plus diphtheria toxoid is a(n)

  1. A) inactivated whole-agent vaccine.
  2. B) attenuated whole-agent vaccine.
  3. C) conjugated vaccine.
  4. D) subunit vaccine.
  5. E) toxoid vaccine.

Answer:  C

Section:  18.1

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  6.3

Learning Outcome:  18.3

Global Outcome:  2

 

 

40) Dead Bordetella pertussis can be used in a(n)

  1. A) inactivated whole-agent vaccine.
  2. B) attenuated whole-agent vaccine.
  3. C) conjugated vaccine.
  4. D) subunit vaccine.
  5. E) toxoid vaccine.

Answer:  A

Section:  18.1

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  6.3

Learning Outcome:  18.3

Global Outcome:  2

 

41) Isolated and purified hepatitis B virus surface antigen can be used in a(n)

  1. A) inactivated whole-agent vaccine.
  2. B) attenuated whole-agent vaccine.
  3. C) conjugated vaccine.
  4. D) subunit vaccine.
  5. E) toxoid vaccine.

Answer:  D

Section:  18.1

Bloom’s Taxonomy:  Analyzing

ASMcue Outcome:  6.3

Learning Outcome:  18.3

Global Outcome:  2

 

42) In a direct ELISA test to screen for drugs in a patient’s urine, what is the third step in the test process?

  1. A) Substrate for the enzyme is added.
  2. B) Enzyme-labeled antibody against the drug being tested is added.
  3. C) The patient’s urine sample is diluted.
  4. D) Antibody against the drug being tested is added.

Answer:  B

Section:  18.2

Bloom’s Taxonomy:  Understanding

Learning Outcome:  18.18

 

43) Which item is from the patient in a direct ELISA test?

  1. A) substrate for the enzyme
  2. B) antigen
  3. C) antihuman immune serum
  4. D) antibodies against the antigen

Answer:  B

Section:  18.2

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  8.3

Learning Outcome:  18.18

 

44) Which of the following tests is MOST useful in determining the presence of AIDS antibodies?

  1. A) agglutination
  2. B) complement fixation
  3. C) neutralization
  4. D) indirect ELISA
  5. E) direct fluorescent-antibody

Answer:  D

Section:  18.2

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  8.3

Learning Outcome:  18.18

Global Outcome:  2

 

45) Which of the following uses fluorescent-labeled antibodies?

  1. A) agglutination
  2. B) complement fixation
  3. C) precipitation
  4. D) flow cytometry
  5. E) neutralization

Answer:  D

Section:  18.2

Bloom’s Taxonomy:  Remembering

Learning Outcome:  18.17

 

46) Which of the following is NOT an advantage of live attenuated vaccine agents?

  1. A) They elicit lifelong immunity.
  2. B) They stimulate by cell-mediated and humoral immune responses.
  3. C) They occasionally revert to virulent forms.
  4. D) They require few or no booster immunizations.
  5. E) The immune response generated by the vaccine closely mimics a real infection.

Answer:  C

Section:  18.1

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  6.3

Learning Outcome:  18.7

Global Outcome:  2

 

 

47) In an immunodiffusion test to diagnose the fungal disease histoplasmosis, a patient’s serum is placed in a well in an agar plate. In a positive test, a precipitate forms as the serum diffuses from the well and meets material diffusing from a second well. In this test process, what is the most likely identity of the material in the second well?

  1. A) antibodies
  2. B) a purified fungal antigen
  3. C) entire fungal cells
  4. D) a purified protozoan antigen
  5. E) red blood cells

Answer:  B

Section:  18.2

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  8.3

Learning Outcome:  18.10

 

48) In an immunodiffusion test to diagnose histoplasmosis, a patient’s serum is placed in a well in an agar plate. In a positive test, a line forms as the serum diffuses from the well and meets material diffusing from a second well. What type of test is this?

  1. A) an agglutination reaction
  2. B) a precipitation reaction
  3. C) a complement-fixation test
  4. D) an indirect ELISA test
  5. E) a direct ELISA test

Answer:  B

Section:  18.2

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  8.3

Learning Outcome:  18.10

 

49) Which of the following statements about measles is FALSE?

  1. A) It is a serious disease.
  2. B) It is preventable by vaccination.
  3. C) Annually, it kills thousands of children worldwide.
  4. D) The disease has been eradicated in the United States.
  5. E) Complications include pneumonia, encephalitis, and death.

Answer:  D

Section:  18.1

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  5.4

Learning Outcome:  18.7

 

 

50) When producing monoclonal antibodies, the use of ________ is critical after fusion to specifically allow only hybrid cells to grow.

  1. A) selective medium
  2. B) neutralizing antibodies
  3. C) enzymatically-labeled antibodies
  4. D) fluorescently-labeled antibodies
  5. E) complement

Answer:  A

Section:  18.2

Bloom’s Taxonomy:  Applying

Learning Outcome:  18.9

 

18.2   True/False Questions

 

1) Vaccines are preparations of organisms or fractions of organisms that are used to induce protective immune responses.

Answer:  TRUE

Section:  18.1

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  6.3

Learning Outcome:  18.1

Global Outcome:  5

 

2) In a vaccine preparation, the term “attenuated” means that the agent does NOT replicate.

Answer:  FALSE

Section:  18.1

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  3.4

Learning Outcome:  18.3

 

3) An injection of “naked” DNA into muscle cells to induce an immune response against the proteins encoded by the DNA is an example of a subunit vaccine.

Answer:  FALSE

Section:  18.1

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  3.4

Learning Outcome:  18.4

 

4) Adjuvants such as aluminum salts are used in vaccines to directly produce highly-specific immune responses.

Answer:  FALSE

Section:  18.1

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  3.4

Learning Outcome:  18.6

Global Outcome:  5

 

5) Blood typing tests are examples of hemagglutination reactions.

Answer:  TRUE

Section:  18.2

Bloom’s Taxonomy:  Remembering

Learning Outcome:  18.13

Global Outcome:  5

 

6) A positive complement-fixation test is indicated by the lysis of the sheep red blood cells added in the indicator phase of the test.

Answer:  FALSE

Section:  18.2

Bloom’s Taxonomy:  Understanding

Learning Outcome:  18.16

Global Outcome:  2

 

7) The home pregnancy test kit is an example of a direct ELISA.

Answer:  TRUE

Section:  18.2

Bloom’s Taxonomy:  Understanding

Learning Outcome:  18.18

Global Outcome:  5

 

8) Western blotting uses DNA probes to detect specific sequences in a mixture of proteins.

Answer:  FALSE

Section:  18.2

Bloom’s Taxonomy:  Understanding

Learning Outcome:  18.19

 

9) A highly specific diagnostic test will be unlikely to indicate a positive result if a specimen being tested is a true negative.

Answer:  TRUE

Section:  18.2

Bloom’s Taxonomy:  Understanding

ASMcue Outcome:  8.3

Learning Outcome:  18.8

Global Outcome:  2

 

10) Agglutination tests use particulate antigens while precipitation tests use soluble antigens.

Answer:  TRUE

Section:  18.2

Bloom’s Taxonomy:  Remembering

ASMcue Outcome:  8.3

Learning Outcome:  18.12

 

 

18.3   Essay Questions

 

1) Describe an ELISA test to detect the presence of HIV antibodies in a patient.

Section:  18.2

Bloom’s Taxonomy:  Applying

ASMcue Outcome:  8.3

Learning Outcome:  18.18

Global Outcome:  8

 

2) Design a serological test to detect botulinum toxin in food.

Section:  18.2

Bloom’s Taxonomy:  Evaluating

ASMcue Outcome:  7.4

Learning Outcome:  18.9

Global Outcome:  8

 

3) In your work in a county public health clinic, you encounter parents of a three-month-old who are considering not vaccinating their child. What concerns might the parents raise regarding vaccine safety? As a microbiology student and public health care worker, what do you say to these parents regarding the risks of not vaccinating their child?

Section:  18.1

Bloom’s Taxonomy:  Evaluating

ASMcue Outcome:  7.4

Learning Outcome:  18.7

Global Outcome:  8

 

4) In a recent influenza epidemic, physicians were utilizing a rapid diagnostic test to determine which patients were infected with influenza type A, type B, or not infected with influenza virus. Such a test was not available in the massive 1918 outbreak that killed millions of people around the globe. How might the availability of such a test have impacted that outbreak, and why?

Section:  18.2

Bloom’s Taxonomy:  Evaluating

ASMcue Outcome:  8.3

Learning Outcome:  18.20

 

5) One of the objections many people have about vaccinations is the amount and variety of chemicals in them. Many people don’t realize how easy the ingredients are to find — they’re required in the product inserts for each vaccination produced. Using the internet, go and find a product insert for a typical vaccine (e.g. influenza vaccine, HPV vaccine, hepatitis B vaccine, etc.). Identify each component contained within it, the amount of that component, and briefly describe what that component does.

Section:  18.1

Bloom’s Taxonomy:  Evaluating

ASMcue Outcome:  7.4

Learning Outcome:  18.5

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