PCB 3063 Exam 3 V1 | PCB 3063 Genetics
| Actual Q&A with Rationale (PCB3063
Exam 3) | University of Central Florida
1. Which enzyme is primarily responsible for unwinding the DNA double helix at the
replication fork in E. coli?
A. Helicase
B. DNA Gyrase
C. Primase
D. Single-stranded binding proteins
Answer: A
Rationale: Helicase utilizes energy from ATP hydrolysis to break the hydrogen bonds
between complementary base pairs. This process allows the two strands to separate and
serve as templates for new DNA synthesis. Without helicase activity, the replication fork
would not be able to progress along the chromosome.
2. What is the function of the enzyme primase during DNA replication?
A. It synthesizes a short RNA primer to provide a 3’-OH group.
B. It removes RNA primers from the lagging strand.
C. It seals the nicks between Okazaki fragments.
D. It proofreads the newly synthesized DNA strand.
,Answer: A
Rationale: DNA polymerases cannot initiate DNA synthesis de novo and require a pre-
existing 3’-OH group. Primase provides this by creating a short RNA sequence that is
complementary to the template strand. This primer serves as the foundation upon which
DNA polymerase III adds subsequent nucleotides.
3. On the lagging strand, DNA is synthesized in short segments known as:
A. Leading fragments
B. Primer sequences
C. TATA boxes
D. Okazaki fragments
Answer: D
Rationale: Because DNA polymerase only works in the 5’ to 3’ direction, the lagging strand
must be synthesized discontinuously. These individual segments are later joined together
to form a continuous DNA strand. This discovery was pivotal in understanding the
asymmetrical nature of the replication fork.
4. Which enzyme removes the RNA primers and fills the gaps with DNA nucleotides in
prokaryotes?
A. DNA Polymerase III
B. Telomerase
, C. Ligase
D. DNA Polymerase I
Answer: D
Rationale: DNA Polymerase I has a unique 5’ to 3’ exonuclease activity that allows it to
excise RNA primers. Once the RNA is removed, its polymerase activity fills the resulting gap
with DNA nucleotides. This is a critical step in finalizing the maturation of the lagging
strand.
5. What is the role of DNA ligase in the replication process?
A. Stabilizing single-stranded DNA
B. Relaxing supercoiled DNA
C. Synthesizing RNA primers
D. Joining the phosphodiester backbone between adjacent DNA fragments.
Answer: D
Rationale: Ligase catalyzes the formation of a phosphodiester bond between the 3’-OH and
5’-phosphate of two DNA segments. This action effectively connects Okazaki fragments into
a single, continuous strand. It is the final enzymatic step required to complete DNA
synthesis on the lagging strand.
6. Which subunit of the E. coli RNA polymerase is responsible for promoter recognition?
A. Alpha subunit
| Actual Q&A with Rationale (PCB3063
Exam 3) | University of Central Florida
1. Which enzyme is primarily responsible for unwinding the DNA double helix at the
replication fork in E. coli?
A. Helicase
B. DNA Gyrase
C. Primase
D. Single-stranded binding proteins
Answer: A
Rationale: Helicase utilizes energy from ATP hydrolysis to break the hydrogen bonds
between complementary base pairs. This process allows the two strands to separate and
serve as templates for new DNA synthesis. Without helicase activity, the replication fork
would not be able to progress along the chromosome.
2. What is the function of the enzyme primase during DNA replication?
A. It synthesizes a short RNA primer to provide a 3’-OH group.
B. It removes RNA primers from the lagging strand.
C. It seals the nicks between Okazaki fragments.
D. It proofreads the newly synthesized DNA strand.
,Answer: A
Rationale: DNA polymerases cannot initiate DNA synthesis de novo and require a pre-
existing 3’-OH group. Primase provides this by creating a short RNA sequence that is
complementary to the template strand. This primer serves as the foundation upon which
DNA polymerase III adds subsequent nucleotides.
3. On the lagging strand, DNA is synthesized in short segments known as:
A. Leading fragments
B. Primer sequences
C. TATA boxes
D. Okazaki fragments
Answer: D
Rationale: Because DNA polymerase only works in the 5’ to 3’ direction, the lagging strand
must be synthesized discontinuously. These individual segments are later joined together
to form a continuous DNA strand. This discovery was pivotal in understanding the
asymmetrical nature of the replication fork.
4. Which enzyme removes the RNA primers and fills the gaps with DNA nucleotides in
prokaryotes?
A. DNA Polymerase III
B. Telomerase
, C. Ligase
D. DNA Polymerase I
Answer: D
Rationale: DNA Polymerase I has a unique 5’ to 3’ exonuclease activity that allows it to
excise RNA primers. Once the RNA is removed, its polymerase activity fills the resulting gap
with DNA nucleotides. This is a critical step in finalizing the maturation of the lagging
strand.
5. What is the role of DNA ligase in the replication process?
A. Stabilizing single-stranded DNA
B. Relaxing supercoiled DNA
C. Synthesizing RNA primers
D. Joining the phosphodiester backbone between adjacent DNA fragments.
Answer: D
Rationale: Ligase catalyzes the formation of a phosphodiester bond between the 3’-OH and
5’-phosphate of two DNA segments. This action effectively connects Okazaki fragments into
a single, continuous strand. It is the final enzymatic step required to complete DNA
synthesis on the lagging strand.
6. Which subunit of the E. coli RNA polymerase is responsible for promoter recognition?
A. Alpha subunit
