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BIOD 210 GENETICS (PORTAGE LEARNING) FINAL
ACTUAL EXAM PREP 2026 ALL QUESTIONS AND
CORRECT DETAILED ANSWERS WITH RATIONALES
ALREADY A GRADED WITH EXPERT FEEDBACK
|NEW AND REVISED
1. A couple has three children: two with type A blood and one with type
O blood. What are the possible genotypes of the parents?
A. IAIA and IAIB
B. IAi and IBi
C. IAIA and IAi
D. IAIB and IBi
Rationale: Type O blood requires two i alleles. For an O child to be
produced, both parents must carry the i allele. The presence of type A
children suggests at least one parent contributes an IA allele. Option B
(IAi x IBi) allows type A (IAi), type B (IBi), and type O (ii) offspring,
consistent with the family. IAIA x IAi (C) cannot produce type O
children. IAIA x IAIB (A) cannot produce O or B children. IAIB x IBi
(D) cannot produce O children.
2. A pregnant woman is concerned about the risk of Down syndrome
based on her age (42 years). The abnormal chromosome number in
Down syndrome results from which meiotic error?
A. Nondisjunction during meiosis I or II
B. Translocation between chromosome 21 and 14
C. Inversion of a segment on chromosome 21
D. Deletion of the q arm of chromosome 21
Rationale: The majority (about 95%) of Down syndrome cases result
from trisomy 21 due to nondisjunction during gamete formation.
Maternal age is a strong risk factor as nondisjunction becomes more
frequent during the later stages of oogenesis. Translocation accounts
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for ~3% of cases but is not associated with advanced maternal age.
Inversion and deletion are not causes of trisomy 21.
3. A patient is diagnosed with Huntington disease, an autosomal
dominant disorder caused by a CAG trinucleotide repeat expansion in
the HTT gene. His mother died of the disease, but his father is
unaffected. What is the probability that his first child will inherit the
disease allele?
A. 0% (the expanded allele cannot be transmitted)
B. 25%
C. 50%
D. 100%
Rationale: Huntington disease is inherited as an autosomal dominant
trait. An affected individual carries one expanded allele and one
normal allele. The risk of transmitting the disease allele to a child is
50% per pregnancy, regardless of the sex of the parent. The
underlying meiotic segregation is independent of paternal or maternal
origin.
4. A research team sequences the entire exome of a patient with a rare
genetic disorder. However, they cannot find any causative variant in the
protein-coding regions. Which genetic element would be most
informative to analyze next?
A. Non-coding regulatory regions (e.g., promoters, enhancers,
untranslated regions)
B. Repetitive elements such as LINEs and SINEs
C. Telomeric repeat sequences
D. Centromeric heterochromatin
Rationale: Many genetic disorders are caused by mutations that affect
gene regulation, including mutations in promoters, enhancers, 5' or 3'
UTRs, or splice sites. Whole-exome sequencing captures only
protein-coding exons. While non-coding variants are more
challenging to interpret, they may explain cases without exonic
,3|Page
findings. Telomeres (C) and centromeric heterochromatin (D) are not
typically associated with single-gene disorders.
5. A genetic counselor presents a pedigree showing an affected male in
every generation, with the disorder transmitted from an affected father to
all his daughters but to none of his sons. Which mode of inheritance is
most consistent?
A. Autosomal dominant
B. X-linked dominant
C. X-linked recessive
D. Y-linked
Rationale: In X-linked dominant inheritance, an affected father passes
his X chromosome to all his daughters (who therefore become
affected) and his Y chromosome to all his sons (who are unaffected).
The pattern of male-to-male transmission is absent. Autosomal
dominant (A) would show male-to-male transmission. X-linked
recessive (C) would rarely produce affected fathers with affected
daughters unless the mother is a carrier. Y-linked (D) would affect all
sons but no daughters.
6. A patient with a family history of early-onset breast and ovarian
cancer is found to have a pathogenic variant in the BRCA1 gene. Based
on current guidelines for hereditary cancer syndromes, which ethical
principle is most relevant when discussing genetic testing with her adult
children?
A. Non-maleficence, because the information could cause psychological
harm
B. Beneficence, because the children would benefit from early cancer
screening if they are carriers
C. The patient has the right to privacy, but offspring have the right
to know their own risk (duty to warn family members)
D. Confidentiality should be maintained regardless of the potential
benefit to relatives
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Rationale: The ethical tension between patient confidentiality and the
potential benefit to at-risk relatives is a key issue in clinical genetics.
Most guidelines encourage the proband to share results with family
members, but they generally do not require the provider to breach
confidentiality. The concept of “duty to warn” applies when there is an
imminent, serious, and preventable harm. For adult-onset conditions
such as hereditary breast cancer, the ethical emphasis is on supporting
the patient in informing relatives.
7. A laboratory technique is used to separate DNA fragments by size,
transfer them to a membrane, and then probe them with a labeled DNA
sequence. What is the name of this method?
A. Southern blot
B. Northern blot
C. Western blot
D. Polymerase chain reaction
Rationale: Southern blotting, named after Edwin Southern, is used for
DNA analysis. It involves restriction digestion, gel electrophoresis,
transfer to a membrane, and hybridization with a labeled DNA probe.
Northern blot (B) is for RNA; Western blot (C) is for protein; PCR (D)
is an amplification method, not a blotting technique.
8. A student examines a DNA sample and notes that 22% of the
nucleotides are thymine. According to Chargaff’s rules, what percentage
of the nucleotides will be cytosine?
A. 22%
B. 28%
C. 28%? Wait, T = 22%, then A = 22%, G+C = 56%, so G = C =
28%.
D. 44%
*Rationale: Chargaff’s rules state that in double-stranded DNA, A = T
and G = C. If T = 22%, then A = 22%, so A+T = 44%, and G+C =
100% – 44% = 56%. Since G = C, then C = 28%. Option C is correct.*
BIOD 210 GENETICS (PORTAGE LEARNING) FINAL
ACTUAL EXAM PREP 2026 ALL QUESTIONS AND
CORRECT DETAILED ANSWERS WITH RATIONALES
ALREADY A GRADED WITH EXPERT FEEDBACK
|NEW AND REVISED
1. A couple has three children: two with type A blood and one with type
O blood. What are the possible genotypes of the parents?
A. IAIA and IAIB
B. IAi and IBi
C. IAIA and IAi
D. IAIB and IBi
Rationale: Type O blood requires two i alleles. For an O child to be
produced, both parents must carry the i allele. The presence of type A
children suggests at least one parent contributes an IA allele. Option B
(IAi x IBi) allows type A (IAi), type B (IBi), and type O (ii) offspring,
consistent with the family. IAIA x IAi (C) cannot produce type O
children. IAIA x IAIB (A) cannot produce O or B children. IAIB x IBi
(D) cannot produce O children.
2. A pregnant woman is concerned about the risk of Down syndrome
based on her age (42 years). The abnormal chromosome number in
Down syndrome results from which meiotic error?
A. Nondisjunction during meiosis I or II
B. Translocation between chromosome 21 and 14
C. Inversion of a segment on chromosome 21
D. Deletion of the q arm of chromosome 21
Rationale: The majority (about 95%) of Down syndrome cases result
from trisomy 21 due to nondisjunction during gamete formation.
Maternal age is a strong risk factor as nondisjunction becomes more
frequent during the later stages of oogenesis. Translocation accounts
,2|Page
for ~3% of cases but is not associated with advanced maternal age.
Inversion and deletion are not causes of trisomy 21.
3. A patient is diagnosed with Huntington disease, an autosomal
dominant disorder caused by a CAG trinucleotide repeat expansion in
the HTT gene. His mother died of the disease, but his father is
unaffected. What is the probability that his first child will inherit the
disease allele?
A. 0% (the expanded allele cannot be transmitted)
B. 25%
C. 50%
D. 100%
Rationale: Huntington disease is inherited as an autosomal dominant
trait. An affected individual carries one expanded allele and one
normal allele. The risk of transmitting the disease allele to a child is
50% per pregnancy, regardless of the sex of the parent. The
underlying meiotic segregation is independent of paternal or maternal
origin.
4. A research team sequences the entire exome of a patient with a rare
genetic disorder. However, they cannot find any causative variant in the
protein-coding regions. Which genetic element would be most
informative to analyze next?
A. Non-coding regulatory regions (e.g., promoters, enhancers,
untranslated regions)
B. Repetitive elements such as LINEs and SINEs
C. Telomeric repeat sequences
D. Centromeric heterochromatin
Rationale: Many genetic disorders are caused by mutations that affect
gene regulation, including mutations in promoters, enhancers, 5' or 3'
UTRs, or splice sites. Whole-exome sequencing captures only
protein-coding exons. While non-coding variants are more
challenging to interpret, they may explain cases without exonic
,3|Page
findings. Telomeres (C) and centromeric heterochromatin (D) are not
typically associated with single-gene disorders.
5. A genetic counselor presents a pedigree showing an affected male in
every generation, with the disorder transmitted from an affected father to
all his daughters but to none of his sons. Which mode of inheritance is
most consistent?
A. Autosomal dominant
B. X-linked dominant
C. X-linked recessive
D. Y-linked
Rationale: In X-linked dominant inheritance, an affected father passes
his X chromosome to all his daughters (who therefore become
affected) and his Y chromosome to all his sons (who are unaffected).
The pattern of male-to-male transmission is absent. Autosomal
dominant (A) would show male-to-male transmission. X-linked
recessive (C) would rarely produce affected fathers with affected
daughters unless the mother is a carrier. Y-linked (D) would affect all
sons but no daughters.
6. A patient with a family history of early-onset breast and ovarian
cancer is found to have a pathogenic variant in the BRCA1 gene. Based
on current guidelines for hereditary cancer syndromes, which ethical
principle is most relevant when discussing genetic testing with her adult
children?
A. Non-maleficence, because the information could cause psychological
harm
B. Beneficence, because the children would benefit from early cancer
screening if they are carriers
C. The patient has the right to privacy, but offspring have the right
to know their own risk (duty to warn family members)
D. Confidentiality should be maintained regardless of the potential
benefit to relatives
, 4|Page
Rationale: The ethical tension between patient confidentiality and the
potential benefit to at-risk relatives is a key issue in clinical genetics.
Most guidelines encourage the proband to share results with family
members, but they generally do not require the provider to breach
confidentiality. The concept of “duty to warn” applies when there is an
imminent, serious, and preventable harm. For adult-onset conditions
such as hereditary breast cancer, the ethical emphasis is on supporting
the patient in informing relatives.
7. A laboratory technique is used to separate DNA fragments by size,
transfer them to a membrane, and then probe them with a labeled DNA
sequence. What is the name of this method?
A. Southern blot
B. Northern blot
C. Western blot
D. Polymerase chain reaction
Rationale: Southern blotting, named after Edwin Southern, is used for
DNA analysis. It involves restriction digestion, gel electrophoresis,
transfer to a membrane, and hybridization with a labeled DNA probe.
Northern blot (B) is for RNA; Western blot (C) is for protein; PCR (D)
is an amplification method, not a blotting technique.
8. A student examines a DNA sample and notes that 22% of the
nucleotides are thymine. According to Chargaff’s rules, what percentage
of the nucleotides will be cytosine?
A. 22%
B. 28%
C. 28%? Wait, T = 22%, then A = 22%, G+C = 56%, so G = C =
28%.
D. 44%
*Rationale: Chargaff’s rules state that in double-stranded DNA, A = T
and G = C. If T = 22%, then A = 22%, so A+T = 44%, and G+C =
100% – 44% = 56%. Since G = C, then C = 28%. Option C is correct.*
