BIOD 210 Module 2 Exam: Genetics 2026/2027 –
Actual Questions with Verified Answers –
Complete Practice Examination – Portage
Learning – Graded A+
Part I: Multiple Choice (30 Questions, 2 points each)
Q1: In humans, widow's peak (pointed hairline) is dominant over straight hairline. If two
heterozygous parents have a child, what is the probability that the child will have a
straight hairline?
A. 0%
B. 25% [CORRECT]
C. 50%
D. 75%
Correct Answer: B
Rationale: Both parents are heterozygous (Ww). A Punnett square shows: 25% WW
(widow's peak), 50% Ww (widow's peak), 25% ww (straight hairline). Only the
homozygous recessive genotype (ww) expresses the straight hairline phenotype. Option
A is incorrect because there is a chance of the recessive phenotype. Option C
represents the probability of a heterozygous genotype, not the straight hairline
phenotype. Option D represents the probability of having widow's peak.
,Q2: A genetic counselor is analyzing a family with a rare autosomal recessive disorder.
Which of the following statements about carrier parents is CORRECT?
A. Both parents must be homozygous dominant
B. Both parents must be heterozygous carriers [CORRECT]
C. One parent must be affected and one must be unaffected
D. Both parents must be affected to have an affected child
Correct Answer: B
Rationale: For two unaffected parents to have a child with an autosomal recessive
disorder, both must be carriers (heterozygous Aa). Each child has a 25% chance of
being affected (aa). Option A is incorrect because homozygous dominant parents (AA)
cannot produce affected offspring. Option C is incorrect because an affected parent
(aa) would have the disorder. Option D is incorrect because if both parents were
affected, they would all have the disorder, not just some children.
Q3: Red-green color blindness is an X-linked recessive trait. A color-blind man and a
woman with normal vision (whose father was color-blind) have a son. What is the
probability that their son will be color-blind?
A. 0%
B. 25%
C. 50% [CORRECT]
D. 100%
,Correct Answer: C
Rationale: The man is XᶜY (color-blind). The woman must be a carrier (XᶜX) because her
father was color-blind (XᶜY) and passed his Xᶜ to all daughters. For a son: he receives Y
from father and either X or Xᶜ from mother (50% chance each). Therefore, 50% chance
of XᶜY (color-blind) or XY (normal). Option A is incorrect because there is a definite
chance. Option B confuses this with the 25% risk for a daughter to be affected. Option D
would only occur if the mother were affected (XᶜXᶜ), which she is not.
Q4: In snapdragons, flower color shows incomplete dominance. Red flowers (RR)
crossed with white flowers (WW) produce all pink flowers (RW). If two pink-flowered
plants are crossed, what phenotypic ratio is expected in the offspring?
A. 1 red : 2 pink : 1 white [CORRECT]
B. 3 red : 1 white
C. 9 red : 3 pink : 3 white : 1 other
D. All pink
Correct Answer: A
Rationale: Incomplete dominance results in a heterozygous phenotype intermediate
between the homozygous parents. The cross RW × RW produces: 25% RR (red), 50% RW
(pink), 25% WW (white), giving a 1:2:1 phenotypic ratio. Option B is the classic
Mendelian ratio for complete dominance. Option C is a dihybrid ratio. Option D would
occur if there were complete dominance or if the trait showed blending inheritance
without segregation.
Q5: Which nitrogenous base is found in RNA but NOT in DNA?
, A. Adenine
B. Guanine
C. Thymine
D. Uracil [CORRECT]
Correct Answer: D
Rationale: RNA contains uracil (U) instead of thymine (T). Both nucleic acids contain
adenine, guanine, and cytosine. During transcription, DNA's thymine is replaced by uracil
in the mRNA transcript. This is a fundamental difference between the two nucleic acids.
Options A, B, and C are all found in DNA.
Q6: During DNA replication, which enzyme is responsible for synthesizing new DNA
strands by adding nucleotides in the 5' to 3' direction?
A. Helicase
B. DNA ligase
C. DNA polymerase [CORRECT]
D. Primase
Correct Answer: C
Rationale: DNA polymerase is the enzyme that catalyzes phosphodiester bond
formation, adding nucleotides only to the 3'-OH end of a growing chain (5' to 3'
synthesis). Helicase (A) unwinds the double helix. DNA ligase (B) joins Okazaki
fragments on the lagging strand. Primase (D) synthesizes RNA primers. The directional
Actual Questions with Verified Answers –
Complete Practice Examination – Portage
Learning – Graded A+
Part I: Multiple Choice (30 Questions, 2 points each)
Q1: In humans, widow's peak (pointed hairline) is dominant over straight hairline. If two
heterozygous parents have a child, what is the probability that the child will have a
straight hairline?
A. 0%
B. 25% [CORRECT]
C. 50%
D. 75%
Correct Answer: B
Rationale: Both parents are heterozygous (Ww). A Punnett square shows: 25% WW
(widow's peak), 50% Ww (widow's peak), 25% ww (straight hairline). Only the
homozygous recessive genotype (ww) expresses the straight hairline phenotype. Option
A is incorrect because there is a chance of the recessive phenotype. Option C
represents the probability of a heterozygous genotype, not the straight hairline
phenotype. Option D represents the probability of having widow's peak.
,Q2: A genetic counselor is analyzing a family with a rare autosomal recessive disorder.
Which of the following statements about carrier parents is CORRECT?
A. Both parents must be homozygous dominant
B. Both parents must be heterozygous carriers [CORRECT]
C. One parent must be affected and one must be unaffected
D. Both parents must be affected to have an affected child
Correct Answer: B
Rationale: For two unaffected parents to have a child with an autosomal recessive
disorder, both must be carriers (heterozygous Aa). Each child has a 25% chance of
being affected (aa). Option A is incorrect because homozygous dominant parents (AA)
cannot produce affected offspring. Option C is incorrect because an affected parent
(aa) would have the disorder. Option D is incorrect because if both parents were
affected, they would all have the disorder, not just some children.
Q3: Red-green color blindness is an X-linked recessive trait. A color-blind man and a
woman with normal vision (whose father was color-blind) have a son. What is the
probability that their son will be color-blind?
A. 0%
B. 25%
C. 50% [CORRECT]
D. 100%
,Correct Answer: C
Rationale: The man is XᶜY (color-blind). The woman must be a carrier (XᶜX) because her
father was color-blind (XᶜY) and passed his Xᶜ to all daughters. For a son: he receives Y
from father and either X or Xᶜ from mother (50% chance each). Therefore, 50% chance
of XᶜY (color-blind) or XY (normal). Option A is incorrect because there is a definite
chance. Option B confuses this with the 25% risk for a daughter to be affected. Option D
would only occur if the mother were affected (XᶜXᶜ), which she is not.
Q4: In snapdragons, flower color shows incomplete dominance. Red flowers (RR)
crossed with white flowers (WW) produce all pink flowers (RW). If two pink-flowered
plants are crossed, what phenotypic ratio is expected in the offspring?
A. 1 red : 2 pink : 1 white [CORRECT]
B. 3 red : 1 white
C. 9 red : 3 pink : 3 white : 1 other
D. All pink
Correct Answer: A
Rationale: Incomplete dominance results in a heterozygous phenotype intermediate
between the homozygous parents. The cross RW × RW produces: 25% RR (red), 50% RW
(pink), 25% WW (white), giving a 1:2:1 phenotypic ratio. Option B is the classic
Mendelian ratio for complete dominance. Option C is a dihybrid ratio. Option D would
occur if there were complete dominance or if the trait showed blending inheritance
without segregation.
Q5: Which nitrogenous base is found in RNA but NOT in DNA?
, A. Adenine
B. Guanine
C. Thymine
D. Uracil [CORRECT]
Correct Answer: D
Rationale: RNA contains uracil (U) instead of thymine (T). Both nucleic acids contain
adenine, guanine, and cytosine. During transcription, DNA's thymine is replaced by uracil
in the mRNA transcript. This is a fundamental difference between the two nucleic acids.
Options A, B, and C are all found in DNA.
Q6: During DNA replication, which enzyme is responsible for synthesizing new DNA
strands by adding nucleotides in the 5' to 3' direction?
A. Helicase
B. DNA ligase
C. DNA polymerase [CORRECT]
D. Primase
Correct Answer: C
Rationale: DNA polymerase is the enzyme that catalyzes phosphodiester bond
formation, adding nucleotides only to the 3'-OH end of a growing chain (5' to 3'
synthesis). Helicase (A) unwinds the double helix. DNA ligase (B) joins Okazaki
fragments on the lagging strand. Primase (D) synthesizes RNA primers. The directional
