PCB 3063 Exam 2 V1 | PCB 3063 Genetics
| Actual Q&A with Rationale (PCB3063
Exam 2 | University of Central Florida
1. In a three-point testcross, which class of offspring is expected to be the least frequent?
A. Non-recombinant (parental) types
B. Double-crossover types
C. Single-crossover types
D. Class with the highest recombination frequency
Answer: B
Rationale: Double-crossover events require two independent breakage and rejoining
events to occur simultaneously between the same three genes. Because the probability of
two independent events occurring together is the product of their individual probabilities,
this class will mathematically be the rarest. This frequency helps geneticists identify the
gene located in the middle of the sequence.
2. Which enzyme is responsible for relieving torsional strain (supercoiling) ahead of the
replication fork in E. coli?
A. DNA Helicase
B. DNA Polymerase III
C. DNA Gyrase (Topoisomerase II)
,D. Primase
Answer: C
Rationale: DNA Gyrase travels ahead of the replication fork and introduces negative
supercoils to counteract the positive supercoiling caused by unwinding. Without this
enzyme, the DNA would become too tightly coiled for replication to continue effectively.
DNA helicase, by contrast, is responsible for the actual breaking of hydrogen bonds
between the bases.
3. According to Chargaff’s rules, if a DNA molecule contains 20% Adenine, what is the
percentage of Cytosine?
A. 20%
B. 60%
C. 40%
D. 30%
Answer: D
Rationale: Chargaff’s rules state that A equals T and G equals C, meaning the total of A+T is
40% in this scenario. This leaves 60% for the G+C content of the DNA molecule. Since G
must equal C, the 60% is divided equally, resulting in 30% Cytosine and 30% Guanine.
4. What is the primary function of DNA Polymerase I in prokaryotic DNA replication?
A. To synthesize the bulk of the DNA on the leading strand
, B. To unwind the double helix
C. To remove RNA primers and fill the gaps with DNA
D. To seal nicks in the sugar-phosphate backbone
Answer: C
Rationale: DNA Polymerase I possesses a unique 5’ to 3’ exonuclease activity that allows it
to excise the RNA primers placed by primase. After removing the RNA, it uses its
polymerase activity to replace those segments with the appropriate DNA nucleotides. DNA
Ligase then follows to seal the final phosphodiester bond between the fragments.
5. In eukaryotic transcription, which RNA polymerase is responsible for synthesizing
messenger RNA (mRNA)?
A. RNA Polymerase I
B. RNA Polymerase IV
C. RNA Polymerase III
D. RNA Polymerase II
Answer: D
Rationale: RNA Polymerase II is the specialized enzyme in eukaryotes that transcribes all
protein-coding genes into pre-mRNA. RNA Polymerase I handles large ribosomal RNAs,
while RNA Polymerase III transcribes tRNAs and small nuclear RNAs. This division of labor
is a key distinction between eukaryotic and prokaryotic transcription systems.
| Actual Q&A with Rationale (PCB3063
Exam 2 | University of Central Florida
1. In a three-point testcross, which class of offspring is expected to be the least frequent?
A. Non-recombinant (parental) types
B. Double-crossover types
C. Single-crossover types
D. Class with the highest recombination frequency
Answer: B
Rationale: Double-crossover events require two independent breakage and rejoining
events to occur simultaneously between the same three genes. Because the probability of
two independent events occurring together is the product of their individual probabilities,
this class will mathematically be the rarest. This frequency helps geneticists identify the
gene located in the middle of the sequence.
2. Which enzyme is responsible for relieving torsional strain (supercoiling) ahead of the
replication fork in E. coli?
A. DNA Helicase
B. DNA Polymerase III
C. DNA Gyrase (Topoisomerase II)
,D. Primase
Answer: C
Rationale: DNA Gyrase travels ahead of the replication fork and introduces negative
supercoils to counteract the positive supercoiling caused by unwinding. Without this
enzyme, the DNA would become too tightly coiled for replication to continue effectively.
DNA helicase, by contrast, is responsible for the actual breaking of hydrogen bonds
between the bases.
3. According to Chargaff’s rules, if a DNA molecule contains 20% Adenine, what is the
percentage of Cytosine?
A. 20%
B. 60%
C. 40%
D. 30%
Answer: D
Rationale: Chargaff’s rules state that A equals T and G equals C, meaning the total of A+T is
40% in this scenario. This leaves 60% for the G+C content of the DNA molecule. Since G
must equal C, the 60% is divided equally, resulting in 30% Cytosine and 30% Guanine.
4. What is the primary function of DNA Polymerase I in prokaryotic DNA replication?
A. To synthesize the bulk of the DNA on the leading strand
, B. To unwind the double helix
C. To remove RNA primers and fill the gaps with DNA
D. To seal nicks in the sugar-phosphate backbone
Answer: C
Rationale: DNA Polymerase I possesses a unique 5’ to 3’ exonuclease activity that allows it
to excise the RNA primers placed by primase. After removing the RNA, it uses its
polymerase activity to replace those segments with the appropriate DNA nucleotides. DNA
Ligase then follows to seal the final phosphodiester bond between the fragments.
5. In eukaryotic transcription, which RNA polymerase is responsible for synthesizing
messenger RNA (mRNA)?
A. RNA Polymerase I
B. RNA Polymerase IV
C. RNA Polymerase III
D. RNA Polymerase II
Answer: D
Rationale: RNA Polymerase II is the specialized enzyme in eukaryotes that transcribes all
protein-coding genes into pre-mRNA. RNA Polymerase I handles large ribosomal RNAs,
while RNA Polymerase III transcribes tRNAs and small nuclear RNAs. This division of labor
is a key distinction between eukaryotic and prokaryotic transcription systems.
