BIOD 171 Microbiology - Module 5 Exam Portage Learning | 2026/2027 Actual Exam
BIOD 171 Microbiology
Module 5 Examination
Portage Learning | 2026/2027 Actual Exam
Verified Answers with Detailed Rationales Aligned to Portage Learning BIOD 171 Course Materials
Total Questions: 100 | Sections: 7 | Format: Multiple Choice (A-D) | Cognitive Mix: 30% Recall / 50% Application /
20% Analysis
SECTION 1: Microbial Genetics - DNA Structure and Replication (Q1-Q15)
Q1: Which of the following correctly describes the sugar-phosphate arrangement in a DNA
nucleotide?
A. A ribose sugar is bonded to a phosphate group at the 2' carbon and to a nitrogenous base at the 1'
carbon.
B. A deoxyribose sugar is bonded to a phosphate group at the 5' carbon and to a nitrogenous base
at the 1' carbon. [CORRECT]
C. A deoxyribose sugar is bonded to a phosphate group at the 3' carbon and to a nitrogenous base at the
5' carbon.
D. A glucose sugar is bonded to a phosphate group at the 6' carbon and to a nitrogenous base at the 1'
carbon.
Correct Answer: B
Rationale: A DNA nucleotide consists of a deoxyribose sugar (lacking the 2'-OH of ribose), a phosphate group
esterified at the 5' carbon, and a nitrogenous base attached at the 1' carbon via an N-glycosidic bond. Choice A
is incorrect because it describes ribose (an RNA feature) and mislocates the phosphate. Choice C reverses the 3'
and 5' positions, and D incorrectly uses glucose. Verification: Portage Learning BIOD 171 Module 5 specifies the
5'-phosphate/1'-base deoxyribose backbone as the universal DNA nucleotide architecture.
Q2: In a double-stranded DNA molecule containing 28% adenine, what is the expected guanine
content?
A. 22% [CORRECT]
B. 28%
C. 44%
D. 56%
Correct Answer: A
Rationale: By Chargaff's rules, adenine pairs with thymine (A = T = 28%), so together A + T = 56%, leaving 44%
for guanine plus cytosine (G + C = 44%), and thus G = C = 22%. Choice B confuses A with G; C incorrectly treats
G + C as if it were G alone; D mistakes the combined A+T for G+C. Verification: Portage Learning BIOD 171
reinforces Chargaff's base-pairing complementarity as the foundation of double-helix stability.
Verified Answers + Detailed Rationales Page 1
,BIOD 171 Microbiology - Module 5 Exam Portage Learning | 2026/2027 Actual Exam
Q3: Which classification correctly groups the nitrogenous bases found in DNA and RNA?
A. Purines: adenine and cytosine; Pyrimidines: guanine, thymine, and uracil
B. Purines: adenine and guanine; Pyrimidines: cytosine, thymine, and uracil [CORRECT]
C. Purines: cytosine and thymine; Pyrimidines: adenine and guanine
D. Purines: adenine and uracil; Pyrimidines: guanine, cytosine, and thymine
Correct Answer: B
Rationale: Purines are the double-ring bases adenine (A) and guanine (G); pyrimidines are the single-ring bases
cytosine (C), thymine (T, in DNA), and uracil (U, in RNA). Choice A misassigns cytosine to purines; C reverses
the categories entirely; D misclassifies uracil as a purine. Verification: Portage Learning BIOD 171 Module 5
anchors base categorization as the prerequisite for understanding complementary base pairing and the central
dogma.
Q4: The antiparallel nature of the DNA double helix means that:
A. The two strands run in the same 5' to 3' direction, allowing simultaneous replication.
B. One strand runs 5' to 3' while the complementary strand runs 3' to 5', enabling complementary
base pairing. [CORRECT]
C. The strands are oriented perpendicular to one another, forming a triple helix.
D. Both strands are composed of alternating purines and pyrimidines, ensuring equal width.
Correct Answer: B
Rationale: Antiparallel orientation means one strand runs 5'->3' while the complementary strand runs 3'->5',
which is essential for complementary A-T/G-C pairing and for directional DNA synthesis by DNA polymerase.
Choice A describes parallel strands (which would not allow proper base pairing); C describes an impossible
geometry; D confuses antiparallelism with the purine-pyrimidine alternating pattern. Verification: Portage
Learning BIOD 171 emphasizes antiparallel orientation as the structural basis for both semiconservative
replication and the leading/lagging strand distinction.
Q5: The Meselson-Stahl experiment supported which model of DNA replication?
A. Conservative replication, where the parental double helix remains intact and a wholly new copy is
synthesized.
B. Dispersive replication, where parental and newly synthesized DNA are randomly interspersed in
segments.
C. Semiconservative replication, where each daughter molecule contains one parental and one
newly synthesized strand. [CORRECT]
D. Random replication, where each daughter molecule is constructed from a random mixture of old and
new nucleotides.
Correct Answer: C
Rationale: Meselson and Stahl grew E. coli in 15N, then shifted to 14N; after one generation the DNA band was
intermediate (15N/14N hybrid), and after two generations two bands appeared (hybrid and light 14N/14N),
proving semiconservative replication in which each daughter helix retains one parental strand paired with one
new strand. Choice A (conservative) would have produced a heavy and a light band after one generation; B
(dispersive) would have produced a single shifting band. Verification: Portage Learning BIOD 171 cites
Meselson-Stahl (1958) as definitive experimental proof of semiconservative replication.
Verified Answers + Detailed Rationales Page 2
,BIOD 171 Microbiology - Module 5 Exam Portage Learning | 2026/2027 Actual Exam
Q6: An E. coli mutant produces a nonfunctional helicase. What would be the most immediate
consequence during DNA replication?
A. DNA polymerase would be unable to add nucleotides to the growing strand.
B. The two parental DNA strands would fail to unwind and separate at the replication fork.
[CORRECT]
C. Okazaki fragments could not be joined together on the lagging strand.
D. RNA primers would be unable to initiate synthesis on the leading strand.
Correct Answer: B
Rationale: Helicase unwinds the double helix at the replication fork by breaking hydrogen bonds between
complementary bases; without it, parental strands cannot separate and the fork cannot advance. Choice A
describes loss of DNA polymerase function; C describes loss of DNA ligase; D describes loss of primase.
Verification: Portage Learning BIOD 171 Module 5 identifies helicase as the initiator of strand separation at oriC,
with subsequent action of single-stranded binding proteins, primase, and polymerase.
Q7: DNA polymerase III synthesizes new DNA only in the 5' to 3' direction. What structural and
mechanistic reason explains this constraint?
A. The enzyme requires a free 5'-OH group on the incoming dNTP to add the next nucleotide.
B. The enzyme requires a free 3'-OH group on the growing strand to which the next dNTP is added,
and proofreading occurs only in the 3' to 5' direction. [CORRECT]
C. The enzyme can only read the template in the 5' to 3' direction.
D. DNA polymerase III requires an RNA primer at both ends of the new strand to function.
Correct Answer: B
Rationale: DNA polymerase III catalyzes phosphodiester bond formation using the free 3'-OH of the terminal
deoxyribose as a nucleophile attacking the 5'-phosphate of the incoming dNTP; therefore synthesis is obligately
5'->3' and the 3'->5' exonuclease proofreading activity operates in the opposite direction. Choice A is incorrect
because it is the 3'-OH that is required, not the 5'-OH; C is wrong because the template is read 3'->5'; D is
incorrect because only one primer per lagging-strand Okazaki fragment is needed. Verification: Portage Learning
BIOD 171 reinforces the 5'->3' synthetic polarity and the requirement for a free 3'-OH (or RNA primer) as
fundamental replication rules.
Q8: Which statement correctly contrasts the leading and lagging strands during DNA replication?
A. The leading strand is synthesized discontinuously as Okazaki fragments; the lagging strand is
synthesized continuously toward the fork.
B. Both strands are synthesized continuously in the same direction toward the replication fork.
C. The leading strand is synthesized continuously in the direction of fork movement; the lagging
strand is synthesized discontinuously as Okazaki fragments away from the fork. [CORRECT]
D. The leading strand requires DNA ligase, while the lagging strand does not.
Correct Answer: C
Rationale: Because DNA polymerase synthesizes only 5'->3' and the two parental strands are antiparallel, the
strand oriented 3'->5' toward the fork (leading strand) is synthesized continuously, while the strand oriented 5'->3'
toward the fork (lagging strand) is synthesized as short Okazaki fragments that are later joined. Choice A
reverses the descriptions; B ignores antiparallel polarity; D is wrong because ligase is required on the lagging
strand (to join Okazaki fragments), not the leading strand. Verification: Portage Learning BIOD 171 stresses this
as the central conceptual contrast in bacterial replication.
Verified Answers + Detailed Rationales Page 3
, BIOD 171 Microbiology - Module 5 Exam Portage Learning | 2026/2027 Actual Exam
Q9: Okazaki fragments in E. coli are typically about 1,000-2,000 nucleotides long. Which enzyme is
directly responsible for joining these fragments into a continuous strand?
A. DNA polymerase I
B. DNA polymerase III
C. DNA ligase [CORRECT]
D. Primase
Correct Answer: C
Rationale: DNA ligase catalyzes the formation of a phosphodiester bond between the 3'-OH of one Okazaki
fragment and the 5'-phosphate of the next, sealing the nick after RNA primers have been removed and replaced
by DNA polymerase I. Choice A (Pol I) removes RNA primers via 5'->3' exonuclease activity and fills the gap but
does not seal the nick; B (Pol III) is the main elongation enzyme; D (primase) synthesizes the RNA primers that
initiate each Okazaki fragment. Verification: Portage Learning BIOD 171 specifies ligase as the final sealing
enzyme of lagging-strand synthesis.
Q10: Which activity of DNA polymerase I is unique among bacterial DNA polymerases and is
essential for removing RNA primers during replication?
A. 5' to 3' polymerase activity
B. 3' to 5' exonuclease (proofreading) activity
C. 5' to 3' exonuclease activity [CORRECT]
D. Helicase activity
Correct Answer: C
Rationale: DNA polymerase I possesses a 5'->3' exonuclease activity that excises RNA primers ahead of its
polymerase activity, allowing simultaneous removal of primers and gap-filling with DNA. Choice A is shared with
Pol III; B (proofreading) is also present on Pol III; D is not a polymerase function. Verification: Portage Learning
BIOD 171 Module 5 distinguishes Pol I by this dual 5'->3' exo/poly activity, essential for primer turnover during
lagging-strand processing.
Q11: Why is primase, rather than DNA polymerase, required to initiate synthesis of every new DNA
strand?
A. Primase synthesizes a short DNA primer that DNA polymerase extends.
B. DNA polymerase cannot initiate synthesis de novo and requires a free 3'-OH provided by a short
RNA primer synthesized by primase. [CORRECT]
C. Primase unwinds the DNA double helix to expose the template strand.
D. Primase relieves supercoiling ahead of the replication fork.
Correct Answer: B
Rationale: DNA polymerases require a pre-existing 3'-OH to which they add nucleotides; they cannot initiate
chains de novo. Primase is an RNA polymerase that synthesizes a short complementary RNA primer providing
the necessary 3'-OH. Choice A is wrong because primase makes RNA, not DNA; C describes helicase; D
describes DNA gyrase. Verification: Portage Learning BIOD 171 emphasizes that primase is the replication
initiator and that primers are later replaced with DNA by Pol I.
Verified Answers + Detailed Rationales Page 4
BIOD 171 Microbiology
Module 5 Examination
Portage Learning | 2026/2027 Actual Exam
Verified Answers with Detailed Rationales Aligned to Portage Learning BIOD 171 Course Materials
Total Questions: 100 | Sections: 7 | Format: Multiple Choice (A-D) | Cognitive Mix: 30% Recall / 50% Application /
20% Analysis
SECTION 1: Microbial Genetics - DNA Structure and Replication (Q1-Q15)
Q1: Which of the following correctly describes the sugar-phosphate arrangement in a DNA
nucleotide?
A. A ribose sugar is bonded to a phosphate group at the 2' carbon and to a nitrogenous base at the 1'
carbon.
B. A deoxyribose sugar is bonded to a phosphate group at the 5' carbon and to a nitrogenous base
at the 1' carbon. [CORRECT]
C. A deoxyribose sugar is bonded to a phosphate group at the 3' carbon and to a nitrogenous base at the
5' carbon.
D. A glucose sugar is bonded to a phosphate group at the 6' carbon and to a nitrogenous base at the 1'
carbon.
Correct Answer: B
Rationale: A DNA nucleotide consists of a deoxyribose sugar (lacking the 2'-OH of ribose), a phosphate group
esterified at the 5' carbon, and a nitrogenous base attached at the 1' carbon via an N-glycosidic bond. Choice A
is incorrect because it describes ribose (an RNA feature) and mislocates the phosphate. Choice C reverses the 3'
and 5' positions, and D incorrectly uses glucose. Verification: Portage Learning BIOD 171 Module 5 specifies the
5'-phosphate/1'-base deoxyribose backbone as the universal DNA nucleotide architecture.
Q2: In a double-stranded DNA molecule containing 28% adenine, what is the expected guanine
content?
A. 22% [CORRECT]
B. 28%
C. 44%
D. 56%
Correct Answer: A
Rationale: By Chargaff's rules, adenine pairs with thymine (A = T = 28%), so together A + T = 56%, leaving 44%
for guanine plus cytosine (G + C = 44%), and thus G = C = 22%. Choice B confuses A with G; C incorrectly treats
G + C as if it were G alone; D mistakes the combined A+T for G+C. Verification: Portage Learning BIOD 171
reinforces Chargaff's base-pairing complementarity as the foundation of double-helix stability.
Verified Answers + Detailed Rationales Page 1
,BIOD 171 Microbiology - Module 5 Exam Portage Learning | 2026/2027 Actual Exam
Q3: Which classification correctly groups the nitrogenous bases found in DNA and RNA?
A. Purines: adenine and cytosine; Pyrimidines: guanine, thymine, and uracil
B. Purines: adenine and guanine; Pyrimidines: cytosine, thymine, and uracil [CORRECT]
C. Purines: cytosine and thymine; Pyrimidines: adenine and guanine
D. Purines: adenine and uracil; Pyrimidines: guanine, cytosine, and thymine
Correct Answer: B
Rationale: Purines are the double-ring bases adenine (A) and guanine (G); pyrimidines are the single-ring bases
cytosine (C), thymine (T, in DNA), and uracil (U, in RNA). Choice A misassigns cytosine to purines; C reverses
the categories entirely; D misclassifies uracil as a purine. Verification: Portage Learning BIOD 171 Module 5
anchors base categorization as the prerequisite for understanding complementary base pairing and the central
dogma.
Q4: The antiparallel nature of the DNA double helix means that:
A. The two strands run in the same 5' to 3' direction, allowing simultaneous replication.
B. One strand runs 5' to 3' while the complementary strand runs 3' to 5', enabling complementary
base pairing. [CORRECT]
C. The strands are oriented perpendicular to one another, forming a triple helix.
D. Both strands are composed of alternating purines and pyrimidines, ensuring equal width.
Correct Answer: B
Rationale: Antiparallel orientation means one strand runs 5'->3' while the complementary strand runs 3'->5',
which is essential for complementary A-T/G-C pairing and for directional DNA synthesis by DNA polymerase.
Choice A describes parallel strands (which would not allow proper base pairing); C describes an impossible
geometry; D confuses antiparallelism with the purine-pyrimidine alternating pattern. Verification: Portage
Learning BIOD 171 emphasizes antiparallel orientation as the structural basis for both semiconservative
replication and the leading/lagging strand distinction.
Q5: The Meselson-Stahl experiment supported which model of DNA replication?
A. Conservative replication, where the parental double helix remains intact and a wholly new copy is
synthesized.
B. Dispersive replication, where parental and newly synthesized DNA are randomly interspersed in
segments.
C. Semiconservative replication, where each daughter molecule contains one parental and one
newly synthesized strand. [CORRECT]
D. Random replication, where each daughter molecule is constructed from a random mixture of old and
new nucleotides.
Correct Answer: C
Rationale: Meselson and Stahl grew E. coli in 15N, then shifted to 14N; after one generation the DNA band was
intermediate (15N/14N hybrid), and after two generations two bands appeared (hybrid and light 14N/14N),
proving semiconservative replication in which each daughter helix retains one parental strand paired with one
new strand. Choice A (conservative) would have produced a heavy and a light band after one generation; B
(dispersive) would have produced a single shifting band. Verification: Portage Learning BIOD 171 cites
Meselson-Stahl (1958) as definitive experimental proof of semiconservative replication.
Verified Answers + Detailed Rationales Page 2
,BIOD 171 Microbiology - Module 5 Exam Portage Learning | 2026/2027 Actual Exam
Q6: An E. coli mutant produces a nonfunctional helicase. What would be the most immediate
consequence during DNA replication?
A. DNA polymerase would be unable to add nucleotides to the growing strand.
B. The two parental DNA strands would fail to unwind and separate at the replication fork.
[CORRECT]
C. Okazaki fragments could not be joined together on the lagging strand.
D. RNA primers would be unable to initiate synthesis on the leading strand.
Correct Answer: B
Rationale: Helicase unwinds the double helix at the replication fork by breaking hydrogen bonds between
complementary bases; without it, parental strands cannot separate and the fork cannot advance. Choice A
describes loss of DNA polymerase function; C describes loss of DNA ligase; D describes loss of primase.
Verification: Portage Learning BIOD 171 Module 5 identifies helicase as the initiator of strand separation at oriC,
with subsequent action of single-stranded binding proteins, primase, and polymerase.
Q7: DNA polymerase III synthesizes new DNA only in the 5' to 3' direction. What structural and
mechanistic reason explains this constraint?
A. The enzyme requires a free 5'-OH group on the incoming dNTP to add the next nucleotide.
B. The enzyme requires a free 3'-OH group on the growing strand to which the next dNTP is added,
and proofreading occurs only in the 3' to 5' direction. [CORRECT]
C. The enzyme can only read the template in the 5' to 3' direction.
D. DNA polymerase III requires an RNA primer at both ends of the new strand to function.
Correct Answer: B
Rationale: DNA polymerase III catalyzes phosphodiester bond formation using the free 3'-OH of the terminal
deoxyribose as a nucleophile attacking the 5'-phosphate of the incoming dNTP; therefore synthesis is obligately
5'->3' and the 3'->5' exonuclease proofreading activity operates in the opposite direction. Choice A is incorrect
because it is the 3'-OH that is required, not the 5'-OH; C is wrong because the template is read 3'->5'; D is
incorrect because only one primer per lagging-strand Okazaki fragment is needed. Verification: Portage Learning
BIOD 171 reinforces the 5'->3' synthetic polarity and the requirement for a free 3'-OH (or RNA primer) as
fundamental replication rules.
Q8: Which statement correctly contrasts the leading and lagging strands during DNA replication?
A. The leading strand is synthesized discontinuously as Okazaki fragments; the lagging strand is
synthesized continuously toward the fork.
B. Both strands are synthesized continuously in the same direction toward the replication fork.
C. The leading strand is synthesized continuously in the direction of fork movement; the lagging
strand is synthesized discontinuously as Okazaki fragments away from the fork. [CORRECT]
D. The leading strand requires DNA ligase, while the lagging strand does not.
Correct Answer: C
Rationale: Because DNA polymerase synthesizes only 5'->3' and the two parental strands are antiparallel, the
strand oriented 3'->5' toward the fork (leading strand) is synthesized continuously, while the strand oriented 5'->3'
toward the fork (lagging strand) is synthesized as short Okazaki fragments that are later joined. Choice A
reverses the descriptions; B ignores antiparallel polarity; D is wrong because ligase is required on the lagging
strand (to join Okazaki fragments), not the leading strand. Verification: Portage Learning BIOD 171 stresses this
as the central conceptual contrast in bacterial replication.
Verified Answers + Detailed Rationales Page 3
, BIOD 171 Microbiology - Module 5 Exam Portage Learning | 2026/2027 Actual Exam
Q9: Okazaki fragments in E. coli are typically about 1,000-2,000 nucleotides long. Which enzyme is
directly responsible for joining these fragments into a continuous strand?
A. DNA polymerase I
B. DNA polymerase III
C. DNA ligase [CORRECT]
D. Primase
Correct Answer: C
Rationale: DNA ligase catalyzes the formation of a phosphodiester bond between the 3'-OH of one Okazaki
fragment and the 5'-phosphate of the next, sealing the nick after RNA primers have been removed and replaced
by DNA polymerase I. Choice A (Pol I) removes RNA primers via 5'->3' exonuclease activity and fills the gap but
does not seal the nick; B (Pol III) is the main elongation enzyme; D (primase) synthesizes the RNA primers that
initiate each Okazaki fragment. Verification: Portage Learning BIOD 171 specifies ligase as the final sealing
enzyme of lagging-strand synthesis.
Q10: Which activity of DNA polymerase I is unique among bacterial DNA polymerases and is
essential for removing RNA primers during replication?
A. 5' to 3' polymerase activity
B. 3' to 5' exonuclease (proofreading) activity
C. 5' to 3' exonuclease activity [CORRECT]
D. Helicase activity
Correct Answer: C
Rationale: DNA polymerase I possesses a 5'->3' exonuclease activity that excises RNA primers ahead of its
polymerase activity, allowing simultaneous removal of primers and gap-filling with DNA. Choice A is shared with
Pol III; B (proofreading) is also present on Pol III; D is not a polymerase function. Verification: Portage Learning
BIOD 171 Module 5 distinguishes Pol I by this dual 5'->3' exo/poly activity, essential for primer turnover during
lagging-strand processing.
Q11: Why is primase, rather than DNA polymerase, required to initiate synthesis of every new DNA
strand?
A. Primase synthesizes a short DNA primer that DNA polymerase extends.
B. DNA polymerase cannot initiate synthesis de novo and requires a free 3'-OH provided by a short
RNA primer synthesized by primase. [CORRECT]
C. Primase unwinds the DNA double helix to expose the template strand.
D. Primase relieves supercoiling ahead of the replication fork.
Correct Answer: B
Rationale: DNA polymerases require a pre-existing 3'-OH to which they add nucleotides; they cannot initiate
chains de novo. Primase is an RNA polymerase that synthesizes a short complementary RNA primer providing
the necessary 3'-OH. Choice A is wrong because primase makes RNA, not DNA; C describes helicase; D
describes DNA gyrase. Verification: Portage Learning BIOD 171 emphasizes that primase is the replication
initiator and that primers are later replaced with DNA by Pol I.
Verified Answers + Detailed Rationales Page 4
