Portage Learning · BIOD 210 Genetics
Final Exam — Latest Update 2026/2027
Questions & Answers — Grade A — 100% Correct
Comprehensive Study Guide — 150-Question Cumulative Exam · 10 Sections
Instructions. This 150-question final exam covers the full BIOD 210 Genetics curriculum across 10 sections: (1)
Molecular Foundations; (2) Mutations, DNA Repair, and Genome Stability; (3) Mendelian and Non-Mendelian
Inheritance; (4) Pedigree Analysis and Genetic Risk; (5) Chromosomal Genetics and Cytogenetics; (6) Linkage,
Recombination, and Gene Mapping; (7) Population and Evolutionary Genetics; (8) Microbial, Viral, and Mobile Genetic
Elements; (9) Cancer Genetics and Developmental Genetics; and (10) Applied, Clinical, and Ethical Genetics. Each
question is followed by the correct answer and a detailed rationale with genetics-specific reasoning, calculation
demonstration, and error analysis aligned with Portage BIOD 210 standards. Cognitive levels are approximately 30%
recall, 50% application, and 20% analysis, with 65% application-based items and 35% direct recall of mechanisms,
disorders, and techniques.
Section 1 — Molecular Foundations (DNA Structure, Replication, Transcription, Translation, &
Gene Expression Regulation)
Questions 1–20 cover DNA structure, semiconservative replication, transcription, RNA processing, translation, and gene
regulation. Aligned with Portage BIOD 210 Module 1.
Q1: In the Watson-Crick DNA double helix, the two strands run antiparallel. This means:
A. Both strands run 5' to 3' in the same direction.
B. One strand runs 5' to 3' and the complementary strand runs 3' to 5', with antiparallel sugar-phosphate
backbones and complementary base pairing (A-T, G-C) [CORRECT]
C. Both strands run 3' to 5'.
D. Strand direction is random.
Correct Answer: B. One strand runs 5' to 3' and the complementary strand runs 3' to 5', with antiparallel
sugar-phosphate backbones and complementary base pairing (A-T, G-C)
Rationale: Antiparallel strands run in opposite directions: one 5'→3', the other 3'→5'. This orientation is essential for
replication (DNA polymerase synthesizes only 5'→3') and for complementary base pairing (A-T, G-C) across the helix.
Parallel strands would not allow proper hydrogen bonding or replication.
,Q2: Adenine pairs with thymine using how many hydrogen bonds, while guanine pairs with cytosine using how
many?
A. A-T: 1; G-C: 2
B. A-T: 2 hydrogen bonds; G-C: 3 hydrogen bonds (G-C is stronger, contributing to higher GC-content DNA
having higher melting temperatures) [CORRECT]
C. A-T: 3; G-C: 2
D. A-T: 4; G-C: 4
Correct Answer: B. A-T: 2 hydrogen bonds; G-C: 3 hydrogen bonds (G-C is stronger, contributing to higher
GC-content DNA having higher melting temperatures)
Rationale: A-T pairs have 2 hydrogen bonds; G-C pairs have 3. The extra bond in G-C makes GC-rich DNA more
thermally stable (higher Tm). This is why PCR primer design and DNA denaturation depend on GC content. Reversing
the bond counts is a common student error.
Q3: DNA replication is described as semiconservative because:
A. Each daughter molecule contains two newly synthesized strands.
B. Each daughter DNA molecule contains one parental (template) strand and one newly synthesized strand,
conserving half of the original molecule [CORRECT]
C. The parental molecule is fully conserved.
D. Only one strand is replicated.
Correct Answer: B. Each daughter DNA molecule contains one parental (template) strand and one newly
synthesized strand, conserving half of the original molecule
Rationale: Semiconservative replication (Meselson-Stahl, 1958) means each daughter DNA molecule contains one
parental and one new strand. This was confirmed by density-labeling experiments. Conservative replication would keep
the original intact; dispersive would mix old and new in patches. Semiconservative is the accepted mechanism.
Q4: Which enzyme unwinds the parental DNA double helix at the replication fork?
A. DNA polymerase
B. Helicase, which breaks the hydrogen bonds between base pairs to separate the two strands [CORRECT]
C. Ligase
D. Primase
Correct Answer: B. Helicase, which breaks the hydrogen bonds between base pairs to separate the two strands
Rationale: Helicase unwinds the double helix by breaking H-bonds between base pairs, separating the two strands to form
the replication fork. DNA polymerase synthesizes new DNA; ligase joins Okazaki fragments; primase synthesizes RNA
primers. Single-strand binding proteins (SSBs) keep the strands apart after unwinding.
,Q5: DNA polymerase can only add nucleotides in which direction, and why?
A. 3' to 5', because of the template orientation.
B. 5' to 3', because it can only add nucleotides to a free 3'-OH group on the growing strand [CORRECT]
C. Either direction.
D. It does not require a direction.
Correct Answer: B. 5' to 3', because it can only add nucleotides to a free 3'-OH group on the growing strand
Rationale: DNA polymerase synthesizes DNA only in the 5'→3' direction, adding nucleotides to a free 3'-OH. This
creates a 'leading strand' (continuous, 5'→3' toward the fork) and a 'lagging strand' (discontinuous Okazaki fragments,
synthesized 5'→3' away from the fork then joined by ligase). This 5'→3' requirement is a fundamental principle of
replication.
Q6: Okazaki fragments are formed on the:
A. Leading strand.
B. Lagging strand, synthesized discontinuously in short 5'→3' fragments away from the replication fork and
later joined by DNA ligase [CORRECT]
C. Template strand.
D. Both leading and lagging strands equally.
Correct Answer: B. Lagging strand, synthesized discontinuously in short 5'→3' fragments away from the
replication fork and later joined by DNA ligase
Rationale: Okazaki fragments are short DNA fragments synthesized on the lagging strand (5'→3' but moving away from
the fork due to antiparallel template). They are initiated by RNA primers (primase), extended by DNA polymerase III, and
joined by DNA ligase. The leading strand is synthesized continuously. Reiji Okazaki discovered these in 1968.
Q7: Which enzyme relieves the torsional strain (supercoiling) ahead of the replication fork by cutting, rotating,
and resealing DNA strands?
A. Helicase
B. Topoisomerase (DNA gyrase in bacteria), which relieves supercoiling by breaking and resealing DNA
strands ahead of the fork [CORRECT]
C. Primase
D. Telomerase
Correct Answer: B. Topoisomerase (DNA gyrase in bacteria), which relieves supercoiling by breaking and
resealing DNA strands ahead of the fork
Rationale: Topoisomerase (DNA gyrase in bacteria) relieves supercoiling by cutting one or both DNA strands, rotating,
and resealing ahead of the fork. Without topoisomerase, the unwinding by helicase would create intolerable torsional
strain. Fluoroquinolone antibiotics target bacterial DNA gyrase. Helicase unwinds; primase makes RNA primers;
telomerase extends telomeres.
, Q8: During transcription in eukaryotes, which enzyme synthesizes messenger RNA (mRNA) from a DNA
template?
A. DNA polymerase
B. RNA polymerase II, which synthesizes mRNA from a DNA template in the 5'→3' direction using
ribonucleotides [CORRECT]
C. Helicase
D. Reverse transcriptase
Correct Answer: B. RNA polymerase II, which synthesizes mRNA from a DNA template in the 5'→3' direction
using ribonucleotides
Rationale: RNA polymerase II synthesizes mRNA in eukaryotes (5'→3') from a DNA template. RNA pol I makes rRNA;
RNA pol III makes tRNA and 5S rRNA. DNA polymerase synthesizes DNA; helicase unwinds; reverse transcriptase
synthesizes DNA from RNA (retroviruses, in lab techniques). RNA pol does not require a primer (unlike DNA pol).
Q9: In eukaryotes, the primary transcript (pre-mRNA) undergoes processing including all of the following
EXCEPT:
A. 5' cap addition.
B. Addition of a poly-A tail.
C. Splicing of introns.
D. Addition of a 3' phosphate group (this does NOT occur; instead, a 5' cap and 3' poly-A tail are added, and
introns are spliced out) [CORRECT]
Correct Answer: D. Addition of a 3' phosphate group (this does NOT occur; instead, a 5' cap and 3' poly-A tail are
added, and introns are spliced out)
Rationale: Eukaryotic pre-mRNA processing includes: 5' cap (7-methylguanosine), 3' poly-A tail, and intron splicing
(snRNPs in spliceosome). A 3' phosphate is NOT added (poly-A tail is added by poly-A polymerase). The 5' cap aids
ribosome binding and mRNA stability; poly-A tail aids nuclear export and stability; splicing joins exons.
Q10: In the genetic code, the start codon AUG codes for which amino acid and also serves as the initiation
signal?
A. Glycine.
B. Methionine — AUG codes for methionine and is the start codon for translation in both prokaryotes
(formyl-methionine) and eukaryotes (methionine) [CORRECT]
C. Alanine.
D. Tryptophan.
Correct Answer: B. Methionine — AUG codes for methionine and is the start codon for translation in both
prokaryotes (formyl-methionine) and eukaryotes (methionine)
Rationale: AUG is the start codon, coding for methionine (Met, M). In prokaryotes, the initiating methionine is
formylated (fMet). AUG also functions as an internal methionine codon. The genetic code is degenerate (multiple codons
per amino acid) but unambiguous (each codon specifies one amino acid). Stop codons: UAA, UAG, UGA.
Final Exam — Latest Update 2026/2027
Questions & Answers — Grade A — 100% Correct
Comprehensive Study Guide — 150-Question Cumulative Exam · 10 Sections
Instructions. This 150-question final exam covers the full BIOD 210 Genetics curriculum across 10 sections: (1)
Molecular Foundations; (2) Mutations, DNA Repair, and Genome Stability; (3) Mendelian and Non-Mendelian
Inheritance; (4) Pedigree Analysis and Genetic Risk; (5) Chromosomal Genetics and Cytogenetics; (6) Linkage,
Recombination, and Gene Mapping; (7) Population and Evolutionary Genetics; (8) Microbial, Viral, and Mobile Genetic
Elements; (9) Cancer Genetics and Developmental Genetics; and (10) Applied, Clinical, and Ethical Genetics. Each
question is followed by the correct answer and a detailed rationale with genetics-specific reasoning, calculation
demonstration, and error analysis aligned with Portage BIOD 210 standards. Cognitive levels are approximately 30%
recall, 50% application, and 20% analysis, with 65% application-based items and 35% direct recall of mechanisms,
disorders, and techniques.
Section 1 — Molecular Foundations (DNA Structure, Replication, Transcription, Translation, &
Gene Expression Regulation)
Questions 1–20 cover DNA structure, semiconservative replication, transcription, RNA processing, translation, and gene
regulation. Aligned with Portage BIOD 210 Module 1.
Q1: In the Watson-Crick DNA double helix, the two strands run antiparallel. This means:
A. Both strands run 5' to 3' in the same direction.
B. One strand runs 5' to 3' and the complementary strand runs 3' to 5', with antiparallel sugar-phosphate
backbones and complementary base pairing (A-T, G-C) [CORRECT]
C. Both strands run 3' to 5'.
D. Strand direction is random.
Correct Answer: B. One strand runs 5' to 3' and the complementary strand runs 3' to 5', with antiparallel
sugar-phosphate backbones and complementary base pairing (A-T, G-C)
Rationale: Antiparallel strands run in opposite directions: one 5'→3', the other 3'→5'. This orientation is essential for
replication (DNA polymerase synthesizes only 5'→3') and for complementary base pairing (A-T, G-C) across the helix.
Parallel strands would not allow proper hydrogen bonding or replication.
,Q2: Adenine pairs with thymine using how many hydrogen bonds, while guanine pairs with cytosine using how
many?
A. A-T: 1; G-C: 2
B. A-T: 2 hydrogen bonds; G-C: 3 hydrogen bonds (G-C is stronger, contributing to higher GC-content DNA
having higher melting temperatures) [CORRECT]
C. A-T: 3; G-C: 2
D. A-T: 4; G-C: 4
Correct Answer: B. A-T: 2 hydrogen bonds; G-C: 3 hydrogen bonds (G-C is stronger, contributing to higher
GC-content DNA having higher melting temperatures)
Rationale: A-T pairs have 2 hydrogen bonds; G-C pairs have 3. The extra bond in G-C makes GC-rich DNA more
thermally stable (higher Tm). This is why PCR primer design and DNA denaturation depend on GC content. Reversing
the bond counts is a common student error.
Q3: DNA replication is described as semiconservative because:
A. Each daughter molecule contains two newly synthesized strands.
B. Each daughter DNA molecule contains one parental (template) strand and one newly synthesized strand,
conserving half of the original molecule [CORRECT]
C. The parental molecule is fully conserved.
D. Only one strand is replicated.
Correct Answer: B. Each daughter DNA molecule contains one parental (template) strand and one newly
synthesized strand, conserving half of the original molecule
Rationale: Semiconservative replication (Meselson-Stahl, 1958) means each daughter DNA molecule contains one
parental and one new strand. This was confirmed by density-labeling experiments. Conservative replication would keep
the original intact; dispersive would mix old and new in patches. Semiconservative is the accepted mechanism.
Q4: Which enzyme unwinds the parental DNA double helix at the replication fork?
A. DNA polymerase
B. Helicase, which breaks the hydrogen bonds between base pairs to separate the two strands [CORRECT]
C. Ligase
D. Primase
Correct Answer: B. Helicase, which breaks the hydrogen bonds between base pairs to separate the two strands
Rationale: Helicase unwinds the double helix by breaking H-bonds between base pairs, separating the two strands to form
the replication fork. DNA polymerase synthesizes new DNA; ligase joins Okazaki fragments; primase synthesizes RNA
primers. Single-strand binding proteins (SSBs) keep the strands apart after unwinding.
,Q5: DNA polymerase can only add nucleotides in which direction, and why?
A. 3' to 5', because of the template orientation.
B. 5' to 3', because it can only add nucleotides to a free 3'-OH group on the growing strand [CORRECT]
C. Either direction.
D. It does not require a direction.
Correct Answer: B. 5' to 3', because it can only add nucleotides to a free 3'-OH group on the growing strand
Rationale: DNA polymerase synthesizes DNA only in the 5'→3' direction, adding nucleotides to a free 3'-OH. This
creates a 'leading strand' (continuous, 5'→3' toward the fork) and a 'lagging strand' (discontinuous Okazaki fragments,
synthesized 5'→3' away from the fork then joined by ligase). This 5'→3' requirement is a fundamental principle of
replication.
Q6: Okazaki fragments are formed on the:
A. Leading strand.
B. Lagging strand, synthesized discontinuously in short 5'→3' fragments away from the replication fork and
later joined by DNA ligase [CORRECT]
C. Template strand.
D. Both leading and lagging strands equally.
Correct Answer: B. Lagging strand, synthesized discontinuously in short 5'→3' fragments away from the
replication fork and later joined by DNA ligase
Rationale: Okazaki fragments are short DNA fragments synthesized on the lagging strand (5'→3' but moving away from
the fork due to antiparallel template). They are initiated by RNA primers (primase), extended by DNA polymerase III, and
joined by DNA ligase. The leading strand is synthesized continuously. Reiji Okazaki discovered these in 1968.
Q7: Which enzyme relieves the torsional strain (supercoiling) ahead of the replication fork by cutting, rotating,
and resealing DNA strands?
A. Helicase
B. Topoisomerase (DNA gyrase in bacteria), which relieves supercoiling by breaking and resealing DNA
strands ahead of the fork [CORRECT]
C. Primase
D. Telomerase
Correct Answer: B. Topoisomerase (DNA gyrase in bacteria), which relieves supercoiling by breaking and
resealing DNA strands ahead of the fork
Rationale: Topoisomerase (DNA gyrase in bacteria) relieves supercoiling by cutting one or both DNA strands, rotating,
and resealing ahead of the fork. Without topoisomerase, the unwinding by helicase would create intolerable torsional
strain. Fluoroquinolone antibiotics target bacterial DNA gyrase. Helicase unwinds; primase makes RNA primers;
telomerase extends telomeres.
, Q8: During transcription in eukaryotes, which enzyme synthesizes messenger RNA (mRNA) from a DNA
template?
A. DNA polymerase
B. RNA polymerase II, which synthesizes mRNA from a DNA template in the 5'→3' direction using
ribonucleotides [CORRECT]
C. Helicase
D. Reverse transcriptase
Correct Answer: B. RNA polymerase II, which synthesizes mRNA from a DNA template in the 5'→3' direction
using ribonucleotides
Rationale: RNA polymerase II synthesizes mRNA in eukaryotes (5'→3') from a DNA template. RNA pol I makes rRNA;
RNA pol III makes tRNA and 5S rRNA. DNA polymerase synthesizes DNA; helicase unwinds; reverse transcriptase
synthesizes DNA from RNA (retroviruses, in lab techniques). RNA pol does not require a primer (unlike DNA pol).
Q9: In eukaryotes, the primary transcript (pre-mRNA) undergoes processing including all of the following
EXCEPT:
A. 5' cap addition.
B. Addition of a poly-A tail.
C. Splicing of introns.
D. Addition of a 3' phosphate group (this does NOT occur; instead, a 5' cap and 3' poly-A tail are added, and
introns are spliced out) [CORRECT]
Correct Answer: D. Addition of a 3' phosphate group (this does NOT occur; instead, a 5' cap and 3' poly-A tail are
added, and introns are spliced out)
Rationale: Eukaryotic pre-mRNA processing includes: 5' cap (7-methylguanosine), 3' poly-A tail, and intron splicing
(snRNPs in spliceosome). A 3' phosphate is NOT added (poly-A tail is added by poly-A polymerase). The 5' cap aids
ribosome binding and mRNA stability; poly-A tail aids nuclear export and stability; splicing joins exons.
Q10: In the genetic code, the start codon AUG codes for which amino acid and also serves as the initiation
signal?
A. Glycine.
B. Methionine — AUG codes for methionine and is the start codon for translation in both prokaryotes
(formyl-methionine) and eukaryotes (methionine) [CORRECT]
C. Alanine.
D. Tryptophan.
Correct Answer: B. Methionine — AUG codes for methionine and is the start codon for translation in both
prokaryotes (formyl-methionine) and eukaryotes (methionine)
Rationale: AUG is the start codon, coding for methionine (Met, M). In prokaryotes, the initiating methionine is
formylated (fMet). AUG also functions as an internal methionine codon. The genetic code is degenerate (multiple codons
per amino acid) but unambiguous (each codon specifies one amino acid). Stop codons: UAA, UAG, UGA.
