ASU BIO 340 Exam Guide: 2026/2027 Session
Instructor: [Name Redacted], Ph.D. Course:
BIO 340 General Genetics Exam |
Comprehensive Final Practice Exam.
Domain 1: Mendelian & Non-Mendelian Inheritance (14 Questions)
1. In a classic dihybrid cross involving two heterozygous parents (RrYy x RrYy), which
phenotypic ratio in the offspring demonstrates the Law of Independent Assortment?
A. 1:1
B. 3:1
C. 9:3:3:1
D. 1:2:1
[CORRECT] C
Rationale: The 9:3:3:1 ratio is the hallmark of a dihybrid cross between two heterozygous
parents for unlinked genes. It demonstrates that the alleles for each gene segregate
independently of one another during gamete formation, resulting in four distinct phenotypic
classes.
2. (Calculation) In a dihybrid cross where both parents are heterozygous (AaBb), what is the
probability of producing an offspring with the dominant phenotype for trait A and the
recessive phenotype for trait B (A_bb)?
A. 9/16
B. 1/16
C. 3/16
D. 1/4
[CORRECT] C
Rationale: The probability of A_ is 3/4. The probability of bb is 1/4. Since the genes assort
independently, the combined probability is (3/4) × (1/4) = 3/16.
,3. Which of the following conditions must be met for a dihybrid cross to yield a 9:3:3:1
phenotypic ratio?
A. The genes must be located on the same chromosome.
B. The genes must assort independently (be unlinked or very far apart).
C. One allele must exhibit complete dominance over the other.
D. Both B and C are correct.
[CORRECT] D
Rationale: The 9:3:3:1 ratio assumes Independent Assortment (unlinked genes) and Complete
Dominance (where heterozygotes look like dominant homozygotes). Without independent
assortment, linkage would skew the ratio.
4. True or False: The 9:3:3:1 phenotypic ratio expected from a dihybrid cross is based on the
genotypic ratio of the F1 generation crossing with a homozygous recessive individual.
[CORRECT] False
Rationale: The 9:3:3:1 ratio results from a monohybrid cross extended to two traits (AaBb x
AaBb). A cross with a homozygous recessive individual (testcross) would yield a 1:1:1:1 ratio if
the genes are unlinked.
5. (Calculation) Hemophilia A is an X-linked recessive disorder. A woman who is a carrier for
hemophilia (X^H X^h) mates with a man who does not have hemophilia (X^H Y). What is the
probability that their first son will have hemophilia?
A. 0%
B. 25%
C. 50%
D. 100%
[CORRECT] C
, Rationale: The mother passes her X chromosome to her sons. She is X^H X^h, so there is a
50% chance she passes X^h (affected) and a 50% chance she passes X^H (unaffected). Sons
inherit the Y chromosome from the father. Therefore, the probability is 1/2 or 50%.
6. (Select-all-that-apply) Select all the statements that accurately apply to X-linked recessive
inheritance patterns like Hemophilia:
A. Females must be homozygous recessive to express the trait.
B. Affected males transmit the allele to all of their daughters.
C. Carrier females have a 50% chance of transmitting the affected allele to each son.
D. The trait skips generations more frequently than autosomal dominant traits.
[CORRECT] A, B, C, D
Rationale: A: Females have two X chromosomes; both must carry the recessive allele for
expression. B: Fathers pass X to daughters; if the father is affected (X^h Y), all daughters get
X^h. C: Carrier females (X^H X^h) produce X^h gametes 50% of the time, determining the
son's status. D: Because carrier females often do not show the trait, the phenotype can
disappear in one generation (daughters) and reappear in the next (grandsons).
7. Why does a male who inherits an X-linked recessive allele for hemophilia from his mother
always express the bleeding disorder?
A. Because the allele is dominant in males.
B. Because he has only one X chromosome and no homologous allele on the Y to mask it.
C. Because testosterone activates the expression of the clotting factor gene.
D. Because the Y chromosome carries a deletion of the normal gene.
[CORRECT] B
Rationale: Males are hemizygous for the X chromosome. They possess only one copy of X-
linked genes. If that single copy carries the recessive mutation (X^h), there is no second copy
(like on the Y chromosome) to provide a normal functional allele, so the phenotype is
expressed.
Instructor: [Name Redacted], Ph.D. Course:
BIO 340 General Genetics Exam |
Comprehensive Final Practice Exam.
Domain 1: Mendelian & Non-Mendelian Inheritance (14 Questions)
1. In a classic dihybrid cross involving two heterozygous parents (RrYy x RrYy), which
phenotypic ratio in the offspring demonstrates the Law of Independent Assortment?
A. 1:1
B. 3:1
C. 9:3:3:1
D. 1:2:1
[CORRECT] C
Rationale: The 9:3:3:1 ratio is the hallmark of a dihybrid cross between two heterozygous
parents for unlinked genes. It demonstrates that the alleles for each gene segregate
independently of one another during gamete formation, resulting in four distinct phenotypic
classes.
2. (Calculation) In a dihybrid cross where both parents are heterozygous (AaBb), what is the
probability of producing an offspring with the dominant phenotype for trait A and the
recessive phenotype for trait B (A_bb)?
A. 9/16
B. 1/16
C. 3/16
D. 1/4
[CORRECT] C
Rationale: The probability of A_ is 3/4. The probability of bb is 1/4. Since the genes assort
independently, the combined probability is (3/4) × (1/4) = 3/16.
,3. Which of the following conditions must be met for a dihybrid cross to yield a 9:3:3:1
phenotypic ratio?
A. The genes must be located on the same chromosome.
B. The genes must assort independently (be unlinked or very far apart).
C. One allele must exhibit complete dominance over the other.
D. Both B and C are correct.
[CORRECT] D
Rationale: The 9:3:3:1 ratio assumes Independent Assortment (unlinked genes) and Complete
Dominance (where heterozygotes look like dominant homozygotes). Without independent
assortment, linkage would skew the ratio.
4. True or False: The 9:3:3:1 phenotypic ratio expected from a dihybrid cross is based on the
genotypic ratio of the F1 generation crossing with a homozygous recessive individual.
[CORRECT] False
Rationale: The 9:3:3:1 ratio results from a monohybrid cross extended to two traits (AaBb x
AaBb). A cross with a homozygous recessive individual (testcross) would yield a 1:1:1:1 ratio if
the genes are unlinked.
5. (Calculation) Hemophilia A is an X-linked recessive disorder. A woman who is a carrier for
hemophilia (X^H X^h) mates with a man who does not have hemophilia (X^H Y). What is the
probability that their first son will have hemophilia?
A. 0%
B. 25%
C. 50%
D. 100%
[CORRECT] C
, Rationale: The mother passes her X chromosome to her sons. She is X^H X^h, so there is a
50% chance she passes X^h (affected) and a 50% chance she passes X^H (unaffected). Sons
inherit the Y chromosome from the father. Therefore, the probability is 1/2 or 50%.
6. (Select-all-that-apply) Select all the statements that accurately apply to X-linked recessive
inheritance patterns like Hemophilia:
A. Females must be homozygous recessive to express the trait.
B. Affected males transmit the allele to all of their daughters.
C. Carrier females have a 50% chance of transmitting the affected allele to each son.
D. The trait skips generations more frequently than autosomal dominant traits.
[CORRECT] A, B, C, D
Rationale: A: Females have two X chromosomes; both must carry the recessive allele for
expression. B: Fathers pass X to daughters; if the father is affected (X^h Y), all daughters get
X^h. C: Carrier females (X^H X^h) produce X^h gametes 50% of the time, determining the
son's status. D: Because carrier females often do not show the trait, the phenotype can
disappear in one generation (daughters) and reappear in the next (grandsons).
7. Why does a male who inherits an X-linked recessive allele for hemophilia from his mother
always express the bleeding disorder?
A. Because the allele is dominant in males.
B. Because he has only one X chromosome and no homologous allele on the Y to mask it.
C. Because testosterone activates the expression of the clotting factor gene.
D. Because the Y chromosome carries a deletion of the normal gene.
[CORRECT] B
Rationale: Males are hemizygous for the X chromosome. They possess only one copy of X-
linked genes. If that single copy carries the recessive mutation (X^h), there is no second copy
(like on the Y chromosome) to provide a normal functional allele, so the phenotype is
expressed.
