BIOD 102 FINAL EXAM 2026/2027 | Essential Biology II with
Lab | Comprehensive All Modules | Verified Q&A | Portage
Learning | Pass Guaranteed - A+ Graded
Section 1: Genetics & Molecular Biology (Modules 1 & 6)
Q1: A researcher isolates a eukaryotic mRNA transcript and observes that it is
significantly shorter than the corresponding DNA template region. Which molecular
mechanisms best explain this discrepancy, and what is the ultimate fate of the removed
sequences?
A. The DNA underwent somatic hypermutation, and the altered sequences are
translated into non-functional peptides that are rapidly degraded.
B. Transcription factors caused premature termination, and the missing sequences are
permanently deleted from the genome.
C. Introns were removed via splicing, and the excised sequences are degraded in the
nucleus. [CORRECT]
D. Exons were spliced out to conserve cellular energy, and the sequences are recycled
for new transcription events.
Correct Answer: C
Rationale: Eukaryotic genes contain non-coding introns that are removed during RNA
splicing, making the mature mRNA shorter than the initial DNA template. The excised
introns are degraded, not recycled or translated. Options A, B, and D incorrectly describe
mutagenesis, genome alteration, and the erroneous removal of coding exons. VERIFIED
✓ - Module 1: Gene Expression & Regulation.
Q2: During a high-fidelity DNA replication process in a human cell, DNA polymerase
encounters a thymine dimer caused by UV radiation. Which cellular response correctly
resolves this issue without introducing mutations?
A. DNA polymerase utilizes its 3' to 5' exonuclease activity to directly excise the dimer
and replace the nucleotides.
B. Nucleotide excision repair (NER) recognizes the distortion, excises a short
oligonucleotide containing the dimer, and DNA polymerase fills the gap. [CORRECT]
C. Mismatch repair proteins detect the abnormal base pairing and remove the newly
synthesized strand entirely.
D. Telomerase extends the 3' end of the chromosome to bypass the damaged region,
preventing replication fork collapse.
Correct Answer: B
,Rationale: Thymine dimers cause significant helical distortion, which is specifically
targeted by the nucleotide excision repair (NER) pathway, not by the exonuclease
proofreading activity of DNA polymerase or mismatch repair. Telomerase is responsible
for extending telomeres, not repairing internal DNA damage. VERIFIED ✓ - Module 1:
DNA Replication & Repair.
Q3: In a Portage Learning lab exercise, students use restriction enzymes to cut a linear
piece of DNA. Enzyme A cuts at one site, and Enzyme B cuts at two sites. If the original
DNA is 10,000 base pairs long, Enzyme A cuts at 3,500 bp, and Enzyme B cuts at 2,000
bp and 7,000 bp, how many fragments will be produced if both enzymes are used in a
single digest?
A. 3
B. 4 [CORRECT]
C. 5
D. 6
Correct Answer: B
Rationale: Enzyme A creates 2 fragments (3500 bp and 6500 bp). Enzyme B cuts within
the 3500 bp fragment (at 2000 bp) and within the 6500 bp fragment (at 7000 bp total,
meaning 5000 bp into the second fragment). This results in four total fragments (2000
bp, 1500 bp, 5000 bp, and 1500 bp). Option C assumes three cut sites always yield five
fragments, ignoring fragment size math. VERIFIED ✓ - Module 6: Molecular Techniques
& Lab Integration.
Q4: A scientist studying gene regulation in E. coli observes that a particular operon is
transcribed continuously regardless of the presence or absence of a specific
metabolite. Which structural feature of this operon most likely accounts for this
constitutive expression?
A. The operator sequence is permanently bound by an active repressor protein.
B. A mutation has deleted the promoter region, preventing RNA polymerase binding.
C. The repressor protein is non-functional due to a mutation in the regulatory gene,
preventing operator binding. [CORRECT]
D. The operon lacks a terminator sequence, causing RNA polymerase to continuously
transcribe past the normal endpoint.
Correct Answer: C
Rationale: Constitutive expression occurs when the operon cannot be turned off. If the
repressor protein is non-functional, it cannot bind to the operator to block transcription,
leading to continuous gene expression. A bound repressor (A) or missing promoter (B)
would prevent transcription, and lacking a terminator would cause read-through but not
,necessarily resolve metabolite-dependent regulation. VERIFIED ✓ - Module 1: Gene
Regulation.
Q5: In a dihybrid cross between two pea plants heterozygous for both seed color (Yy)
and seed shape (Rr), where yellow and round are dominant, what proportion of the
offspring will be homozygous recessive for both traits?
A. 1/16 [CORRECT]
B. 1/8
C. 3/16
D. 9/16
Correct Answer: A
Rationale: According to Mendelian genetics, a dihybrid cross (YyRr x YyRr) yields a
9:3:3:1 phenotypic ratio. The proportion of offspring that are homozygous recessive for
both traits (yyrr) is calculated by multiplying the probability of yy (1/4) by the probability
of rr (1/4), resulting in 1/16. VERIFIED ✓ - Module 1: Mendelian Genetics.
Q6: A population of wildflowers is in Hardy-Weinberg equilibrium for flower color, where
red (R) is dominant over white (r). If the frequency of the recessive allele (r) is 0.3, what
percentage of the population is expected to be heterozygous?
A. 9%
B. 42% [CORRECT]
C. 49%
D. 51%
Correct Answer: B
Rationale: Using the Hardy-Weinberg equations (p + q = 1, p² + 2pq + q² = 1), if q = 0.3,
then p = 0.7. The frequency of heterozygotes is represented by 2pq, which is 2(0.7)(0.3)
= 0.42, or 42%. Option A is q² (homozygous recessive), and C is p² (homozygous
dominant). VERIFIED ✓ - Module 1: Population Genetics & Hardy-Weinberg.
Q7: Which of the following statements correctly differentiates between transcription in
prokaryotes and eukaryotes?
A. Eukaryotic transcription occurs in the cytoplasm, while prokaryotic transcription
occurs in the nucleus.
B. Prokaryotic RNA polymerase requires a poly-A tail to stabilize the mRNA, whereas
eukaryotic mRNA does not.
C. Eukaryotic transcription involves three distinct RNA polymerases, while prokaryotes
use a single RNA polymerase complex. [CORRECT]
D. Prokaryotic transcription produces monocistronic mRNA, whereas eukaryotic
transcription produces polycistronic mRNA.
Correct Answer: C
, Rationale: Eukaryotes utilize RNA polymerases I, II, and III for different RNA types, while
prokaryotes rely on a single RNA polymerase composed of a core enzyme and sigma
factor. Transcription occurs in the nucleus for eukaryotes (not cytoplasm), polycistronic
mRNA is a prokaryotic feature (not eukaryotic), and poly-A tails are added to eukaryotic
mRNA (not prokaryotic). VERIFIED ✓ - Module 1: Central Dogma - Transcription.
Q8: A point mutation occurs that changes a codon from UAU to UAC. Both codons code
for the amino acid tyrosine. What type of mutation is this, and what is its likely
phenotypic effect?
A. Missense mutation; it will likely alter the protein's tertiary structure and function.
B. Nonsense mutation; it will cause premature termination of translation.
C. Silent mutation; it will likely have no effect on the protein's function. [CORRECT]
D. Frameshift mutation; it will alter the reading frame and completely change the
downstream amino acid sequence.
Correct Answer: C
Rationale: A silent mutation changes a nucleotide without changing the encoded amino
acid due to the redundancy of the genetic code. Because the amino acid sequence
remains unaltered, the protein's structure and function are unlikely to be affected.
Missense changes the amino acid, nonsense creates a stop codon, and frameshifts
alter the entire downstream reading frame. VERIFIED ✓ - Module 1: Mutations.
Q9: During translation, what is the direct role of transfer RNA (tRNA)?
A. To form the ribosomal subunits that catalyze peptide bond formation.
B. To carry specific amino acids to the ribosome and match them to the correct mRNA
codons via the anticodon. [CORRECT]
C. To transcribe the genetic code from DNA into a complementary RNA sequence.
D. To unwind the DNA double helix in preparation for ribosomal binding.
Correct Answer: B
Rationale: tRNA molecules act as adaptors, bringing specific amino acids to the
ribosome and ensuring they are incorporated into the growing polypeptide chain in the
correct order by base-pairing their anticodon with the mRNA codon. Ribosomal RNA
(rRNA) forms ribosomes, and RNA polymerase handles transcription. VERIFIED ✓ -
Module 1: Central Dogma - Translation.
Q10: A geneticist is analyzing a pedigree for a rare human disease and notices that the
trait appears in every generation, affects both males and females equally, and affected
fathers pass the trait to all of their daughters but none of their sons. What is the most
likely mode of inheritance?
A. Autosomal recessive
B. Autosomal dominant
Lab | Comprehensive All Modules | Verified Q&A | Portage
Learning | Pass Guaranteed - A+ Graded
Section 1: Genetics & Molecular Biology (Modules 1 & 6)
Q1: A researcher isolates a eukaryotic mRNA transcript and observes that it is
significantly shorter than the corresponding DNA template region. Which molecular
mechanisms best explain this discrepancy, and what is the ultimate fate of the removed
sequences?
A. The DNA underwent somatic hypermutation, and the altered sequences are
translated into non-functional peptides that are rapidly degraded.
B. Transcription factors caused premature termination, and the missing sequences are
permanently deleted from the genome.
C. Introns were removed via splicing, and the excised sequences are degraded in the
nucleus. [CORRECT]
D. Exons were spliced out to conserve cellular energy, and the sequences are recycled
for new transcription events.
Correct Answer: C
Rationale: Eukaryotic genes contain non-coding introns that are removed during RNA
splicing, making the mature mRNA shorter than the initial DNA template. The excised
introns are degraded, not recycled or translated. Options A, B, and D incorrectly describe
mutagenesis, genome alteration, and the erroneous removal of coding exons. VERIFIED
✓ - Module 1: Gene Expression & Regulation.
Q2: During a high-fidelity DNA replication process in a human cell, DNA polymerase
encounters a thymine dimer caused by UV radiation. Which cellular response correctly
resolves this issue without introducing mutations?
A. DNA polymerase utilizes its 3' to 5' exonuclease activity to directly excise the dimer
and replace the nucleotides.
B. Nucleotide excision repair (NER) recognizes the distortion, excises a short
oligonucleotide containing the dimer, and DNA polymerase fills the gap. [CORRECT]
C. Mismatch repair proteins detect the abnormal base pairing and remove the newly
synthesized strand entirely.
D. Telomerase extends the 3' end of the chromosome to bypass the damaged region,
preventing replication fork collapse.
Correct Answer: B
,Rationale: Thymine dimers cause significant helical distortion, which is specifically
targeted by the nucleotide excision repair (NER) pathway, not by the exonuclease
proofreading activity of DNA polymerase or mismatch repair. Telomerase is responsible
for extending telomeres, not repairing internal DNA damage. VERIFIED ✓ - Module 1:
DNA Replication & Repair.
Q3: In a Portage Learning lab exercise, students use restriction enzymes to cut a linear
piece of DNA. Enzyme A cuts at one site, and Enzyme B cuts at two sites. If the original
DNA is 10,000 base pairs long, Enzyme A cuts at 3,500 bp, and Enzyme B cuts at 2,000
bp and 7,000 bp, how many fragments will be produced if both enzymes are used in a
single digest?
A. 3
B. 4 [CORRECT]
C. 5
D. 6
Correct Answer: B
Rationale: Enzyme A creates 2 fragments (3500 bp and 6500 bp). Enzyme B cuts within
the 3500 bp fragment (at 2000 bp) and within the 6500 bp fragment (at 7000 bp total,
meaning 5000 bp into the second fragment). This results in four total fragments (2000
bp, 1500 bp, 5000 bp, and 1500 bp). Option C assumes three cut sites always yield five
fragments, ignoring fragment size math. VERIFIED ✓ - Module 6: Molecular Techniques
& Lab Integration.
Q4: A scientist studying gene regulation in E. coli observes that a particular operon is
transcribed continuously regardless of the presence or absence of a specific
metabolite. Which structural feature of this operon most likely accounts for this
constitutive expression?
A. The operator sequence is permanently bound by an active repressor protein.
B. A mutation has deleted the promoter region, preventing RNA polymerase binding.
C. The repressor protein is non-functional due to a mutation in the regulatory gene,
preventing operator binding. [CORRECT]
D. The operon lacks a terminator sequence, causing RNA polymerase to continuously
transcribe past the normal endpoint.
Correct Answer: C
Rationale: Constitutive expression occurs when the operon cannot be turned off. If the
repressor protein is non-functional, it cannot bind to the operator to block transcription,
leading to continuous gene expression. A bound repressor (A) or missing promoter (B)
would prevent transcription, and lacking a terminator would cause read-through but not
,necessarily resolve metabolite-dependent regulation. VERIFIED ✓ - Module 1: Gene
Regulation.
Q5: In a dihybrid cross between two pea plants heterozygous for both seed color (Yy)
and seed shape (Rr), where yellow and round are dominant, what proportion of the
offspring will be homozygous recessive for both traits?
A. 1/16 [CORRECT]
B. 1/8
C. 3/16
D. 9/16
Correct Answer: A
Rationale: According to Mendelian genetics, a dihybrid cross (YyRr x YyRr) yields a
9:3:3:1 phenotypic ratio. The proportion of offspring that are homozygous recessive for
both traits (yyrr) is calculated by multiplying the probability of yy (1/4) by the probability
of rr (1/4), resulting in 1/16. VERIFIED ✓ - Module 1: Mendelian Genetics.
Q6: A population of wildflowers is in Hardy-Weinberg equilibrium for flower color, where
red (R) is dominant over white (r). If the frequency of the recessive allele (r) is 0.3, what
percentage of the population is expected to be heterozygous?
A. 9%
B. 42% [CORRECT]
C. 49%
D. 51%
Correct Answer: B
Rationale: Using the Hardy-Weinberg equations (p + q = 1, p² + 2pq + q² = 1), if q = 0.3,
then p = 0.7. The frequency of heterozygotes is represented by 2pq, which is 2(0.7)(0.3)
= 0.42, or 42%. Option A is q² (homozygous recessive), and C is p² (homozygous
dominant). VERIFIED ✓ - Module 1: Population Genetics & Hardy-Weinberg.
Q7: Which of the following statements correctly differentiates between transcription in
prokaryotes and eukaryotes?
A. Eukaryotic transcription occurs in the cytoplasm, while prokaryotic transcription
occurs in the nucleus.
B. Prokaryotic RNA polymerase requires a poly-A tail to stabilize the mRNA, whereas
eukaryotic mRNA does not.
C. Eukaryotic transcription involves three distinct RNA polymerases, while prokaryotes
use a single RNA polymerase complex. [CORRECT]
D. Prokaryotic transcription produces monocistronic mRNA, whereas eukaryotic
transcription produces polycistronic mRNA.
Correct Answer: C
, Rationale: Eukaryotes utilize RNA polymerases I, II, and III for different RNA types, while
prokaryotes rely on a single RNA polymerase composed of a core enzyme and sigma
factor. Transcription occurs in the nucleus for eukaryotes (not cytoplasm), polycistronic
mRNA is a prokaryotic feature (not eukaryotic), and poly-A tails are added to eukaryotic
mRNA (not prokaryotic). VERIFIED ✓ - Module 1: Central Dogma - Transcription.
Q8: A point mutation occurs that changes a codon from UAU to UAC. Both codons code
for the amino acid tyrosine. What type of mutation is this, and what is its likely
phenotypic effect?
A. Missense mutation; it will likely alter the protein's tertiary structure and function.
B. Nonsense mutation; it will cause premature termination of translation.
C. Silent mutation; it will likely have no effect on the protein's function. [CORRECT]
D. Frameshift mutation; it will alter the reading frame and completely change the
downstream amino acid sequence.
Correct Answer: C
Rationale: A silent mutation changes a nucleotide without changing the encoded amino
acid due to the redundancy of the genetic code. Because the amino acid sequence
remains unaltered, the protein's structure and function are unlikely to be affected.
Missense changes the amino acid, nonsense creates a stop codon, and frameshifts
alter the entire downstream reading frame. VERIFIED ✓ - Module 1: Mutations.
Q9: During translation, what is the direct role of transfer RNA (tRNA)?
A. To form the ribosomal subunits that catalyze peptide bond formation.
B. To carry specific amino acids to the ribosome and match them to the correct mRNA
codons via the anticodon. [CORRECT]
C. To transcribe the genetic code from DNA into a complementary RNA sequence.
D. To unwind the DNA double helix in preparation for ribosomal binding.
Correct Answer: B
Rationale: tRNA molecules act as adaptors, bringing specific amino acids to the
ribosome and ensuring they are incorporated into the growing polypeptide chain in the
correct order by base-pairing their anticodon with the mRNA codon. Ribosomal RNA
(rRNA) forms ribosomes, and RNA polymerase handles transcription. VERIFIED ✓ -
Module 1: Central Dogma - Translation.
Q10: A geneticist is analyzing a pedigree for a rare human disease and notices that the
trait appears in every generation, affects both males and females equally, and affected
fathers pass the trait to all of their daughters but none of their sons. What is the most
likely mode of inheritance?
A. Autosomal recessive
B. Autosomal dominant
