BIOD 210 Genetics Final Exam Actual Exam 2026/2027 – Comprehensive
Genetics Assessment with Detailed Rationales | 100% Verified | Pass Guaranteed
– A+ Graded
Section 1: Mendelian Genetics & Inheritance Patterns
Q1: A true-breeding pea plant with purple flowers is crossed with a true-breeding pea
plant with white flowers. All F1 offspring have purple flowers. When the F1 plants are
self-fertilized, the F2 generation shows a 3:1 ratio of purple to white flowers. This result
supports which of Mendel's principles?
A. Law of Independent Assortment
B. Law of Segregation [CORRECT]
C. Law of Dominance only
D. Law of Unit Characters
Correct Answer: B
Rationale: The 3:1 phenotypic ratio in the F2 generation demonstrates the Law of
Segregation, which states that during gamete formation, the two alleles for a trait
separate (segregate) so that each gamete receives only one allele. The reappearance of
the white phenotype in F2 confirms that the allele was present but masked in F1
heterozygotes.
,Q2: In humans, the ability to taste phenylthiocarbamide (PTC) is a dominant trait (T),
while the inability to taste it is recessive (t). If a heterozygous taster (Tt) has children
with a non-taster (tt), what is the probability that their first child will be a taster?
A. 0%
B. 25%
C. 50% [CORRECT]
D. 75%
Correct Answer: C
Rationale: A Tt × tt cross produces offspring with genotypes Tt and tt in a 1:1 ratio (50%
Tt tasters, 50% tt non-tasters). This is a test cross that reveals the heterozygous
genotype of the taster parent.
Q3: A couple has four children, all with blood type A. The father has blood type AB and
the mother has blood type O. Which statement about the children's genotypes is
correct?
A. All children must be heterozygous (IAi) [CORRECT]
B. All children must be homozygous (IAIA)
C. The children could be either IAIA or IAi
D. The children's blood type indicates non-paternity
Correct Answer: A
,Rationale: The father with blood type AB has genotype IAIB, and the mother with blood
type O has genotype ii. All children must inherit IB from the father and i from the mother,
resulting in genotype IAi (blood type A). They cannot be IAIA because the mother can
only contribute an i allele.
Q4: In snapdragons, red flower color (RR) is incompletely dominant over white flower
color (rr), with heterozygotes (Rr) showing pink flowers. If two pink snapdragons are
crossed, what phenotypic ratio will appear in the offspring?
A. 3 red : 1 white
B. 1 red : 2 pink : 1 white [CORRECT]
C. 9 red : 3 pink : 4 white
D. All pink flowers
Correct Answer: B
Rationale: An Rr × Rr cross produces genotypes RR (1/4), Rr (1/2), and rr (1/4). With
incomplete dominance, each genotype corresponds to a distinct phenotype: red (RR),
pink (Rr), and white (rr), yielding a 1:2:1 phenotypic ratio.
Q5: A man with blood type B (whose mother had blood type O) marries a woman with
blood type A (whose father had blood type O). What is the probability that their first
child will have blood type O?
A. 0%
B. 25% [CORRECT]
C. 50%
, D. 75%
Correct Answer: B
Rationale: The man with blood type B and an O mother must be genotype IBi. The
woman with blood type A and an O father must be genotype IAi. An IBi × IAi cross
produces offspring with genotypes IAIB (AB, 25%), IAi (A, 25%), IBi (B, 25%), and ii (O,
25%). The probability of blood type O is 25%.
Q6: Which of the following human genetic disorders is inherited in an autosomal
recessive pattern?
A. Huntington disease
B. Cystic fibrosis [CORRECT]
C. Neurofibromatosis type 1
D. Familial hypercholesterolemia
Correct Answer: B
Rationale: Cystic fibrosis is caused by mutations in the CFTR gene and follows
autosomal recessive inheritance. Both parents of an affected child are typically carriers
(heterozygotes). Huntington disease is autosomal dominant, while neurofibromatosis
type 1 and familial hypercholesterolemia are also autosomal dominant disorders.
Q7: A child is born with albinism (autosomal recessive). Both parents have normal
pigmentation. What is the probability that their next child will also have albinism?
A. 0%
Genetics Assessment with Detailed Rationales | 100% Verified | Pass Guaranteed
– A+ Graded
Section 1: Mendelian Genetics & Inheritance Patterns
Q1: A true-breeding pea plant with purple flowers is crossed with a true-breeding pea
plant with white flowers. All F1 offspring have purple flowers. When the F1 plants are
self-fertilized, the F2 generation shows a 3:1 ratio of purple to white flowers. This result
supports which of Mendel's principles?
A. Law of Independent Assortment
B. Law of Segregation [CORRECT]
C. Law of Dominance only
D. Law of Unit Characters
Correct Answer: B
Rationale: The 3:1 phenotypic ratio in the F2 generation demonstrates the Law of
Segregation, which states that during gamete formation, the two alleles for a trait
separate (segregate) so that each gamete receives only one allele. The reappearance of
the white phenotype in F2 confirms that the allele was present but masked in F1
heterozygotes.
,Q2: In humans, the ability to taste phenylthiocarbamide (PTC) is a dominant trait (T),
while the inability to taste it is recessive (t). If a heterozygous taster (Tt) has children
with a non-taster (tt), what is the probability that their first child will be a taster?
A. 0%
B. 25%
C. 50% [CORRECT]
D. 75%
Correct Answer: C
Rationale: A Tt × tt cross produces offspring with genotypes Tt and tt in a 1:1 ratio (50%
Tt tasters, 50% tt non-tasters). This is a test cross that reveals the heterozygous
genotype of the taster parent.
Q3: A couple has four children, all with blood type A. The father has blood type AB and
the mother has blood type O. Which statement about the children's genotypes is
correct?
A. All children must be heterozygous (IAi) [CORRECT]
B. All children must be homozygous (IAIA)
C. The children could be either IAIA or IAi
D. The children's blood type indicates non-paternity
Correct Answer: A
,Rationale: The father with blood type AB has genotype IAIB, and the mother with blood
type O has genotype ii. All children must inherit IB from the father and i from the mother,
resulting in genotype IAi (blood type A). They cannot be IAIA because the mother can
only contribute an i allele.
Q4: In snapdragons, red flower color (RR) is incompletely dominant over white flower
color (rr), with heterozygotes (Rr) showing pink flowers. If two pink snapdragons are
crossed, what phenotypic ratio will appear in the offspring?
A. 3 red : 1 white
B. 1 red : 2 pink : 1 white [CORRECT]
C. 9 red : 3 pink : 4 white
D. All pink flowers
Correct Answer: B
Rationale: An Rr × Rr cross produces genotypes RR (1/4), Rr (1/2), and rr (1/4). With
incomplete dominance, each genotype corresponds to a distinct phenotype: red (RR),
pink (Rr), and white (rr), yielding a 1:2:1 phenotypic ratio.
Q5: A man with blood type B (whose mother had blood type O) marries a woman with
blood type A (whose father had blood type O). What is the probability that their first
child will have blood type O?
A. 0%
B. 25% [CORRECT]
C. 50%
, D. 75%
Correct Answer: B
Rationale: The man with blood type B and an O mother must be genotype IBi. The
woman with blood type A and an O father must be genotype IAi. An IBi × IAi cross
produces offspring with genotypes IAIB (AB, 25%), IAi (A, 25%), IBi (B, 25%), and ii (O,
25%). The probability of blood type O is 25%.
Q6: Which of the following human genetic disorders is inherited in an autosomal
recessive pattern?
A. Huntington disease
B. Cystic fibrosis [CORRECT]
C. Neurofibromatosis type 1
D. Familial hypercholesterolemia
Correct Answer: B
Rationale: Cystic fibrosis is caused by mutations in the CFTR gene and follows
autosomal recessive inheritance. Both parents of an affected child are typically carriers
(heterozygotes). Huntington disease is autosomal dominant, while neurofibromatosis
type 1 and familial hypercholesterolemia are also autosomal dominant disorders.
Q7: A child is born with albinism (autosomal recessive). Both parents have normal
pigmentation. What is the probability that their next child will also have albinism?
A. 0%
