PORTAGE LEARNING — ESSENTIAL BIOLOGY II
Module 1: BIOD 102/ BIOD102 (Latest 2026/2027 Update)
A+
Essential Biology II W/Lab | Questions & Answers| Grade A
100% Correct (Verified Solutions)- Portage Learning
2026/2027 Official Exam
A+ 5 100%
QUESTIONS VERIFIED EXAM DOMAINS COVERED RATIONALES INCLUDED
CATEGORIES
■ 1. DNA Structure & Replication
■ 2. Transcription & RNA Processing
■ 3. Translation & the Genetic Code
■ 4. Gene Regulation (Prokaryotic & Eukaryotic)
■ 5. Mutations & DNA Repair Mechanisms
STUVIAACTUALEXAM
Original practice examination aligned to Portage Learning BIOD 102 Module 1 learning outcomes (DNA replication & protein synthesis). Not affiliated with Portage Learning.
, SECTION 1 — DNA Structure & Replication
Q1. A researcher isolates DNA from a bacterial culture and determines that 28 % of the bases are adenine. Using Chargaff’s rules, the
percentage of thymine in this DNA sample should be approximately:
A. 28 %
B. 14 %
C. 50 %
D. 22 %
Correct Answer: A
Rationale: Chargaff’s rules state that the amount of adenine equals the amount of thymine (A = T) and the amount of guanine equals the amount of
cytosine (G = C). Therefore a DNA sample containing 28 % adenine also contains 28 % thymine.
Q2. During DNA replication the two strands of the double helix are separated by which enzyme, allowing each strand to serve as a
template?
A. DNA polymerase I
B. Helicase
C. Primase
D. DNA ligase
Correct Answer: B
Rationale: Helicase unwinds the double helix by breaking the hydrogen bonds between complementary bases, creating the replication fork. The other
listed enzymes perform different roles in replication.
Q3. Okazaki fragments are synthesized on the lagging strand because:
A. DNA polymerase can add nucleotides only to a 3′ end and the lagging strand is oriented 5′→3′ overall relative to the fork movement
B. Helicase works exclusively on the lagging strand
C. The lagging strand contains only RNA
D. Ligase synthesizes the lagging strand continuously
Correct Answer: A
Rationale: DNA polymerases synthesize DNA exclusively in the 5′→3′ direction. Because the two template strands are antiparallel, one new strand (the
lagging strand) must be made discontinuously as short Okazaki fragments.
Q4. The enzyme that joins the sugar-phosphate backbones of adjacent Okazaki fragments is:
A. Helicase
B. Primase
C. Topoisomerase
D. DNA ligase
Correct Answer: D
Rationale: After the RNA primers of Okazaki fragments are removed and the gaps filled with DNA, DNA ligase seals the remaining nicks by forming
phosphodiester bonds, producing a continuous lagging strand.
Q5. A mutation disables the 3′→5′ exonuclease activity of DNA polymerase III. The most immediate consequence is:
A. Failure to unwind the parental double helix
B. Reduced proofreading and a higher rate of replication errors
C. Inability to initiate primer synthesis
D. Loss of the ability to remove RNA primers
Correct Answer: B
Rationale: The 3′→5′ exonuclease activity of DNA polymerase III provides the proofreading function that removes incorrectly incorporated nucleotides.
Loss of this activity elevates the mutation rate during replication.
Q6. In eukaryotes, the ends of linear chromosomes are maintained by the enzyme telomerase. Telomerase is a specialized:
A. Ligase that seals telomeric nicks
B. Helicase that unwinds telomeric DNA
C. Reverse transcriptase that extends the 3′ overhang using an internal RNA template
D. DNA polymerase that fills gaps left by primer removal
Correct Answer: C
Rationale: Telomerase carries its own RNA template and adds telomeric repeats to the 3′ end of the chromosome by reverse transcription, compensating
for the end-replication problem of linear DNA.
Q7. Single-strand binding proteins (SSBs) function during replication by:
A. Coating exposed single-stranded DNA to prevent re-annealing and protect it from nucleases
B. Relieving supercoiling ahead of the fork
C. Sealing nicks between Okazaki fragments
D. Synthesizing RNA primers
Correct Answer: A
Rationale: Once helicase separates the strands, SSBs bind the single-stranded DNA, stabilizing it in an extended conformation and preventing
secondary-structure formation or re-annealing until a new complementary strand is synthesized.
STUVIAACTUALEXAM | BIOD 102 Module 1 2026/2027 | Page 2
Module 1: BIOD 102/ BIOD102 (Latest 2026/2027 Update)
A+
Essential Biology II W/Lab | Questions & Answers| Grade A
100% Correct (Verified Solutions)- Portage Learning
2026/2027 Official Exam
A+ 5 100%
QUESTIONS VERIFIED EXAM DOMAINS COVERED RATIONALES INCLUDED
CATEGORIES
■ 1. DNA Structure & Replication
■ 2. Transcription & RNA Processing
■ 3. Translation & the Genetic Code
■ 4. Gene Regulation (Prokaryotic & Eukaryotic)
■ 5. Mutations & DNA Repair Mechanisms
STUVIAACTUALEXAM
Original practice examination aligned to Portage Learning BIOD 102 Module 1 learning outcomes (DNA replication & protein synthesis). Not affiliated with Portage Learning.
, SECTION 1 — DNA Structure & Replication
Q1. A researcher isolates DNA from a bacterial culture and determines that 28 % of the bases are adenine. Using Chargaff’s rules, the
percentage of thymine in this DNA sample should be approximately:
A. 28 %
B. 14 %
C. 50 %
D. 22 %
Correct Answer: A
Rationale: Chargaff’s rules state that the amount of adenine equals the amount of thymine (A = T) and the amount of guanine equals the amount of
cytosine (G = C). Therefore a DNA sample containing 28 % adenine also contains 28 % thymine.
Q2. During DNA replication the two strands of the double helix are separated by which enzyme, allowing each strand to serve as a
template?
A. DNA polymerase I
B. Helicase
C. Primase
D. DNA ligase
Correct Answer: B
Rationale: Helicase unwinds the double helix by breaking the hydrogen bonds between complementary bases, creating the replication fork. The other
listed enzymes perform different roles in replication.
Q3. Okazaki fragments are synthesized on the lagging strand because:
A. DNA polymerase can add nucleotides only to a 3′ end and the lagging strand is oriented 5′→3′ overall relative to the fork movement
B. Helicase works exclusively on the lagging strand
C. The lagging strand contains only RNA
D. Ligase synthesizes the lagging strand continuously
Correct Answer: A
Rationale: DNA polymerases synthesize DNA exclusively in the 5′→3′ direction. Because the two template strands are antiparallel, one new strand (the
lagging strand) must be made discontinuously as short Okazaki fragments.
Q4. The enzyme that joins the sugar-phosphate backbones of adjacent Okazaki fragments is:
A. Helicase
B. Primase
C. Topoisomerase
D. DNA ligase
Correct Answer: D
Rationale: After the RNA primers of Okazaki fragments are removed and the gaps filled with DNA, DNA ligase seals the remaining nicks by forming
phosphodiester bonds, producing a continuous lagging strand.
Q5. A mutation disables the 3′→5′ exonuclease activity of DNA polymerase III. The most immediate consequence is:
A. Failure to unwind the parental double helix
B. Reduced proofreading and a higher rate of replication errors
C. Inability to initiate primer synthesis
D. Loss of the ability to remove RNA primers
Correct Answer: B
Rationale: The 3′→5′ exonuclease activity of DNA polymerase III provides the proofreading function that removes incorrectly incorporated nucleotides.
Loss of this activity elevates the mutation rate during replication.
Q6. In eukaryotes, the ends of linear chromosomes are maintained by the enzyme telomerase. Telomerase is a specialized:
A. Ligase that seals telomeric nicks
B. Helicase that unwinds telomeric DNA
C. Reverse transcriptase that extends the 3′ overhang using an internal RNA template
D. DNA polymerase that fills gaps left by primer removal
Correct Answer: C
Rationale: Telomerase carries its own RNA template and adds telomeric repeats to the 3′ end of the chromosome by reverse transcription, compensating
for the end-replication problem of linear DNA.
Q7. Single-strand binding proteins (SSBs) function during replication by:
A. Coating exposed single-stranded DNA to prevent re-annealing and protect it from nucleases
B. Relieving supercoiling ahead of the fork
C. Sealing nicks between Okazaki fragments
D. Synthesizing RNA primers
Correct Answer: A
Rationale: Once helicase separates the strands, SSBs bind the single-stranded DNA, stabilizing it in an extended conformation and preventing
secondary-structure formation or re-annealing until a new complementary strand is synthesized.
STUVIAACTUALEXAM | BIOD 102 Module 1 2026/2027 | Page 2
