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BIOD 102 MODULE 2 ACTUAL EXAM 2026/2027 | Portage Learning Essential Biology II | Verified Questions & Answers | Pass Guaranteed - A+ Graded

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Pass the Portage Learning BIOD 102 Biology II Module 2 Exam on your first attempt with this complete 2026/2027 guide featuring verified questions and answers. This A+ Graded resource covers all Module 2 domains including plant evolution and diversity, alternation of generations, plant tissue systems (dermal, ground, vascular), primary and secondary growth, and reproductive strategies . Each answer is carefully verified and aligned with the latest Portage Learning BIOD 102 course objectives for 2026/2027 . Perfect for nursing and pre-health students seeking comprehensive Module 2 exam preparation. With our Pass Guarantee, you can confidently prepare for your BIOD 102 Module 2 assessment. Download your complete verified Q&A guide instantly!

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BIOD 102 Biology II - Module 2 Examination
Portage Learning | Updated Edition

Total Questions: 75 | Sections: 6 | Format: Multiple Choice (A–D, single best answer)
Time Allotted: 110 minutes | Passing Score: 70%

Instructions to the Candidate: Read each question carefully and select the single best answer from options
A–D. The correct option in this study edition is annotated with [CORRECT]; a separate answer key line and a
microbiology-specific rationale follow each question. Topics covered include viral structure and replication, prokaryotic
diversity and genetics, protist and fungal biology, and microbial ecology, disease, and biotechnology. Cognitive level
distribution: approximately 30% recall, 50% application, and 20% analysis. Calculators are not required.


Section 1: Viral Structure, Classification, and Replication
(Capsids, Envelopes, Genomes, Lytic/Lysogenic Cycles, & Retroviruses) - Questions 1-12


Q1: A researcher isolates a virus and observes under electron microscopy that its protein
coat is composed of repeating capsomere subunits arranged in a polyhedral geometry with
20 triangular faces. Which capsid morphology is being described, and what is the primary
evolutionary advantage of this symmetry?
A. Helical capsid; allows flexible genome packaging of varying lengths
B. Complex capsid; permits modular attachment of tail fibers for host recognition
C. Icosahedral capsid; minimizes surface-area-to-volume ratio, maximizing genome
capacity per protein subunit [CORRECT]
D. Envelope-derived capsid; provides antigenic variability through membrane
glycoproteins
Correct Answer: C
Rationale: An icosahedral capsid has 20 triangular faces built from repeated capsomeres, providing maximum interior
volume for genome packaging while requiring the smallest number of distinct protein subunits, an efficient evolutionary
strategy. Helical capsids (A) form rod-shaped structures wrapping the genome in a spiral, not polyhedral. Complex
capsids (B), such as bacteriophage T4, include head-and-tail assemblages, not pure icosahedra. Envelopes (D) are lipid
bilayers derived from host membranes, not protein capsid structures.




BIOD 102 Module 2 Examination Page 1

,BIOD 102 Biology II - Module 2 Examination | Portage Learning (2026/2027 Updated)



Q2: A patient presents with recurrent respiratory infections, and laboratory analysis
identifies an enveloped respiratory virus. Which structural featuremost directly enables
the virus to enter host cells via membrane fusion, and what is the consequence of removing
this structure experimentally?
A. Capsomere proteins; removal prevents genome uncoating but not attachment
B. Spikes (peplomers) embedded in the lipid envelope; removal blocks
receptor-mediated attachment and fusion, abolishing infectivity [CORRECT]
C. Nucleocapsid core; removal destroys genome integrity but permits attachment
D. Lipid bilayer without proteins; removal has no effect because the bilayer is purely
structural
Correct Answer: B
Rationale: Enveloped viruses carry viral glycoprotein spikes (peplomers) inserted into their host-derived lipid
envelope; these spikes mediate both host receptor recognition and membrane fusion, enabling entry. Stripping the
envelope or spikes renders the virion noninfectious because it cannot attach or fuse. Capsomeres (A) form the protein
shell but do not mediate membrane fusion. The nucleocapsid (C) packages the genome but is interior and not the entry
apparatus. The lipid bilayer (D) alone, without spikes, is functionally inert for entry.


Q3: A novel virus is isolated from a tick vector. Its genome is a single-stranded RNA
molecule that must serve directly as mRNA upon entry into the host cytoplasm. According
to the Baltimore classification, which class does this virus belong to, and what is the
immediate biosynthetic consequence?
A. Class I (dsDNA); the genome must enter the nucleus for host RNA polymerase II
transcription
B. Class III (dsRNA); the genome is segmented and requires a viral RNA-dependent RNA
polymerase
C. Class IV (+)ssRNA); the genome is directly translated by host ribosomes to produce
viral proteins [CORRECT]
D. Class V (-)ssRNA; the genome must first be copied into complementary mRNA by viral
RNA polymerase
Correct Answer: C
Rationale: A positive-sense single-stranded RNA virus (Baltimore Class IV) has a genome that is functionally
equivalent to host mRNA, so it can be immediately translated by host ribosomes upon entry, producing the viral
polyprotein or nonstructural proteins including the RNA-dependent RNA polymerase. Class I (A) is double-stranded DNA,
requiring nuclear transcription. Class III (B) requires a virion-associated polymerase because dsRNA is not
mRNA-equivalent. Class V (D) is negative-sense RNA, which is the antisense complement of mRNA and must be converted
before translation.




BIOD 102 Module 2 Examination Page 2

,BIOD 102 Biology II - Module 2 Examination | Portage Learning (2026/2027 Updated)



Q4: During an in vitro bacteriophage infection, a researcher observes the following
sequential events: (1) phage tail fibers contact host receptor, (2) base plate conformational
change, (3) sheath contraction, (4) genome transit through tail tube, and (5) immediate
shutdown of host macromolecular synthesis. Which replication cycle is occurring, and
what is the typical outcome for the host cell?
A. Lysogenic cycle; the host survives and divides, carrying the integrated prophage
B. Lytic cycle; the host undergoes lysis and releases progeny virions [CORRECT]
C. Chronic release cycle; the host buds virions continuously without lysis
D. Retroviral cycle; the genome is reverse-transcribed and integrated prior to expression
Correct Answer: B
Rationale: The sequence described, tail fiber attachment, base plate rearrangement, sheath contraction, genome
injection, and host takeover, is the canonical lytic cycle of a T-even bacteriophage, culminating in host lysis and release of
progeny virions. The lysogenic cycle (A) instead features integration of the genome as a prophage without host shutdown.
Chronic release (C) is characteristic of some enveloped animal viruses that bud, not bacteriophages. Retroviral
replication (D) involves reverse transcription and integration, not sheath contraction.


Q5: A temperate bacteriophage integrates its circular genome into the host chromosome at
a specific attachment site (attB) and persists as a prophage. Under which environmental
trigger will the prophage most commonly exit the lysogenic state and enter the lytic cycle?
A. Excess glucose in the medium, which represses the cI repressor
B. UV irradiation or DNA-damaging agents that induce the SOS response, cleaving the
cI repressor [CORRECT]
C. Anaerobic conditions that activate integrase expression
D. High multiplicity of infection by an unrelated bacteriophage
Correct Answer: B
Rationale: Lysogenic maintenance depends on the cI repressor silencing lytic genes. DNA damage, such as UV
irradiation, activates the host SOS response, leading RecA to catalyze autocleavage of cI; loss of cI repression permits
transcription of early lytic genes and excision of the prophage. Glucose (A) affects catabolite repression, not cI.
Anaerobiosis (C) does not trigger integrase for excision. Coinfection (D) does not provide a direct trigger for induction in
temperate phages.




BIOD 102 Module 2 Examination Page 3

, BIOD 102 Biology II - Module 2 Examination | Portage Learning (2026/2027 Updated)



Q6: HIV-1, the causative agent of AIDS, carries two copies of its (+)ssRNA genome along
with a virion-associated enzyme essential for its replication. Which enzyme copies the viral
RNA into DNA, and what is the immediate fate of this DNA in the host cell?
A. RNA-dependent RNA polymerase; the new RNA is translated directly in the cytoplasm
B. Reverse transcriptase; the resulting double-stranded DNA is transported into the
nucleus and integrated into the host chromosome by viral integrase [CORRECT]
C. Host DNA polymerase; the DNA remains episomal in the cytoplasm
D. Ribonuclease H; the RNA genome is degraded and translation ceases
Correct Answer: B
Rationale: HIV packages reverse transcriptase, a retroviral RNA-dependent DNA polymerase that synthesizes a
complementary DNA strand and then degrades the RNA template to produce double-stranded viral DNA. This DNA is
imported into the nucleus and integrated into the host chromosome by viral integrase, forming the provirus.
RNA-dependent RNA polymerase (A) is used by (+)ssRNA viruses like poliovirus, not retroviruses. Host DNA polymerase
(C) replicates cellular DNA, not retroviral reverse transcripts. RNase H (D) is a domain activity within reverse
transcriptase, not a standalone enzyme that halts replication.


Q7: Electron micrographs of an unidentified bacterial virus reveal an icosahedral head, a
contractile sheathed tail, tail fibers, and a base plate. Which phage is most consistent with
this description, and what does the tail sheath's contractile mechanism accomplish?
A. Lambda phage; the sheath injects a protein toxin directly into the cytoplasm
B. T4 bacteriophage; the contractile sheath acts as a molecular syringe, driving the
tail tube through the cell envelope to deliver the genome [CORRECT]
C. M13 filamentous phage; the sheath mediates budding from the host envelope
D. MS2 phage; the sheath packages the capsid during maturation
Correct Answer: B
Rationale: T4 is the textbook example of a complex bacteriophage with an icosahedral head, contractile sheathed tail,
base plate, and tail fibers; upon attachment the sheath contracts and forces the tail tube through the bacterial envelope,
delivering DNA into the host. Lambda (A) is temperate with a noncontractile tail. M13 (C) is filamentous and releases by
budding without lysis, lacking a contractile sheath. MS2 (D) is a small icosahedral RNA phage with no tail or sheath at
all.


Q8: An HIV-infected cell is treated with a drug that specifically inhibits the viral protease.
What is the most direct consequence for the viral progeny released from this cell, and why
are they noninfectious?
A. The viral genome cannot be reverse transcribed, so no provirus is formed
B. The Gag-Pol polyprotein is not cleaved into mature structural proteins and
enzymes, producing immature, noninfectious virions [CORRECT]
C. The envelope glycoprotein gp120 is internalized and the virus cannot attach to CD4
D. Host ribosomes cannot translate the viral mRNA, halting protein synthesis
Correct Answer: B
Rationale: HIV protease cleaves the Gag and Gag-Pol polyproteins during virion maturation after budding; without
these cleavages, virions retain an immature morphology with uncleaved polyproteins and are noninfectious. Reverse
transcriptase inhibitors (A) act upstream on provirus formation, not protease. Protease does not affect gp120 trafficking
(C). Protease does not interfere with host translation (D), which occurs during viral protein expression.




BIOD 102 Module 2 Examination Page 4

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