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MCB 3020C Exam 3 V1 | MCB 3020C General Microbiology | Actual Q&A with Rationale (MCB3020C Exam 3) | University of Central Florida

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MCB 3020C Exam 3 V1 | MCB 3020C General Microbiology | Actual Q&A with Rationale (MCB3020C Exam 3) | University of Central Florida

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MCB 3020C Exam 3 V1 | MCB 3020C General Microbiology | Actual
Q&A with Rationale (MCB3020C Exam 3) | University of Central
Florida
1. Which enzyme is primarily responsible for synthesizing the new DNA strand during
replication in E. coli?
A. DNA Polymerase I

B. DNA Polymerase III

C. RNA Polymerase

D. DNA Ligase

Answer: B
Explanation: DNA Polymerase III is the main enzyme involved in the polymerization of the
nascent DNA strand. It requires a primer with a free 3’-OH group to begin synthesis and
works in the 5’ to 3’ direction. This enzyme possesses high processivity and proofreading
capabilities to ensure genomic integrity.

2. What is the role of the sigma factor in bacterial transcription?
A. Termination of the mRNA transcript

B. Catalyzing the formation of phosphodiester bonds

C. Unwinding the DNA double helix

D. Recognition of the promoter sequence

Answer: D
Explanation: The sigma factor is a protein needed only for initiation of RNA synthesis in
bacteria. It enables specific binding of RNA polymerase to gene promoters, ensuring
transcription starts at the correct site. Once the transcription bubble forms and a short
stretch of RNA is synthesized, the sigma factor is typically released.

3. In the lac operon, what happens when lactose is present and glucose is absent?
A. The repressor remains bound to the operator

B. cAMP levels are low, preventing activation

C. The CAP protein cannot bind to the promoter

D. Transcription of the structural genes is induced
Answer: D

,Explanation: When lactose is present, allolactose acts as an inducer by binding to the
repressor and preventing it from binding to the operator. Simultaneously, the absence of
glucose leads to high cAMP levels, which activate the Catabolite Activator Protein (CAP).
This dual condition results in high levels of transcription for the lactose metabolism genes.

4. Which type of horizontal gene transfer involves the uptake of naked DNA from the
environment?
A. Conjugation

B. Transduction

C. Transformation

D. Transposition

Answer: C
Explanation: Transformation is the process by which a competent bacterial cell takes up
free DNA fragments from its surroundings. This DNA may then be integrated into the host
genome through homologous recombination. This mechanism was famously demonstrated
by Frederick Griffith in his experiments with Streptococcus pneumoniae.

5. A mutation that results in a premature stop codon is known as a:
A. Silent mutation

B. Nonsense mutation

C. Missense mutation

D. Frameshift mutation

Answer: B
Explanation: A nonsense mutation is a point mutation in a sequence of DNA that results in
a premature stop codon. This leads to the production of a truncated, and usually
nonfunctional, protein product. Such mutations often have severe phenotypic
consequences compared to silent or missense mutations.

6. During DNA replication, which enzyme relieves the torsional stress (supercoiling) ahead of
the replication fork?
A. Helicase

B. Primase

C. DNA Gyrase (Topoisomerase II)

D. Single-strand binding proteins
Answer: C

, Explanation: DNA Gyrase is a type of topoisomerase that introduces negative supercoils
into the DNA to neutralize the positive supercoiling caused by helicase unwinding. Without
this enzyme, the DNA would become overwound and the replication fork would stall. It is a
critical target for certain antibiotics like ciprofloxacin.

7. What is the function of the Shine-Dalgarno sequence in prokaryotic translation?
A. It marks the site where transcription ends

B. It aligns the ribosome with the start codon

C. It serves as the binding site for RNA polymerase

D. It signals the start of DNA replication

Answer: B
Explanation: The Shine-Dalgarno sequence is a ribosomal binding site in bacterial
messenger RNA, generally located around 8 bases upstream of the start codon AUG. It is
complementary to a sequence at the 3’ end of the 16S rRNA component of the small
ribosomal subunit. This base-pairing ensures that the ribosome is correctly positioned to
initiate protein synthesis.

8. Which of the following describes a Hfr (High Frequency of Recombination) cell?
A. A cell that lacks the F plasmid

B. A cell with a free-floating F plasmid

C. A cell where the F plasmid has integrated into the chromosome

D. A cell that can only undergo transduction

Answer: C
Explanation: An Hfr cell is created when the F (fertility) factor integrates into the bacterial
chromosome through site-specific recombination. When this cell undergoes conjugation, it
attempts to transfer the entire chromosome to a recipient cell. However, the conjugation
bridge usually breaks before the entire chromosome is transferred, resulting in the transfer
of some chromosomal genes.

9. In the Ames test, a high number of colonies growing on the histidine-deficient medium
indicates:
A. The substance is not a mutagen

B. The substance caused back-mutations in the Salmonella strain

C. The bacteria have died due to toxicity

D. The bacteria were already wild-type
Answer: B

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