PORTAGE LEARNING • ESSENTIAL BIOLOGY II
Final Exam: BIOD 102/ BIOD102
(Latest 2026/2027 Update)
Essential Biology II W/Lab
A+
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. Molecular Genetics & Gene Expression (DNA replication, transcription, translation, regulation, repair)
2. Endocrine System & Hormone Signaling (glands, feedback, disease states)
3. Neurophysiology, Muscle & Reflexes (action potentials, EC coupling, reflex arcs)
4. Homeostatic Mechanisms (calcium, glucose, osmoregulation, blood pressure, thermoregulation)
5. Pharmacology, Epidemiology & Laboratory Applications (PK/PD, trials, outbreak investigation, labs)
STUVIAACTUALEXAM
Original practice content aligned to public course blueprint • Not an official secured exam
, Section 1: Molecular Genetics & Gene Expression
Q1. A molecular biologist is studying a newly discovered DNA polymerase isolated from a thermophilic bacterium.
During an in vitro replication assay at 72°C, the enzyme continues synthesizing DNA even after the template strand is
briefly heated and cooled. The researcher notes that the polymerase retains activity after exposure to temperatures
that would denature most mesophilic enzymes. Based on these observations, which property most likely explains the
enzyme's performance in the assay?
A. High processivity conferred by a sliding clamp homolog that remains bound at elevated temperatures
B. Intrinsic thermostability of the catalytic domain allowing continuous polymerization without dissociation
C. Preference for RNA primers that remain annealed longer than DNA primers at high temperature
D. Ability to initiate replication de novo without requiring an RNA primer or existing 3'-OH
Correct Answer: B
Rationale: Thermophilic DNA polymerases such as Taq retain catalytic activity at high temperatures due to structural adaptations
that stabilize the protein fold. The scenario describes continued synthesis after thermal cycling, pointing primarily to intrinsic
thermostability rather than accessory clamp proteins, primer preference, or primer-independent initiation (which is characteristic of
primases or certain viral enzymes, not typical DNA polymerases).
Q2. During transcription of a eukaryotic gene encoding a membrane receptor, RNA polymerase II pauses shortly after
initiation. Chromatin immunoprecipitation reveals enrichment of a multiprotein complex at the promoter-proximal region.
Subsequent treatment with a kinase inhibitor that blocks phosphorylation of the CTD of RNA polymerase II prevents
release of the polymerase into productive elongation. Which molecular event is most directly impaired by the kinase
inhibitor in this experimental system?
A. Recruitment of the mediator complex to the enhancer region upstream of the gene
B. Splicing of the first intron by the U1 and U2 snRNPs of the spliceosome
C. Cleavage and polyadenylation of the nascent transcript at the poly-A signal
D. Transition from initiation to elongation via phosphorylation of serine residues on the CTD
Correct Answer: D
Rationale: Phosphorylation of the C-terminal domain (CTD) of RNA polymerase II by kinases such as CDK7/CDK9 is required for the
transition from promoter-proximal pausing to productive elongation. Blocking this phosphorylation keeps the polymerase stalled near
the promoter. Mediator recruitment, polyadenylation, and splicing occur at different stages and are not the primary direct targets of
CTD kinase inhibition in this context.
Q3. A patient presents with a rare autosomal recessive disorder characterized by ultraviolet sensitivity and defective
nucleotide excision repair. Cultured fibroblasts from the patient show markedly reduced repair of UV-induced thymine
dimers. Sequencing identifies biallelic loss-of-function variants in a gene encoding a helicase that unwinds DNA around
the lesion. Which step of nucleotide excision repair is most likely compromised in this patient's cells?
A. Recognition of the helix distortion by XPC or the UV-DDB complex
B. Unwinding of the DNA duplex flanking the lesion to create a single-stranded bubble
C. Dual incision of the damaged strand by the XPF-ERCC1 and XPG endonucleases
D. Resynthesis of the excised oligonucleotide by DNA polymerase δ or ε and ligation
Correct Answer: B
Rationale: The helicase activity (typically TFIIH components XPB/XPD) is required to open the DNA around the damage site after
initial recognition. Loss of this helicase function prevents formation of the repair bubble needed for subsequent dual incision.
Recognition, incision, and resynthesis steps remain intact in principle but cannot proceed efficiently without proper unwinding.
STUVIAACTUALEXAM —1— BIOD 102 Final Exam 2026/2027
, Q4. In a bacterial operon under negative control, a repressor protein binds an operator sequence and blocks
transcription. Addition of an inducer molecule causes a conformational change in the repressor, reducing its affinity for
the operator. A researcher introduces a mutation that locks the repressor in the high-affinity DNA-binding conformation
regardless of inducer presence. What transcriptional outcome is expected for the structural genes of this operon under
both inducing and non-inducing conditions?
A. Complete absence of transcription under both inducing and non-inducing conditions
B. Constitutive high-level expression independent of inducer concentration
C. Inducible expression that remains fully responsive to the inducer molecule
D. Transient expression only during the transition from non-inducing to inducing conditions
Correct Answer: A
Rationale: A repressor locked in the DNA-binding conformation will continuously occupy the operator and prevent RNA polymerase
from initiating transcription, irrespective of inducer levels. This produces a permanent 'off' state (uninducible phenotype), the opposite
of constitutive expression that would result from a non-binding repressor mutation.
Q5. A graduate student is translating a synthetic mRNA in a cell-free system containing all necessary amino acids,
tRNAs, and ribosomes. The mRNA sequence begins with an AUG codon followed by a series of alternating C and U
nucleotides. The resulting polypeptide contains alternating proline and serine residues. Which feature of the genetic
code is most clearly illustrated by this experimental result?
A. Codons are read in non-overlapping triplets beginning from a fixed start site
B. The code is degenerate, with multiple codons specifying the same amino acid
C. The code is overlapping, allowing each nucleotide to participate in multiple codons
D. Stop codons can be suppressed by mutated tRNAs under certain conditions
Correct Answer: A
Rationale: The production of a regular alternating Pro-Ser polypeptide from an alternating CU sequence demonstrates that the
mRNA is decoded in successive, non-overlapping triplet codons (CUC = Pro, UCU = Ser, etc.) starting from the AUG initiator.
Overlapping reading would produce a more complex mixture; degeneracy and stop-codon suppression are not the primary
phenomena illustrated here.
Q6. During DNA replication in a eukaryotic cell, the lagging strand is synthesized discontinuously. After an Okazaki
fragment is completed, the RNA primer must be removed and the resulting gap filled before ligation can occur. Which
enzymatic activities are sequentially required to process the Okazaki fragment and join it to the previous fragment?
A. DNA polymerase α removes the primer; DNA ligase seals the nick without further polymerization
B. Exonuclease III trims the 3' end; DNA polymerase β fills from the 5' end; ligase joins fragments
C. Helicase unwinds the primer; primase resynthesizes DNA; topoisomerase religates the backbone
D. RNase H or FEN1 removes the RNA primer; DNA polymerase δ fills the gap; DNA ligase seals the nick
Correct Answer: D
Rationale: In eukaryotes, RNase H and/or flap endonuclease 1 (FEN1) remove the RNA primer. DNA polymerase δ then fills the
resulting gap with DNA, and DNA ligase I seals the remaining nick. DNA polymerase α is primarily involved in primer synthesis, not
removal; the other listed combinations misassign the enzymatic roles.
Q7. A frameshift mutation is introduced by insertion of a single nucleotide within the coding sequence of a gene. The
resulting mRNA is translated until a premature stop codon is encountered. In eukaryotic cells, this aberrant mRNA is
rapidly degraded by a surveillance pathway. Which cellular process is primarily responsible for eliminating this
defective transcript?
A. RNA interference directed by endogenous siRNAs complementary to the mutated region
B. Exosome-mediated degradation initiated by deadenylation of the poly-A tail
C. Nonsense-mediated mRNA decay triggered by the premature termination codon
D. Transcriptional silencing via histone deacetylation at the mutated gene locus
Correct Answer: C
Rationale: Nonsense-mediated decay (NMD) specifically recognizes mRNAs containing premature termination codons (often
marked by exon-junction complexes downstream of the stop) and targets them for degradation. While the exosome participates in
general mRNA turnover, NMD is the specialized pathway activated by frameshift-induced premature stops. RNAi and transcriptional
silencing act at different levels.
STUVIAACTUALEXAM —2— BIOD 102 Final Exam 2026/2027
Final Exam: BIOD 102/ BIOD102
(Latest 2026/2027 Update)
Essential Biology II W/Lab
A+
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. Molecular Genetics & Gene Expression (DNA replication, transcription, translation, regulation, repair)
2. Endocrine System & Hormone Signaling (glands, feedback, disease states)
3. Neurophysiology, Muscle & Reflexes (action potentials, EC coupling, reflex arcs)
4. Homeostatic Mechanisms (calcium, glucose, osmoregulation, blood pressure, thermoregulation)
5. Pharmacology, Epidemiology & Laboratory Applications (PK/PD, trials, outbreak investigation, labs)
STUVIAACTUALEXAM
Original practice content aligned to public course blueprint • Not an official secured exam
, Section 1: Molecular Genetics & Gene Expression
Q1. A molecular biologist is studying a newly discovered DNA polymerase isolated from a thermophilic bacterium.
During an in vitro replication assay at 72°C, the enzyme continues synthesizing DNA even after the template strand is
briefly heated and cooled. The researcher notes that the polymerase retains activity after exposure to temperatures
that would denature most mesophilic enzymes. Based on these observations, which property most likely explains the
enzyme's performance in the assay?
A. High processivity conferred by a sliding clamp homolog that remains bound at elevated temperatures
B. Intrinsic thermostability of the catalytic domain allowing continuous polymerization without dissociation
C. Preference for RNA primers that remain annealed longer than DNA primers at high temperature
D. Ability to initiate replication de novo without requiring an RNA primer or existing 3'-OH
Correct Answer: B
Rationale: Thermophilic DNA polymerases such as Taq retain catalytic activity at high temperatures due to structural adaptations
that stabilize the protein fold. The scenario describes continued synthesis after thermal cycling, pointing primarily to intrinsic
thermostability rather than accessory clamp proteins, primer preference, or primer-independent initiation (which is characteristic of
primases or certain viral enzymes, not typical DNA polymerases).
Q2. During transcription of a eukaryotic gene encoding a membrane receptor, RNA polymerase II pauses shortly after
initiation. Chromatin immunoprecipitation reveals enrichment of a multiprotein complex at the promoter-proximal region.
Subsequent treatment with a kinase inhibitor that blocks phosphorylation of the CTD of RNA polymerase II prevents
release of the polymerase into productive elongation. Which molecular event is most directly impaired by the kinase
inhibitor in this experimental system?
A. Recruitment of the mediator complex to the enhancer region upstream of the gene
B. Splicing of the first intron by the U1 and U2 snRNPs of the spliceosome
C. Cleavage and polyadenylation of the nascent transcript at the poly-A signal
D. Transition from initiation to elongation via phosphorylation of serine residues on the CTD
Correct Answer: D
Rationale: Phosphorylation of the C-terminal domain (CTD) of RNA polymerase II by kinases such as CDK7/CDK9 is required for the
transition from promoter-proximal pausing to productive elongation. Blocking this phosphorylation keeps the polymerase stalled near
the promoter. Mediator recruitment, polyadenylation, and splicing occur at different stages and are not the primary direct targets of
CTD kinase inhibition in this context.
Q3. A patient presents with a rare autosomal recessive disorder characterized by ultraviolet sensitivity and defective
nucleotide excision repair. Cultured fibroblasts from the patient show markedly reduced repair of UV-induced thymine
dimers. Sequencing identifies biallelic loss-of-function variants in a gene encoding a helicase that unwinds DNA around
the lesion. Which step of nucleotide excision repair is most likely compromised in this patient's cells?
A. Recognition of the helix distortion by XPC or the UV-DDB complex
B. Unwinding of the DNA duplex flanking the lesion to create a single-stranded bubble
C. Dual incision of the damaged strand by the XPF-ERCC1 and XPG endonucleases
D. Resynthesis of the excised oligonucleotide by DNA polymerase δ or ε and ligation
Correct Answer: B
Rationale: The helicase activity (typically TFIIH components XPB/XPD) is required to open the DNA around the damage site after
initial recognition. Loss of this helicase function prevents formation of the repair bubble needed for subsequent dual incision.
Recognition, incision, and resynthesis steps remain intact in principle but cannot proceed efficiently without proper unwinding.
STUVIAACTUALEXAM —1— BIOD 102 Final Exam 2026/2027
, Q4. In a bacterial operon under negative control, a repressor protein binds an operator sequence and blocks
transcription. Addition of an inducer molecule causes a conformational change in the repressor, reducing its affinity for
the operator. A researcher introduces a mutation that locks the repressor in the high-affinity DNA-binding conformation
regardless of inducer presence. What transcriptional outcome is expected for the structural genes of this operon under
both inducing and non-inducing conditions?
A. Complete absence of transcription under both inducing and non-inducing conditions
B. Constitutive high-level expression independent of inducer concentration
C. Inducible expression that remains fully responsive to the inducer molecule
D. Transient expression only during the transition from non-inducing to inducing conditions
Correct Answer: A
Rationale: A repressor locked in the DNA-binding conformation will continuously occupy the operator and prevent RNA polymerase
from initiating transcription, irrespective of inducer levels. This produces a permanent 'off' state (uninducible phenotype), the opposite
of constitutive expression that would result from a non-binding repressor mutation.
Q5. A graduate student is translating a synthetic mRNA in a cell-free system containing all necessary amino acids,
tRNAs, and ribosomes. The mRNA sequence begins with an AUG codon followed by a series of alternating C and U
nucleotides. The resulting polypeptide contains alternating proline and serine residues. Which feature of the genetic
code is most clearly illustrated by this experimental result?
A. Codons are read in non-overlapping triplets beginning from a fixed start site
B. The code is degenerate, with multiple codons specifying the same amino acid
C. The code is overlapping, allowing each nucleotide to participate in multiple codons
D. Stop codons can be suppressed by mutated tRNAs under certain conditions
Correct Answer: A
Rationale: The production of a regular alternating Pro-Ser polypeptide from an alternating CU sequence demonstrates that the
mRNA is decoded in successive, non-overlapping triplet codons (CUC = Pro, UCU = Ser, etc.) starting from the AUG initiator.
Overlapping reading would produce a more complex mixture; degeneracy and stop-codon suppression are not the primary
phenomena illustrated here.
Q6. During DNA replication in a eukaryotic cell, the lagging strand is synthesized discontinuously. After an Okazaki
fragment is completed, the RNA primer must be removed and the resulting gap filled before ligation can occur. Which
enzymatic activities are sequentially required to process the Okazaki fragment and join it to the previous fragment?
A. DNA polymerase α removes the primer; DNA ligase seals the nick without further polymerization
B. Exonuclease III trims the 3' end; DNA polymerase β fills from the 5' end; ligase joins fragments
C. Helicase unwinds the primer; primase resynthesizes DNA; topoisomerase religates the backbone
D. RNase H or FEN1 removes the RNA primer; DNA polymerase δ fills the gap; DNA ligase seals the nick
Correct Answer: D
Rationale: In eukaryotes, RNase H and/or flap endonuclease 1 (FEN1) remove the RNA primer. DNA polymerase δ then fills the
resulting gap with DNA, and DNA ligase I seals the remaining nick. DNA polymerase α is primarily involved in primer synthesis, not
removal; the other listed combinations misassign the enzymatic roles.
Q7. A frameshift mutation is introduced by insertion of a single nucleotide within the coding sequence of a gene. The
resulting mRNA is translated until a premature stop codon is encountered. In eukaryotic cells, this aberrant mRNA is
rapidly degraded by a surveillance pathway. Which cellular process is primarily responsible for eliminating this
defective transcript?
A. RNA interference directed by endogenous siRNAs complementary to the mutated region
B. Exosome-mediated degradation initiated by deadenylation of the poly-A tail
C. Nonsense-mediated mRNA decay triggered by the premature termination codon
D. Transcriptional silencing via histone deacetylation at the mutated gene locus
Correct Answer: C
Rationale: Nonsense-mediated decay (NMD) specifically recognizes mRNAs containing premature termination codons (often
marked by exon-junction complexes downstream of the stop) and targets them for degradation. While the exosome participates in
general mRNA turnover, NMD is the specialized pathway activated by frameshift-induced premature stops. RNAi and transcriptional
silencing act at different levels.
STUVIAACTUALEXAM —2— BIOD 102 Final Exam 2026/2027
