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MCB 3020C Exam 2 V2 | MCB 3020C General Microbiology | Actual Q&A with Rationale (MCB3020C Exam 2) | University of Central Florida

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MCB 3020C Exam 2 V2 | MCB 3020C General Microbiology | Actual Q&A with Rationale (MCB3020C Exam 2) | University of Central Florida

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MCB 3020C Exam 2 V2 | MCB 3020C General Microbiology | Actual
Q&A with Rationale (MCB3020C Exam 2) | University of Central
Florida
1. Which of the following describes the net energy yield from one molecule of glucose
undergoing glycolysis via the Embden-Meyerhof-Parnas pathway?
A. 2 ATP and 2 NADH

B. 4 ATP and 2 NADH

C. 2 ATP and 4 NADH

D. 38 ATP and 10 NADH

Answer: A
Explanation: Glycolysis involves an initial energy investment phase where two ATP
molecules are consumed to phosphorylate glucose. During the energy payoff phase, four
ATP molecules are generated through substrate-level phosphorylation. Therefore, the net
gain for the cell is two ATP and two reduced NADH molecules per glucose.

2. In aerobic respiration, what is the final electron acceptor in the electron transport chain?
A. Nitrate (NO3-)

B. Oxygen (O2)

C. Sulfate (SO4 2-)

D. Pyruvate

Answer: B
Explanation: Aerobic respiration utilizes oxygen as the terminal electron acceptor at the
end of the electron transport chain. Oxygen has a high reduction potential, allowing for the
maximum extraction of energy from electrons. In the absence of oxygen, microbes must
switch to anaerobic respiration or fermentation.

3. Which enzyme is responsible for unwinding the DNA double helix at the replication fork
during bacterial DNA replication?
A. DNA Helicase

B. DNA Polymerase I

C. DNA Ligase
D. Primase

Answer: A

,Explanation: DNA Helicase breaks the hydrogen bonds between the nitrogenous bases of
the two DNA strands. This action creates a replication fork, allowing other enzymes to
access the single-stranded templates. Without helicase, the replication machinery would be
unable to progress along the chromosome.

4. A bacterial culture that grows best at a temperature of 15 degrees Celsius and cannot grow
above 20 degrees Celsius is classified as a:
A. Mesophile

B. Thermophile

C. Psychrotroph

D. Psychrophile

Answer: D
Explanation: Psychrophiles are extremophilic organisms that are capable of growth and
reproduction in cold temperatures. They typically have membranes rich in unsaturated
fatty acids to maintain fluidity at low temperatures. In contrast, mesophiles prefer
moderate temperatures closer to 37 degrees Celsius.

5. The time required for a bacterial population to double in number is referred to as the:
A. Lag phase

B. Generation time

C. Log phase

D. Stationary phase

Answer: B
Explanation: Generation time, also known as doubling time, is a constant value during the
exponential growth phase for a specific organism under specific conditions. It varies
greatly between species, ranging from minutes to days. This value is critical for calculating
bacterial growth rates in clinical and industrial settings.

6. Which component of the RNA polymerase holoenzyme is responsible for recognizing the
specific promoter sequence on the DNA?
A. Alpha subunit

B. Core enzyme

C. Beta subunit

D. Sigma factor
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 to ensure
transcription starts at the correct site. Once transcription begins, the sigma factor usually
dissociates from the core enzyme.

7. Which of the following methods is considered a form of sterilization?
A. Pasteurization

B. Autoclaving

C. Disinfection with alcohol

D. Sanitization

Answer: B
Explanation: Autoclaving uses saturated steam under high pressure to achieve
temperatures above the boiling point of water, typically 121 degrees Celsius. This process
is effective at destroying all microbial life, including highly resistant bacterial endospores.
Pasteurization and disinfection only reduce microbial loads but do not ensure total
sterility.

8. In the lac operon, what happens when lactose is present and glucose is absent?
A. Allolactose binds the repressor, and cAMP levels are high, leading to high transcription.

B. The repressor binds the operator and transcription is blocked.

C. The promoter is degraded by enzymes.

D. Transcription occurs at a basal, very low level.

Answer: A
Explanation: When lactose is present, allolactose acts as an inducer by binding to the lac
repressor, preventing it from blocking the operator. Simultaneously, low glucose levels lead
to high cAMP, which activates the Catabolite Activator Protein (CAP) to enhance
transcription. This dual control ensures the cell only invests energy in lactose metabolism
when necessary.

9. Which phase of the bacterial growth curve is characterized by a high rate of metabolic
activity and binary fission?
A. Log (Exponential) phase

B. Stationary phase

C. Lag phase

D. Death phase
Answer: A

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