Chapter 8 Microbial Genetics My Nursing Test Banks

Microbiology: An Introduction, 12e (Tortora)

Chapter 8   Microbial Genetics

8.1   Multiple-Choice Questions

1) A gene is best defined as

A) any random segment of DNA.

B) three nucleotides that code for an amino acid.

C) a sequence of nucleotides in DNA that codes for a functional product.

D) a sequence of nucleotides in RNA that codes for a functional product.

E) the RNA product of a transcribed section of DNA.

Answer:  C

Section:  8.1

Blooms Taxonomy:  Knowledge

Learning Outcome:  8.1

2) Which of the following pairs is mismatched?

A) DNA polymerase makes a molecule of DNA from a DNA template

B) RNA polymerase makes a molecule of RNA from an RNA template

C) DNA ligase joins segments of DNA

D) transposase insertion of DNA segments into DNA

E) DNA gyrase coils and twists DNA

Answer:  B

Section:  8.1

Blooms Taxonomy:  Knowledge

Learning Outcome:  8.3

3) Which of the following statements is FALSE?

A) DNA polymerase joins nucleotides in one direction (5 to 3) only.

B) The leading strand of DNA is made continuously.

C) The lagging strand of DNA is started by an RNA primer.

D) DNA replication proceeds in only one direction around the bacterial chromosome.

E) Multiple replication forks are possible on a bacterial chromosome.

Answer:  D

Section:  8.1

Blooms Taxonomy:  Comprehension

Learning Outcome:  8.3

Global Outcome:  2

4) DNA is constructed of

A) a single strand of nucleotides with internal hydrogen bonding.

B) two complementary strands of nucleotides bonded AC and GT.

C) two strands of nucleotides running in an antiparallel configuration.

D) two strands of identical nucleotides in a parallel configuration with hydrogen bonds between them.

E) None of the answers is correct.

Answer:  C

Section:  8.1

Blooms Taxonomy:  Comprehension

Learning Outcome:  8.2

Global Outcome:  2

5) Which of the following is NOT a product of transcription?

A) a new strand of DNA

B) rRNA

C) tRNA

D) mRNA

E) None of the answers are correct; all of these are products of transcription.

Answer:  A

Section:  8.1

Blooms Taxonomy:  Comprehension

Learning Outcome:  8.4

6) Which of the following statements about bacteriocins is FALSE?

A) The genes coding for them are on plasmids.

B) They cause food-poisoning symptoms.

C) Nisin is a bacteriocin used as a food preservative.

D) They can be used to identify certain bacteria.

E) Bacteriocins kill bacteria.

Answer:  B

Section:  8.4

Blooms Taxonomy:  Comprehension

Learning Outcome:  8.16

Global Outcome:  7

7) Figure 8.1

In Figure 8.1, which colonies are streptomycin-resistant and leucine-requiring?

A) 1, 2, 3, and 9

B) 3 and 9

C) 4, 6, and 8

D) 4 and 8

E) 5 and 6

Answer:  D

Section:  8.3

Blooms Taxonomy:  Analysis

ASMcue Outcome:  4.1

Learning Outcome:  8.12

Global Outcome:  2

Table 8.1

Culture 1:  F+, leucine+, histidine+

Culture 2:  F, leucine, histidine

8) In Table 8.1, what will be the result of conjugation between cultures 1 and 2 (reminder: F+ has a different meaning than Hfr)?

A) 1 will remain the same;

2 will become F+, leucine, histidine

B) 1 will become F, leu+, his+;

2 will become F+, leu, his

C) 1 will become F, leu, his;

2 will remain the same

D) 1 will remain the same;

2 will become F+, leu+, his+

E) 1 will remain the same;

2 will become F+ and recombination may occur

Answer:  A

Section:  8.4

Blooms Taxonomy:  Analysis

ASMcue Outcome:  4.1

Learning Outcome:  8.15

Global Outcome:  2

9) In Table 8.1, if culture 1 mutates to Hfr, what will be the result of conjugation between the two cultures?

A) They will both remain the same.

B) 1 will become F+, leu+, his+;

2 will become F+, leu+, his+

C) 1 will remain the same;

recombination will occur in 2

D) 1 will become F, leu+, his+;

2 will become Hfr, leu+, his+

E) The answer cannot be determined based on the information provided.

Answer:  C

Section:  8.4

Blooms Taxonomy:  Analysis

ASMcue Outcome:  4.1

Learning Outcome:  8.15

Global Outcome:  2

10) An enzyme produced in response to the presence of a substrate is called a(n)

A) inducible enzyme.

B) repressible enzyme.

C) restriction enzyme.

D) operator.

E) promoter.

Answer:  A

Section:  8.2

Blooms Taxonomy:  Knowledge

ASMcue Outcome:  4.3

Learning Outcome:  8.7

11) Transformation is the transfer of DNA from a donor to a recipient cell

A) by a bacteriophage.

B) as naked DNA in solution.

C) by cell-to-cell contact.

D) by crossing over.

E) by sexual reproduction.

Answer:  B

Section:  8.4

Blooms Taxonomy:  Knowledge

ASMcue Outcome:  4.1

Learning Outcome:  8.15

12) Genetic change in bacteria can be brought about by

A) mutation.

B) conjugation.

C) transduction.

D) transformation.

E) All of the answers are correct.

Answer:  E

Section:  8.4

Blooms Taxonomy:  Comprehension

ASMcue Outcome:  4.1

Learning Outcome:  8.15

13) Which of the following statements regarding a bacterium that is R+ is FALSE?

A) It possesses a plasmid.

B) R+ can be transferred to a cell of the same species.

C) It is resistant to certain drugs and heavy metals.

D) It is F+.

E) R+ can be transferred to a different species.

Answer:  D

Section:  8.4

Blooms Taxonomy:  Comprehension

ASMcue Outcome:  4.1

Learning Outcome:  8.16

14) The initial effect of ionizing radiation on a cell is that it causes

A) DNA to break.

B) bonding between adjacent thymines.

C) base substitutions.

D) the formation of highly reactive ions.

E) the cells to get hot.

Answer:  D

Section:  8.3

Blooms Taxonomy:  Knowledge

Learning Outcome:  8.9

15) According to the operon model, for the synthesis of an inducible enzyme to occur, the

A) end-product must not be in excess.

B) substrate must bind to the enzyme.

C) substrate must bind to the repressor.

D) repressor must bind to the operator.

E) repressor must not be synthesized.

Answer:  C

Section:  8.2

Blooms Taxonomy:  Comprehension

ASMcue Outcome:  4.3

Learning Outcome:  8.7

Global Outcome:  2

16) Synthesis of a repressible enzyme is stopped by the

A) allosteric transition.

B) substrate binding to the repressor.

C) corepressor binding to the operator.

D) corepressor-repressor complex binding to the operator.

E) end product binding to the promoter.

Answer:  D

Section:  8.2

Blooms Taxonomy:  Comprehension

ASMcue Outcome:  4.3

Learning Outcome:  8.7

Global Outcome:  2

Figure 8.2

17) In Figure 8.2, if base 4 is thymine, what is base 4?

A) adenine

B) thymine

C) cytosine

D) guanine

E) uracil

Answer:  A

Section:  8.1

Blooms Taxonomy:  Analysis

Learning Outcome:  8.3

Global Outcome:  2

18) In Figure 8.2, if base 4 is thymine, what is base 11 (remember the complimentary configuration of bases in DNA)?

A) adenine

B) thymine

C) cytosine

D) guanine

E) uracil

Answer:  B

Section:  8.1

Blooms Taxonomy:  Analysis

Learning Outcome:  8.3

Global Outcome:  2

19) In Figure 8.2, base 2 (and ONLY the base) is covalently bound/attached to

A) ribose.

B) phosphate.

C) deoxyribose.

D) thymine.

E) The answer cannot be determined based on the information provided.

Answer:  C

Section:  8.1

Blooms Taxonomy:  Knowledge

Learning Outcome:  8.3

20) The damage caused by ultraviolet radiation is

A) never repaired.

B) repaired during transcription.

C) repaired during translation.

D) cut out and replaced.

E) repaired by DNA replication.

Answer:  D

Section:  8.3

Blooms Taxonomy:  Knowledge

Learning Outcome:  8.10

Table 8.2

Codon on mRNA and corresponding amino acid

UUA

leucine

UAA

nonsense

GCA

alanine

AAU

asparagine

AAG

lysine

UGC

cysteine

GUU

valine

UCG, UCU

serine

21) Refer to Table 8.2. If the sequence of amino acids encoded by a strand of DNA is

serine-alanine-lysine-leucine, what is the order of bases in the sense strand of DNA?

A) 3 UGUGCAAAGUUA

B) 3 AGACGTTTCAAT

C) 3 TCTCGTTTGTTA

D) 5 TGTGCTTTCTTA

E) 5 AGAGCTTTGAAT

Answer:  B

Section:  8.1

Blooms Taxonomy:  Analysis

Learning Outcome:  8.4

Global Outcome:  2

22) Refer to Table 8.2. If the sequence of amino acids encoded by a strand of DNA is

serine-alanine-lysine-leucine, the coding for the antisense strand of DNA is

A) 5 ACAGTTTCAAT.

B) 5 TCTGCAAAGTTA.

C) 3 UGUGCAAAGUUA.

D) 3 UCUCGAAAGUUA.

E) 3 TCACGUUUCAAU.

Answer:  B

Section:  8.1

Blooms Taxonomy:  Analysis

Learning Outcome:  8.4

Global Outcome:  2

23) Refer to Table 8.2 The anticodon for valine is

A) GUU.

B) CUU.

C) CTT.

D) CAA.

E) GTA.

Answer:  D

Section:  8.1

Blooms Taxonomy:  Comprehension

Learning Outcome:  8.4

Global Outcome:  2

24) Refer to Table 8.2. What is the sequence of amino acids encoded by the following sequence of bases in a strand of DNA (pay attention to the polarity of the DNA here)?

3 ATTACGCTTTGC

A) leucine-arginine-lysine-alanine

B) asparagine-arginine-lysine-alanine

C) asparagine-cysteine-valine-serine

D) Translation would stop at the first codon.

E) The answer cannot be determined based on the information provided.

Answer:  D

Section:  8.1

Blooms Taxonomy:  Analysis

Learning Outcome:  8.4

Global Outcome:  2

25) Refer to Table 8.2. If an indeterminate frameshift mutation occurred in the sequence of bases shown below, what would be the sequence of amino acids coded for?

3 ATTACGCTTTGC

A) leucine-arginine-lysine-alanine

B) asparagine-arginine-lysine-alanine

C) asparagine-cysteine-valine-serine

D) Translation would stop at the first codon.

E) The answer cannot be determined based on the information provided.

Answer:  E

Section:  8.3

Blooms Taxonomy:  Comprehension

Learning Outcome:  8.9

Global Outcome:  2

Figure 8.3 Metabolic Pathway

26) In Figure 8.3, if compound C reacts with the allosteric site of enzyme A, this would exemplify

A) a mutation.

B) repression.

C) feedback inhibition.

D) competitive inhibition.

E) transcription.

Answer:  C

Section:  8.2

Blooms Taxonomy:  Comprehension

ASMcue Outcome:  4.3

Learning Outcome:  8.6

Global Outcome:  2

27) In Figure 8.3, if enzyme A is a repressible enzyme, compound C would

A) always be in excess.

B) bind to Enzyme A.

C) bind to the corepressor for Gene a.

D) bind to RNA polymerase.

E) bind directly to gene a.

Answer:  C

Section:  8.2

Blooms Taxonomy:  Comprehension

ASMcue Outcome:  4.3

Learning Outcome:  8.7

Global Outcome:  2

28) In Figure 8.3, if enzyme A is an inducible enzyme,

A) compound C would bind to the repressor for Gene a.

B) compound A would bind to the repressor for Gene a.

C) compound B would bind to enzyme A directly.

D) compound A would react with enzyme B directly.

E) compound C would react with gene a directly.

Answer:  B

Section:  8.2

Blooms Taxonomy:  Comprehension

ASMcue Outcome:  4.3

Learning Outcome:  8.7

Global Outcome:  2

29) Conjugation differs from reproduction because conjugation

A) replicates DNA.

B) transfers DNA vertically, to new cells.

C) transfers DNA horizontally, to nearby cells without those cells undergoing replication.

D) transcribes DNA to RNA.

E) copies RNA to make DNA.

Answer:  C

Section:  8.4

Blooms Taxonomy:  Comprehension

ASMcue Outcome:  4.1

Learning Outcome:  8.15

30) The necessary ingredients for DNA synthesis can be mixed together in a test tube. The DNA polymerase is from Thermus aquaticus, and the template is from a human cell. The DNA synthesized would be most similar to

A) human DNA.

B) T. aquaticus DNA.

C) a mixture of human and T. aquaticus DNA.

D) human RNA.

E) T. aquaticus RNA.

Answer:  A

Section:  8.1

Blooms Taxonomy:  Comprehension

ASMcue Outcome:  6.3

Learning Outcome:  8.3

Global Outcome:  7

31) Table 8.3

Amino Acids Encoded by the Human p53 Gene

Based on the information in Table 8.3, prostate cancer is probably the result of which kind of mutation?

A) analog

B) frameshift

C) missense

D) nonsense

E) None of the answers is correct.

Answer:  D

Section:  8.3

Blooms Taxonomy:  Analysis

Learning Outcome:  8.9

Global Outcome:  7

32) Figure 8.4

In Figure 8.4, the antibiotic chloramphenicol binds the 50S large subunit of a ribosome as shown (the light gray area is the large subunit, while the black shape is the drug). From this information you can conclude that chloramphenicol

A) prevents transcription in eukaryotes.

B) prevents translation in eukaryotes.

C) prevents transcription in prokaryotes.

D) prevents translation in prokaryotes.

E) prevents mRNA-ribosome binding.

Answer:  D

Section:  8.1

Blooms Taxonomy:  Analysis

ASMcue Outcome:  4.2

Learning Outcome:  8.5

Global Outcome:  7

33) The mechanism by which the presence of glucose inhibits the lac operon is

A) catabolite repression.

B) translation.

C) DNA polymerase.

D) repression.

E) induction.

Answer:  A

Section:  8.2

Blooms Taxonomy:  Knowledge

ASMcue Outcome:  4.3

Learning Outcome:  8.7

34) If you knew the sequence of nucleotides within a gene, which one of the following could you determine with the most accuracy?

A) the primary structure of the protein

B) the secondary structure of the protein

C) the tertiary structure of the protein

D) the quaternary structure of the protein

E) The answer cannot be determined based on the information provided.

Answer:  A

Section:  8.1

Blooms Taxonomy:  Comprehension

Learning Outcome:  8.4

Global Outcome:  2

35) An enzyme that makes covalent bonds between the sugar of one nucleotide and the phosphate groups in another nucleotide in DNA is

A) RNA polymerase.

B) DNA ligase

C) DNA helicase.

D) transposase.

E) DNA polymerase.

Answer:  B

Section:  8.1

Blooms Taxonomy:  Knowledge

Learning Outcome:  8.3

36) An enzyme that copies DNA to make a molecule of RNA is

A) RNA polymerase.

B) DNA ligase.

C) DNA helicase.

D) transposase.

E) DNA polymerase.

Answer:  A

Section:  8.1

Blooms Taxonomy:  Knowledge

Learning Outcome:  8.4

37) An enzyme that catalyzes the cutting and resealing of DNA, and is translated from insertion sequences, is

A) RNA polymerase.

B) DNA ligase.

C) DNA helicase.

D) transposase.

E) DNA polymerase.

Answer:  D

Section:  8.4

Blooms Taxonomy:  Knowledge

Learning Outcome:  8.16

38) Repair of damaged DNA, in some instances and mechanisms, might be viewed as a race between an endonuclease and

A) DNA ligase.

B) DNA polymerase.

C) helicase.

D) methylase.

E) primase.

Answer:  D

Section:  8.3

Blooms Taxonomy:  Comprehension

Learning Outcome:  8.10

Global Outcome:  2

39) The cancer gene ras produces mRNA containing an extra exon that includes a number of UAA codons. Cancer cells produce ras mRNA missing this exon. This mistake most likely is due to a mistake by

A) a chemical mutagen.

B) DNA polymerase.

C) photolyases.

D) snRNPs.

E) UV radiation.

Answer:  D

Section:  8.1

Blooms Taxonomy:  Comprehension

Learning Outcome:  8.5

Global Outcome:  7

40) Figure 8.5

In Figure 8.5, which model of the lac operon correctly shows RNA polymerase, lactose, and repressor protein when the structural genes are being transcribed?

A) a

B) b

C) c

D) d

E) e

Answer:  D

Section:  8.2

Blooms Taxonomy:  Analysis

ASMcue Outcome:  4.3

Learning Outcome:  8.7

Global Outcome:  2

41) The miRNAs in a cell

A) are found in prokaryotic cells.

B) are a part of the prokaryotic ribosome.

C) are a part of the eukaryotic ribosome.

D) allow different cells to produce different proteins.

E) are responsible for inducing operons.

Answer:  D

Section:  8.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  8.8

42) Assume the two E.coli strains shown below are allowed to conjugate.

Hfr: pro+, arg+, his+, lys+, met+, ampicillin-sensitive

F-: pro, arg, his, lys, met, ampicillin-resistant

What supplements would you add to glucose minimal salts agar to select for a recombinant cell that is lys+, arg+, amp-resistant?

A) ampicillin, lysine, arginine

B) lysine, arginine

C) ampicillin, proline, histidine, methionine

D) proline, histidine, methionine

E) ampicillin, proline, histidine, lysine

Answer:  C

Section:  8.3

Blooms Taxonomy:  Analysis

ASMcue Outcome:  4.1

Learning Outcome:  8.12

Global Outcome:  2

43) Protein synthesis in eukaryotes is similar to the process in prokaryotes in that both eukaryotes and prokaryotes

A) have exons.

B) have introns.

C) require snRNPS.

D) use methionine as the start amino acid.

E) use codons to determine polypeptide sequences.

Answer:  E

Section:  8.1

Blooms Taxonomy:  Comprehension

ASMcue Outcome:  4.2

Learning Outcome:  8.5

Global Outcome:  7

8.2   True/False Questions

1) Recombination will always alter a cells genotype.

Answer:  TRUE

Section:  8.4

Blooms Taxonomy:  Comprehension

ASMcue Outcome:  4.5

Learning Outcome:  8.15

2) Open-reading frames are segments of DNA in which both start and stop codons are found.

Answer:  TRUE

Section:  8.1

Blooms Taxonomy:  Comprehension

Learning Outcome:  8.1

3) Bacteria typically contain multiple chromosomes.

Answer:  FALSE

Section:  8.1

Blooms Taxonomy:  Knowledge

ASMcue Outcome:  4.2

Learning Outcome:  8.1

Global Outcome:  7

4) Mutations that are harmful to cells occur more frequently than those that benefit cells.

Answer:  TRUE

Section:  8.3

Blooms Taxonomy:  Knowledge

Learning Outcome:  8.10

Global Outcome:  7

5) The miRNAs in a cell inhibit protein synthesis by forming complementary bonds with rRNA.

Answer:  FALSE

Section:  8.2

Blooms Taxonomy:  Knowledge

Learning Outcome:  8.8

6) Some cells may contain multiple genomes.

Answer:  FALSE

Section:  8.1

Blooms Taxonomy:  Knowledge

Learning Outcome:  8.1

Global Outcome:  7

7) Both base substitution and frameshift mutations can result in the formation of premature stop codons.

Answer:  TRUE

Section:  8.3

Blooms Taxonomy:  Knowledge

Learning Outcome:  8.9

8) In the Ames test, any colonies that form on the control should be the result of spontaneous mutations.

Answer:  TRUE

Section:  8.3

Blooms Taxonomy:  Knowledge

Learning Outcome:  8.13

9) Transposition (insertion of a transposon into a DNA sequence) results in the formation of base substitution mutations in a cells DNA.

Answer:  FALSE

Section:  8.3

Blooms Taxonomy:  Knowledge

Learning Outcome:  8.9

Global Outcome:  7

10) Cell-to-cell contact is required for transduction to occur.

Answer:  FALSE

Section:  8.4

Blooms Taxonomy:  Comprehension

Learning Outcome:  8.15

8.3   Essay Questions

1) What is the survival value of the semiconservative replication of DNA?

Section:  8.1

Blooms Taxonomy:  Synthesis

Learning Outcome:  8.3

Global Outcome:  8

2) Scientists are concerned that bacteria will be resistant to all antibiotics within the next decade. Using your knowledge of genetics, describe how bacterial populations can develop drug resistance in such a short time frame.

Section:  8.4

Blooms Taxonomy:  Synthesis

ASMcue Outcome:  4.1

Learning Outcome:  8.15

Global Outcome:  8

3) Explain why the following statement is false: Sexual reproduction is the only mechanism for genetic change.

Section:  8.4

Blooms Taxonomy:  Synthesis

Learning Outcome:  8.14

Global Outcome:  8

4) Why must the cultures used in the Ames test be auxotrophic? Explain the necessity of using a strain that is mutated in this experiment.

Section:  8.3

Blooms Taxonomy:  Synthesis

Learning Outcome:  8.13

Global Outcome:  8

5) What is the survival value of the degeneracy of the genetic code?

Section:  8.1

Blooms Taxonomy:  Synthesis

Learning Outcome:  8.1

Global Outcome:  8

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