BIOD 210 Genetics Module 7 - Portage Learning - Latest 2026/2027
BIOD 210 Genetics Module 7
Portage Learning | Questions & Answers | Grade A
Latest Update 2026/2027 | 110 Questions | 100% Correct | Grade A | Human Genome | Mitochondrial | Epigenetics |
Clinical Genetics | All Answers Verified
This examination assesses mastery of BIOD 210 Genetics Module 7 content aligned with Portage Learning course
objectives and current clinical genetics practice standards. Coverage spans human genome organization and variation
(repetitive DNA, copy number variation, SNP analysis), mitochondrial genetics and disease (mtDNA inheritance,
heteroplasmy, threshold effect), epigenetics and chromatin regulation (DNA methylation, histone modification, chromatin
remodeling, imprinting, X-inactivation), genetics of immunity and blood groups (antibody gene rearrangement, ABO and
Rh systems, hemolytic disease of the newborn), inborn errors of metabolism (PKU, galactosemia, Tay-Sachs, MCAD,
newborn screening), genetic disorders of hemoglobin and clotting (sickle cell, thalassemias, hemophilia, von Willebrand
disease), reproductive genetics and prenatal diagnosis (prenatal screening, NIPT, amniocentesis, CVS, preimplantation
testing), and genetics of aging, longevity, and emerging areas (telomere biology, genomic instability, cellular senescence,
epigenetic clocks). Each question includes a verified rationale with clinical genetics reasoning and error analysis.
Cognitive level distribution: approximately 30% recall, 50% application, and 20% analysis. Question style: 65%
application-based, 35% direct recall.
SECTION 1: Human Genome Organization and Variation
Genome Structure, Repetitive DNA, Copy Number Variation, & SNP Analysis | Questions 1-15 | 15 Questions
This section covers human genome organization, gene distribution, repetitive DNA classes (LINEs, SINEs,
microsatellites), copy number variation, structural variants, and SNP analysis.
Q1. The human genome contains approximately how many base pairs, and roughly what percentage encodes
protein-coding genes?
A. 3 billion base pairs; ~50% protein-coding
B. 3 billion base pairs; ~1.5% protein-coding, with the remainder consisting of regulatory regions, introns,
repetitive DNA, and non-coding RNA genes [CORRECT]
C. 300 million base pairs; ~10% protein-coding
D. 30 billion base pairs; ~25% protein-coding
Correct Answer: B — 3 billion base pairs; ~1.5% protein-coding, with the remainder consisting of
regulatory regions, introns, repetitive DNA, and non-coding RNA genes
Rationale: The human genome contains approximately 3 billion base pairs, but only about 1.5% encodes
protein-coding genes (~20,000 genes). Option B is correct. The remainder includes regulatory sequences, introns,
repetitive DNA (LINEs, SINEs, satellite DNA), and non-coding RNA genes. Options A, C, and D either
overstate the protein-coding portion or use incorrect genome sizes.
BIOD 210 Module 7 - Grade A | 100% Correct | Verified Answers Page 1
,BIOD 210 Genetics Module 7 - Portage Learning - Latest 2026/2027
Q2. Which class of repetitive DNA makes up the largest portion of the human genome and includes elements
such as L1 (LINE-1)?
A. Short interspersed nuclear elements (SINEs)
B. Long interspersed nuclear elements (LINEs), which are retrotransposons that comprise about 17% of the
human genome [CORRECT]
C. Microsatellites (short tandem repeats)
D. Ribosomal RNA gene clusters
Correct Answer: B — Long interspersed nuclear elements (LINEs), which are retrotransposons that
comprise about 17% of the human genome
Rationale: LINEs (Long Interspersed Nuclear Elements), particularly L1, are the most abundant
retrotransposons in the human genome, comprising about 17% of total genomic DNA. Option B is correct.
SINEs (Option A), including Alu elements, are also common but less abundant (~11%). Microsatellites (Option
C) are short tandem repeats that make up a small fraction. rRNA clusters (Option D) are tandemly repeated but
represent a tiny portion. LINEs can autonomously retrotranspose and cause disease when inserted into genes.
Q3. An Alu element is an example of which type of repetitive DNA?
A. LINE (long interspersed nuclear element)
B. SINE (short interspersed nuclear element), a non-autonomous retrotransposon that depends on LINE
machinery for transposition [CORRECT]
C. Satellite DNA (tandem repeat)
D. Telomeric repeat
Correct Answer: B — SINE (short interspersed nuclear element), a non-autonomous retrotransposon
that depends on LINE machinery for transposition
Rationale: Alu elements are SINEs (Short Interspersed Nuclear Elements) - approximately 300 bp
non-autonomous retrotransposons that rely on LINE-1 (L1) machinery for transposition. Option B is correct. Alu
elements are the most abundant SINEs in primates, with over 1 million copies (~11% of the genome). Option A
is wrong - Alu is a SINE, not LINE. Satellite DNA (Option C) consists of tandem repeats at
centromeres/telomeres. Telomeric repeats (Option D) are a specific tandem repeat (TTAGGG).
Q4. A researcher is analyzing a region of the genome where a 3-base-pair sequence (e.g., CAG) is repeated
10-30 times. This type of repetitive DNA is called:
A. A LINE element
B. A microsatellite (short tandem repeat, STR) [CORRECT]
C. A copy number variant
D. An Alu element
Correct Answer: B — A microsatellite (short tandem repeat, STR)
Rationale: Microsatellites, also called short tandem repeats (STRs), are 2-6 base pair sequences repeated
typically 5-50 times. A 3-bp CAG repeat is a classic microsatellite. Option B is correct. LINEs (Option A) and
Alu elements (Option D) are interspersed repeats, not tandem repeats. Copy number variants (Option C) involve
larger deletions/duplications (>1 kb), not short tandem repeats. Expansion of CAG repeats causes Huntington
disease and other triplet repeat disorders.
BIOD 210 Module 7 - Grade A | 100% Correct | Verified Answers Page 2
,BIOD 210 Genetics Module 7 - Portage Learning - Latest 2026/2027
Q5. A healthy individual is found to have three copies of a 1.5-million-base-pair region on chromosome 15
instead of the usual two copies. This variation is BEST classified as:
A. A single nucleotide polymorphism (SNP)
B. A copy number variant (CNV) - a duplication of a genomic segment >1 kilobase [CORRECT]
C. A microsatellite expansion
D. A point mutation
Correct Answer: B — A copy number variant (CNV) - a duplication of a genomic segment >1 kilobase
Rationale: Copy number variants (CNVs) are deletions or duplications of genomic segments larger than 1
kilobase. A 1.5-million-bp (1.5 Mb) duplication producing three copies instead of two is a CNV. Option B is
correct. SNPs (Option A) and point mutations (Option D) involve single nucleotide changes, not large segments.
Microsatellite expansions (Option C) involve short tandem repeats, not large duplications. CNVs are common
and contribute significantly to human genetic variation and disease.
Q6. A single nucleotide polymorphism (SNP) is BEST defined as:
A. A large chromosomal rearrangement
B. A variation at a single nucleotide position present in >1% of the population, the most common type of
human genetic variation [CORRECT]
C. A deletion of an entire gene
D. An expansion of a trinucleotide repeat
Correct Answer: B — A variation at a single nucleotide position present in >1% of the population, the
most common type of human genetic variation
Rationale: A SNP (single nucleotide polymorphism) is a variation at a single nucleotide position present in
greater than 1% of the population - the most common form of human genetic variation (~10 million SNPs in the
human genome). Option B is correct. SNPs are not large rearrangements (Option A), gene deletions (Option C),
or trinucleotide expansions (Option D). SNPs serve as markers in genome-wide association studies (GWAS) and
can affect disease risk, drug response, and other traits.
Q7. A genome-wide association study (GWAS) identifies a SNP significantly associated with type 2 diabetes.
The SNP is located in an intron of a gene. Which statement BEST explains the most likely functional
implication?
A. The SNP definitively causes diabetes by altering the protein sequence
B. The SNP may affect gene regulation (splicing, transcription factor binding, or expression level) or may be in
linkage disequilibrium with a nearby causal variant; further functional studies are needed [CORRECT]
C. The SNP cannot have any functional effect because it is in an intron
D. The SNP must be the direct cause of all type 2 diabetes cases
Correct Answer: B — The SNP may affect gene regulation (splicing, transcription factor binding, or
expression level) or may be in linkage disequilibrium with a nearby causal variant; further functional
studies are needed
Rationale: Intronic SNPs identified by GWAS may affect gene regulation (splicing, transcription factor binding,
enhancer function, expression level) or may be in linkage disequilibrium with the true causal variant nearby.
Option B is correct. GWAS identifies associations, not causation (Option A is premature). Intronic SNPs CAN
have functional effects (Option C is wrong). A single SNP cannot explain all cases of a complex disease (Option
D is wrong). Functional studies are needed to confirm the mechanism.
BIOD 210 Module 7 - Grade A | 100% Correct | Verified Answers Page 3
, BIOD 210 Genetics Module 7 - Portage Learning - Latest 2026/2027
Q8. Two genes located close together on the same chromosome tend to be inherited together. This
phenomenon is called:
A. Independent assortment
B. Linkage - genes in close physical proximity tend to be co-inherited because recombination between them is
rare [CORRECT]
C. Genomic imprinting
D. X-inactivation
Correct Answer: B — Linkage - genes in close physical proximity tend to be co-inherited because
recombination between them is rare
Rationale: Genetic linkage occurs when genes are located close together on the same chromosome and tend to
be inherited together because recombination between them is rare. Option B is correct. Independent assortment
(Option A) applies to genes on different chromosomes or far apart on the same chromosome. Genomic
imprinting (Option C) is parent-of-origin gene expression. X-inactivation (Option D) is dosage compensation in
females. Linkage analysis is used in gene mapping.
Q9. Linkage disequilibrium (LD) refers to:
A. Random association of alleles at different loci
B. Non-random association of alleles at different loci, where certain allele combinations occur more or less
frequently than expected by chance, often due to physical proximity and shared evolutionary history
[CORRECT]
C. The breaking apart of linked genes through recombination
D. The presence of disease-causing mutations
Correct Answer: B — Non-random association of alleles at different loci, where certain allele
combinations occur more or less frequently than expected by chance, often due to physical proximity
and shared evolutionary history
Rationale: Linkage disequilibrium (LD) is the non-random association of alleles at different loci - certain allele
combinations occur more or less frequently than expected by chance, typically due to physical proximity and
shared evolutionary history. Option B is correct. Random association (Option A) is the absence of LD.
Recombination (Option C) reduces LD over time. Disease mutations (Option D) are unrelated to LD definition.
LD enables GWAS and haplotype mapping.
Q10. Which of the following is an example of a structural variant in the human genome?
A. A single nucleotide change from A to G
B. An inversion of a 500-kb segment on chromosome 9 [CORRECT]
C. A CAG trinucleotide repeat expansion from 20 to 50 copies
D. A change in DNA methylation at a CpG site
Correct Answer: B — An inversion of a 500-kb segment on chromosome 9
Rationale: Structural variants include inversions, translocations, large deletions/duplications (>1 kb), and
insertions. A 500-kb inversion is a structural variant. Option B is correct. A single nucleotide change (Option A)
is a point mutation/SNP. A trinucleotide repeat expansion (Option C) is a microsatellite expansion. DNA
methylation (Option D) is an epigenetic change, not a DNA sequence structural variant.
BIOD 210 Module 7 - Grade A | 100% Correct | Verified Answers Page 4
BIOD 210 Genetics Module 7
Portage Learning | Questions & Answers | Grade A
Latest Update 2026/2027 | 110 Questions | 100% Correct | Grade A | Human Genome | Mitochondrial | Epigenetics |
Clinical Genetics | All Answers Verified
This examination assesses mastery of BIOD 210 Genetics Module 7 content aligned with Portage Learning course
objectives and current clinical genetics practice standards. Coverage spans human genome organization and variation
(repetitive DNA, copy number variation, SNP analysis), mitochondrial genetics and disease (mtDNA inheritance,
heteroplasmy, threshold effect), epigenetics and chromatin regulation (DNA methylation, histone modification, chromatin
remodeling, imprinting, X-inactivation), genetics of immunity and blood groups (antibody gene rearrangement, ABO and
Rh systems, hemolytic disease of the newborn), inborn errors of metabolism (PKU, galactosemia, Tay-Sachs, MCAD,
newborn screening), genetic disorders of hemoglobin and clotting (sickle cell, thalassemias, hemophilia, von Willebrand
disease), reproductive genetics and prenatal diagnosis (prenatal screening, NIPT, amniocentesis, CVS, preimplantation
testing), and genetics of aging, longevity, and emerging areas (telomere biology, genomic instability, cellular senescence,
epigenetic clocks). Each question includes a verified rationale with clinical genetics reasoning and error analysis.
Cognitive level distribution: approximately 30% recall, 50% application, and 20% analysis. Question style: 65%
application-based, 35% direct recall.
SECTION 1: Human Genome Organization and Variation
Genome Structure, Repetitive DNA, Copy Number Variation, & SNP Analysis | Questions 1-15 | 15 Questions
This section covers human genome organization, gene distribution, repetitive DNA classes (LINEs, SINEs,
microsatellites), copy number variation, structural variants, and SNP analysis.
Q1. The human genome contains approximately how many base pairs, and roughly what percentage encodes
protein-coding genes?
A. 3 billion base pairs; ~50% protein-coding
B. 3 billion base pairs; ~1.5% protein-coding, with the remainder consisting of regulatory regions, introns,
repetitive DNA, and non-coding RNA genes [CORRECT]
C. 300 million base pairs; ~10% protein-coding
D. 30 billion base pairs; ~25% protein-coding
Correct Answer: B — 3 billion base pairs; ~1.5% protein-coding, with the remainder consisting of
regulatory regions, introns, repetitive DNA, and non-coding RNA genes
Rationale: The human genome contains approximately 3 billion base pairs, but only about 1.5% encodes
protein-coding genes (~20,000 genes). Option B is correct. The remainder includes regulatory sequences, introns,
repetitive DNA (LINEs, SINEs, satellite DNA), and non-coding RNA genes. Options A, C, and D either
overstate the protein-coding portion or use incorrect genome sizes.
BIOD 210 Module 7 - Grade A | 100% Correct | Verified Answers Page 1
,BIOD 210 Genetics Module 7 - Portage Learning - Latest 2026/2027
Q2. Which class of repetitive DNA makes up the largest portion of the human genome and includes elements
such as L1 (LINE-1)?
A. Short interspersed nuclear elements (SINEs)
B. Long interspersed nuclear elements (LINEs), which are retrotransposons that comprise about 17% of the
human genome [CORRECT]
C. Microsatellites (short tandem repeats)
D. Ribosomal RNA gene clusters
Correct Answer: B — Long interspersed nuclear elements (LINEs), which are retrotransposons that
comprise about 17% of the human genome
Rationale: LINEs (Long Interspersed Nuclear Elements), particularly L1, are the most abundant
retrotransposons in the human genome, comprising about 17% of total genomic DNA. Option B is correct.
SINEs (Option A), including Alu elements, are also common but less abundant (~11%). Microsatellites (Option
C) are short tandem repeats that make up a small fraction. rRNA clusters (Option D) are tandemly repeated but
represent a tiny portion. LINEs can autonomously retrotranspose and cause disease when inserted into genes.
Q3. An Alu element is an example of which type of repetitive DNA?
A. LINE (long interspersed nuclear element)
B. SINE (short interspersed nuclear element), a non-autonomous retrotransposon that depends on LINE
machinery for transposition [CORRECT]
C. Satellite DNA (tandem repeat)
D. Telomeric repeat
Correct Answer: B — SINE (short interspersed nuclear element), a non-autonomous retrotransposon
that depends on LINE machinery for transposition
Rationale: Alu elements are SINEs (Short Interspersed Nuclear Elements) - approximately 300 bp
non-autonomous retrotransposons that rely on LINE-1 (L1) machinery for transposition. Option B is correct. Alu
elements are the most abundant SINEs in primates, with over 1 million copies (~11% of the genome). Option A
is wrong - Alu is a SINE, not LINE. Satellite DNA (Option C) consists of tandem repeats at
centromeres/telomeres. Telomeric repeats (Option D) are a specific tandem repeat (TTAGGG).
Q4. A researcher is analyzing a region of the genome where a 3-base-pair sequence (e.g., CAG) is repeated
10-30 times. This type of repetitive DNA is called:
A. A LINE element
B. A microsatellite (short tandem repeat, STR) [CORRECT]
C. A copy number variant
D. An Alu element
Correct Answer: B — A microsatellite (short tandem repeat, STR)
Rationale: Microsatellites, also called short tandem repeats (STRs), are 2-6 base pair sequences repeated
typically 5-50 times. A 3-bp CAG repeat is a classic microsatellite. Option B is correct. LINEs (Option A) and
Alu elements (Option D) are interspersed repeats, not tandem repeats. Copy number variants (Option C) involve
larger deletions/duplications (>1 kb), not short tandem repeats. Expansion of CAG repeats causes Huntington
disease and other triplet repeat disorders.
BIOD 210 Module 7 - Grade A | 100% Correct | Verified Answers Page 2
,BIOD 210 Genetics Module 7 - Portage Learning - Latest 2026/2027
Q5. A healthy individual is found to have three copies of a 1.5-million-base-pair region on chromosome 15
instead of the usual two copies. This variation is BEST classified as:
A. A single nucleotide polymorphism (SNP)
B. A copy number variant (CNV) - a duplication of a genomic segment >1 kilobase [CORRECT]
C. A microsatellite expansion
D. A point mutation
Correct Answer: B — A copy number variant (CNV) - a duplication of a genomic segment >1 kilobase
Rationale: Copy number variants (CNVs) are deletions or duplications of genomic segments larger than 1
kilobase. A 1.5-million-bp (1.5 Mb) duplication producing three copies instead of two is a CNV. Option B is
correct. SNPs (Option A) and point mutations (Option D) involve single nucleotide changes, not large segments.
Microsatellite expansions (Option C) involve short tandem repeats, not large duplications. CNVs are common
and contribute significantly to human genetic variation and disease.
Q6. A single nucleotide polymorphism (SNP) is BEST defined as:
A. A large chromosomal rearrangement
B. A variation at a single nucleotide position present in >1% of the population, the most common type of
human genetic variation [CORRECT]
C. A deletion of an entire gene
D. An expansion of a trinucleotide repeat
Correct Answer: B — A variation at a single nucleotide position present in >1% of the population, the
most common type of human genetic variation
Rationale: A SNP (single nucleotide polymorphism) is a variation at a single nucleotide position present in
greater than 1% of the population - the most common form of human genetic variation (~10 million SNPs in the
human genome). Option B is correct. SNPs are not large rearrangements (Option A), gene deletions (Option C),
or trinucleotide expansions (Option D). SNPs serve as markers in genome-wide association studies (GWAS) and
can affect disease risk, drug response, and other traits.
Q7. A genome-wide association study (GWAS) identifies a SNP significantly associated with type 2 diabetes.
The SNP is located in an intron of a gene. Which statement BEST explains the most likely functional
implication?
A. The SNP definitively causes diabetes by altering the protein sequence
B. The SNP may affect gene regulation (splicing, transcription factor binding, or expression level) or may be in
linkage disequilibrium with a nearby causal variant; further functional studies are needed [CORRECT]
C. The SNP cannot have any functional effect because it is in an intron
D. The SNP must be the direct cause of all type 2 diabetes cases
Correct Answer: B — The SNP may affect gene regulation (splicing, transcription factor binding, or
expression level) or may be in linkage disequilibrium with a nearby causal variant; further functional
studies are needed
Rationale: Intronic SNPs identified by GWAS may affect gene regulation (splicing, transcription factor binding,
enhancer function, expression level) or may be in linkage disequilibrium with the true causal variant nearby.
Option B is correct. GWAS identifies associations, not causation (Option A is premature). Intronic SNPs CAN
have functional effects (Option C is wrong). A single SNP cannot explain all cases of a complex disease (Option
D is wrong). Functional studies are needed to confirm the mechanism.
BIOD 210 Module 7 - Grade A | 100% Correct | Verified Answers Page 3
, BIOD 210 Genetics Module 7 - Portage Learning - Latest 2026/2027
Q8. Two genes located close together on the same chromosome tend to be inherited together. This
phenomenon is called:
A. Independent assortment
B. Linkage - genes in close physical proximity tend to be co-inherited because recombination between them is
rare [CORRECT]
C. Genomic imprinting
D. X-inactivation
Correct Answer: B — Linkage - genes in close physical proximity tend to be co-inherited because
recombination between them is rare
Rationale: Genetic linkage occurs when genes are located close together on the same chromosome and tend to
be inherited together because recombination between them is rare. Option B is correct. Independent assortment
(Option A) applies to genes on different chromosomes or far apart on the same chromosome. Genomic
imprinting (Option C) is parent-of-origin gene expression. X-inactivation (Option D) is dosage compensation in
females. Linkage analysis is used in gene mapping.
Q9. Linkage disequilibrium (LD) refers to:
A. Random association of alleles at different loci
B. Non-random association of alleles at different loci, where certain allele combinations occur more or less
frequently than expected by chance, often due to physical proximity and shared evolutionary history
[CORRECT]
C. The breaking apart of linked genes through recombination
D. The presence of disease-causing mutations
Correct Answer: B — Non-random association of alleles at different loci, where certain allele
combinations occur more or less frequently than expected by chance, often due to physical proximity
and shared evolutionary history
Rationale: Linkage disequilibrium (LD) is the non-random association of alleles at different loci - certain allele
combinations occur more or less frequently than expected by chance, typically due to physical proximity and
shared evolutionary history. Option B is correct. Random association (Option A) is the absence of LD.
Recombination (Option C) reduces LD over time. Disease mutations (Option D) are unrelated to LD definition.
LD enables GWAS and haplotype mapping.
Q10. Which of the following is an example of a structural variant in the human genome?
A. A single nucleotide change from A to G
B. An inversion of a 500-kb segment on chromosome 9 [CORRECT]
C. A CAG trinucleotide repeat expansion from 20 to 50 copies
D. A change in DNA methylation at a CpG site
Correct Answer: B — An inversion of a 500-kb segment on chromosome 9
Rationale: Structural variants include inversions, translocations, large deletions/duplications (>1 kb), and
insertions. A 500-kb inversion is a structural variant. Option B is correct. A single nucleotide change (Option A)
is a point mutation/SNP. A trinucleotide repeat expansion (Option C) is a microsatellite expansion. DNA
methylation (Option D) is an epigenetic change, not a DNA sequence structural variant.
BIOD 210 Module 7 - Grade A | 100% Correct | Verified Answers Page 4
