Question 1

Recombination will always alter a cell's genotype.

Accepted Answer

A) True 
 B)False

Question 2

Figure 8.4
  
-In Figure 8.4, the antibiotic chloramphenicol binds the 50S ribosome as shown. From this information you can conclude that chloramphenicol

Accepted Answer

A)  prevents translation in eukaryotes. 
B)  prevents mRNA- ribosome binding. 
C)  prevents translation in prokaryotes. 
D)  prevents transcription in prokaryotes. 
E)  prevents transcription in eukaryotes. 
A)  prevents translation in eukaryotes. 
B)  prevents mRNA- ribosome binding. 
C)  prevents translation in prokaryotes. 
D)  prevents transcription in prokaryotes. 
E)  prevents transcription in eukaryotes.

Question 3

Repair of damaged DNA might be viewed as a race between an endonuclease and

Accepted Answer

A)  DNA ligase. 
B)  DNA polymerase. 
C)  helicase. 
D)  primase. 
E)  methylase. 
A)  DNA ligase. 
B)  DNA polymerase. 
C)  helicase. 
D)  primase. 
E)  methylase.

Question 4

A gene is best defined as

Accepted Answer

A)  three nucleotides that code for an amino acid. 
B)  a transcribed unit of DNA. 
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)  a segment of DNA. 
A)  three nucleotides that code for an amino acid. 
B)  a transcribed unit of DNA. 
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)  a segment of DNA.

Question 5

Synthesis of a repressible enzyme is stopped by the

Accepted Answer

A)  substrate binding to the repressor. 
B)  end product binding to the promoter. 
C)  corepressor binding to the operator. 
D)  corepressor- repressor complex binding to the operator. 
E)  allosteric transition. 
A)  substrate binding to the repressor. 
B)  end product binding to the promoter. 
C)  corepressor binding to the operator. 
D)  corepressor- repressor complex binding to the operator. 
E)  allosteric transition.

Question 6

Which of the following is NOT a product of transcription?

Accepted Answer

A)  mRNA 
B)  tRNA 
C)  rRNA 
D)  a new strand of DNA 
E)  None of the answers are correct; all of these are products of transcription. 
A)  mRNA 
B)  tRNA 
C)  rRNA 
D)  a new strand of DNA 
E)  None of the answers are correct; all of these are products of transcription.

Question 7

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

Accepted Answer

A) True 
 B)False

Question 8

The mechanism by which the presence of arabinose controls the arabinose operon is

Accepted Answer

A)  repression. 
B)  catabolite repression. 
C)  DNA polymerase. 
D)  translation. 
E)  induction. 
A)  repression. 
B)  catabolite repression. 
C)  DNA polymerase. 
D)  translation. 
E)  induction.

Question 9

Figure 8.2
   REPLICATION FORK
-In Figure 8.2, base 2 is attached to

Accepted Answer

A)  deoxyribose. 
B)  thymine. 
C)  phosphate. 
D)  ribose. 
E)  The answer cannot be determined based on the information provided. 
A)  deoxyribose. 
B)  thymine. 
C)  phosphate. 
D)  ribose. 
E)  The answer cannot be determined based on the information provided.

Question 10

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

Accepted Answer

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. 
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.

Question 11

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

Accepted Answer

A)  RNA polymerase. 
B)  DNA polymerase. 
C)  transposase. 
D)  DNA ligase. 
E)  DNA helicase. 
A)  RNA polymerase. 
B)  DNA polymerase. 
C)  transposase. 
D)  DNA ligase. 
E)  DNA helicase.

Question 12

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

Accepted Answer

A) True 
 B)False

Question 13

Table 8.1 $\text { Culture 1: F+, leucinet+, histidinet +}$
$\text { Culture 2: F-, leucine-, histidine- }$ 
-In Table 8.1, what will be the result of conjugation between cultures 1 and 2?

Accepted Answer

A)  1 will remain the same; 2 will become F+, leucine- , histidine- 
B)  1 will become F- , leu- , his- ; 2 will remain the same 
C)  1 will become F- , leu+, his+; 2 will become F+, leu- , his- 
D)  1 will remain the same; 2 will become F+ and recombination may occur 
E)  1 will remain the same; 2 will become F+, leu+, his+ 
A)  1 will remain the same; 2 will become F+, leucine- , histidine- 
B)  1 will become F- , leu- , his- ; 2 will remain the same 
C)  1 will become F- , leu+, his+; 2 will become F+, leu- , his- 
D)  1 will remain the same; 2 will become F+ and recombination may occur 
E)  1 will remain the same; 2 will become F+, leu+, his+

Question 14

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

Accepted Answer

A) True 
 B)False

Question 15

Table 8.2
Codon on mRNA and corresponding amino acid $\begin{array} { | l | l | l | l | } 
\hline \text { UUA } & \text { leucine } & \text { UAA } & \text { nonsense } \
\hline \text { GCA } & \text { alanine } & \text { AAU } & \text { sparagine } \
\hline \text { AAG } & \text { lysine } & \text { UGC } & \text { cysteine } \
\hline \text { GUU } & \text { valine } & \begin{array} { l } 
\text { UCG, } \
\text { UCU }
\end{array} & \text { serine } \
\hline
\end{array}$
-Refer to Table 8.2. What is the sequence of amino acids encoded by the following sequence of bases in a strand of DNA? 3' ATTACGCTTTGC

Accepted Answer

A)  asparagine- arginine- lysine- alanine 
B)  asparagine- cysteine- valine- serine 
C)  leucine- arginine- lysine- alanine 
D)  Translation would stop at the first codon. 
E)  The answer cannot be determined based on the information provided. 
A)  asparagine- arginine- lysine- alanine 
B)  asparagine- cysteine- valine- serine 
C)  leucine- arginine- lysine- alanine 
D)  Translation would stop at the first codon. 
E)  The answer cannot be determined based on the information provided.

Question 16

Table 8.2
Codon on mRNA and corresponding amino acid $\begin{array} { | l | l | l | l | } 
\hline \text { UUA } & \text { leucine } & \text { UAA } & \text { nonsense } \
\hline \text { GCA } & \text { alanine } & \text { AAU } & \text { sparagine } \
\hline \text { AAG } & \text { lysine } & \text { UGC } & \text { cysteine } \
\hline \text { GUU } & \text { valine } & \begin{array} { l } 
\text { UCG, } \
\text { UCU }
\end{array} & \text { serine } \
\hline
\end{array}$
-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

Accepted Answer

A)  5' ACAGTTTCAAT. 
B)  5' TCTGCAAAGTTA. 
C)  3' TCACGUUUCAAU. 
D)  3' UGUGCAAAGUUA. 
E)  3' UCUCGAAAGUUA. 
A)  5' ACAGTTTCAAT. 
B)  5' TCTGCAAAGTTA. 
C)  3' TCACGUUUCAAU. 
D)  3' UGUGCAAAGUUA. 
E)  3' UCUCGAAAGUUA.

Exam 8: Microbial Genetics

Recombination will always alter a cell's genotype.

Figure 8.4 -In Figure 8.4, the antibiotic chloramphenicol binds the 50S ribosome as shown. From this information you can conclude that chloramphenicol

Repair of damaged DNA might be viewed as a race between an endonuclease and

A gene is best defined as

Synthesis of a repressible enzyme is stopped by the

Which of the following is NOT a product of transcription?

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

The mechanism by which the presence of arabinose controls the arabinose operon is

Figure 8.2 REPLICATION FORK -In Figure 8.2, base 2 is attached to

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

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

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

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

Table 8.2 Codon on mRNA and corresponding amino acid UUA leucine UAA nonsense GCA alanine AAU sparagine AAG lysine UGC cysteine GUU valine UCG, UCU serine -Refer to Table 8.2. What is the sequence of amino acids encoded by the following sequence of bases in a strand of DNA? 3' ATTACGCTTTGC

The Microbial World and You

Chemical Principles

Observing Microorganisms Through a Microscope

Functional Anatomy of Prokaryotic and Eukaryotic Cells

Microbial Metabolism

Microbial Growth

The Control of Microbial Growth

Biotechnology and Recombinant Dna

Classification of Microorganisms

The Prokaryotes: Domains Bacteria and Archaea

The Eukaryotes: Fungi, Algae, Protozoa, and Helminths

Viruses, Viroids, and Prions

Principles of Disease and Epidemiology

Microbial Mechanisms of Pathogenicity

Innate Immunity: Nonspecific Defenses of the Host

Adaptive Immunity: Specific Defenses of the Host

Practical Applications of Immunology

Disorders Associated With the Immune System

Antimicrobial Drugs

Microbial Diseases of the Skin and Eyes

Microbial Diseases of the Nervous System

Microbial Diseases of the Cardiovascular and Lymphatic Systems

Microbial Diseases of the Respiratory System

Microbial Diseases of the Digestive System

Microbial Diseases of the Urinary and Reproductive Systems

Environmental Microbiology

Applied and Industrial Microbiology

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