PCB 3063 Final Exam V2 | PCB 3063
Genetics | Actual Q&A with Rationale
(PCB3063 Final Exam) | University of
Central Florida
1. In a standard Mendelian dihybrid cross (AaBb x AaBb), what is the expected phenotypic
ratio assuming independent assortment and complete dominance?
A. 3:1
B. 9:3:3:1
C. 1:2:1
D. 1:1:1:1
Answer: B
Rationale: The 9:3:3:1 ratio represents the four possible phenotypic combinations
resulting from independent assortment of two genes. This ratio assumes that both parents
are heterozygous and that there is no interaction between the two gene loci. It is a
fundamental benchmark in classical genetics for predicting offspring variation.
2. Which enzyme is primarily responsible for unwinding the DNA double helix at the
replication fork in E. coli?
A. DNA Polymerase I
B. Helicase
,C. Topoisomerase
D. Primase
Answer: B
Rationale: Helicase uses ATP hydrolysis to break the hydrogen bonds between the
nitrogenous bases of the two strands. By separating the strands, it creates the replication
fork necessary for polymerase enzymes to access the template. This process is essential for
the initiation and progression of DNA synthesis.
3. If a sample of DNA is found to contain 20% Cytosine, what percentage of the DNA is
composed of Adenine?
A. 20%
B. 30%
C. 40%
D. 60%
Answer: B
Rationale: According to Chargaff’s rules, the amount of Cytosine equals Guanine, so G is
also 20%, totaling 40% for G+C. The remaining 60% must be split equally between Adenine
and Thymine. Therefore, Adenine makes up exactly 30% of the total base composition.
, 4. During the Lac operon regulation, what happens when both Glucose and Lactose are
present in the environment?
A. Transcription is at its maximum level.
B. Transcription occurs at a very low basal level.
C. The repressor remains bound to the operator.
D. cAMP levels are high, activating CAP.
Answer: B
Rationale: The presence of lactose removes the repressor, but high glucose levels keep
cAMP levels low, preventing the CAP activator from binding. Without the CAP-cAMP
complex, RNA polymerase binds inefficiently to the promoter. Consequently, the operon is
only transcribed at a minimal, basal level.
5. A true-breeding plant with red flowers is crossed with a true-breeding plant with white
flowers. If all F1 offspring have pink flowers, what inheritance pattern is exhibited?
A. Codominance
B. Incomplete dominance
C. Complete dominance
D. Epistasis
Answer: B
Genetics | Actual Q&A with Rationale
(PCB3063 Final Exam) | University of
Central Florida
1. In a standard Mendelian dihybrid cross (AaBb x AaBb), what is the expected phenotypic
ratio assuming independent assortment and complete dominance?
A. 3:1
B. 9:3:3:1
C. 1:2:1
D. 1:1:1:1
Answer: B
Rationale: The 9:3:3:1 ratio represents the four possible phenotypic combinations
resulting from independent assortment of two genes. This ratio assumes that both parents
are heterozygous and that there is no interaction between the two gene loci. It is a
fundamental benchmark in classical genetics for predicting offspring variation.
2. Which enzyme is primarily responsible for unwinding the DNA double helix at the
replication fork in E. coli?
A. DNA Polymerase I
B. Helicase
,C. Topoisomerase
D. Primase
Answer: B
Rationale: Helicase uses ATP hydrolysis to break the hydrogen bonds between the
nitrogenous bases of the two strands. By separating the strands, it creates the replication
fork necessary for polymerase enzymes to access the template. This process is essential for
the initiation and progression of DNA synthesis.
3. If a sample of DNA is found to contain 20% Cytosine, what percentage of the DNA is
composed of Adenine?
A. 20%
B. 30%
C. 40%
D. 60%
Answer: B
Rationale: According to Chargaff’s rules, the amount of Cytosine equals Guanine, so G is
also 20%, totaling 40% for G+C. The remaining 60% must be split equally between Adenine
and Thymine. Therefore, Adenine makes up exactly 30% of the total base composition.
, 4. During the Lac operon regulation, what happens when both Glucose and Lactose are
present in the environment?
A. Transcription is at its maximum level.
B. Transcription occurs at a very low basal level.
C. The repressor remains bound to the operator.
D. cAMP levels are high, activating CAP.
Answer: B
Rationale: The presence of lactose removes the repressor, but high glucose levels keep
cAMP levels low, preventing the CAP activator from binding. Without the CAP-cAMP
complex, RNA polymerase binds inefficiently to the promoter. Consequently, the operon is
only transcribed at a minimal, basal level.
5. A true-breeding plant with red flowers is crossed with a true-breeding plant with white
flowers. If all F1 offspring have pink flowers, what inheritance pattern is exhibited?
A. Codominance
B. Incomplete dominance
C. Complete dominance
D. Epistasis
Answer: B
