BIOD 210
Genetics Module 2
Questions and Verified Answers
GRADE A | 100% CORRECT
Portage Learning | Genetics Course
Comprehensive 105-Question Module 2 Examination Guide
Detailed Rationales with Cross Demonstrations and Probability Calculations
Total Questions 105 MCQs (4 options, one correct)
Sections 8 genetics content domains
Cognitive Mix 30% recall / 50% application / 20% analysis
Format 65% application-based / 35% direct recall
Rationale Depth 2-4 sentences with genetic reasoning, cross demonstration, and error analysis
Portage Learning BIOD 210 | Genetics Module 2 | Z.ai Academic Series Confidential - Educational Use
,BIOD 210 Module 2 - Genetics - Questions and Verified Answers Latest 2026/2027 | Grade A | 100% Correct
Examination Overview
Purpose: This 105-question examination guide is designed to prepare Portage Learning
students for the BIOD 210 Genetics Module 2 assessment. The questions cover the eight core
content domains: Mendelian genetics and monohybrid crosses; dihybrid crosses and
independent assortment; non-Mendelian inheritance patterns; sex determination and
sex-linked inheritance; pedigree analysis and inheritance pattern recognition; linkage,
crossing over, and recombination; extranuclear inheritance and epigenetics; and
multifactorial traits, population genetics basics, and applied genetics. Each question includes
a detailed rationale with genetic reasoning, cross demonstration, and error analysis to support
Grade A preparation.
How to use this document: Attempt all 105 questions across the 8 sections. Verify each answer
against the rationale. Pay special attention to "What is the probability that..." probability
calculation questions, "What is the most likely mode of inheritance?" pedigree interpretation
questions, blood group (multiple allele) application questions, and recombination frequency
calculations — these are highest-yield on the BIOD 210 Module 2 assessment. The rationales
are designed to double as a genetics study guide with cross demonstrations and common error
analysis.
Section Title Q Range # Qs
Section 1 Mendelian Genetics and Monohybrid Crosses Q1-Q16 16
Section 2 Dihybrid Crosses and Independent Assortment Q17-Q28 12
Section 3 Non-Mendelian Inheritance Patterns Q29-Q42 14
Section 4 Sex Determination and Sex-Linked Inheritance Q43-Q55 13
Section 5 Pedigree Analysis and Inheritance Pattern Recognition Q56-Q69 14
Section 6 Linkage, Crossing Over, and Recombination Q70-Q80 11
Section 7 Extranuclear Inheritance and Epigenetics Q81-Q89 9
Section 8 Multifactorial Traits, Population Genetics Basics, and Applied Genetics
Q90-Q105 16
TOTAL 105
BIOD 210 Module 2 Page 2 105 Questions
, Section 1: Mendelian Genetics and Monohybrid Crosses
Alleles, Genotypes, Phenotypes, Punnett Squares, & Probability Rules
Q1. In pea plants, purple flower color (P) is dominant over white (p). If two
heterozygous plants (Pp × Pp) are crossed, what is the expected phenotypic ratio of
the offspring?
A. 1:1 purple:white
B. 1:2:1 (PP:Pp:pp)
C. 3:1 purple:white [CORRECT]
D. 9:3:3:1
Correct Answer: C
Rationale: Pp × Pp yields a 1 PP : 2 Pp : 1 pp genotypic ratio. Because P is completely
dominant, both PP and Pp plants are purple (3/4) while only pp is white (1/4). The phenotypic
ratio is therefore 3 purple : 1 white. The 1:2:1 ratio is the genotypic ratio (incorrect as a
phenotypic answer), 1:1 occurs in a testcross (Pp × pp), and 9:3:3:1 is the dihybrid phenotypic
ratio — not applicable to a single-gene cross.
Q2. A homozygous dominant (TT) tall pea plant is crossed with a homozygous
recessive (tt) short plant. What proportion of the F1 generation will be tall?
A. 25%
B. 100% [CORRECT]
C. 50%
D. 75%
Correct Answer: B
Rationale: TT × tt produces F1 offspring that are all Tt (heterozygous). Because T (tall) is
dominant over t (short), every F1 plant expresses the tall phenotype. The Punnett square yields
four Tt combinations out of four — 100% tall. The 25%, 50%, and 75% values would apply to
other cross types (e.g., a Tt × Tt cross gives 75% tall).
Q3. Mendel's Law of Segregation states that:
A. The two alleles for a trait separate (segregate) during gamete formation, with each
gamete receiving only one allele [CORRECT]
B. Alleles of different genes assort independently during gamete formation
C. Dominant alleles always mask recessive alleles in the phenotype
D. Genes located on the same chromosome are always inherited together
Correct Answer: A
Rationale: The Law of Segregation: diploid individuals carry two alleles per gene, and these
alleles separate during meiosis so that each gamete receives only one allele. This is the
molecular basis of the 3:1 monohybrid ratio. The first option describes the Law of Independent
Assortment; the third describes dominance (a separate principle); the fourth describes
complete linkage (a violation of independent assortment).
, Q4. A plant with purple flowers of unknown genotype is test-crossed with a
white-flowered plant. If the offspring are 50% purple and 50% white, what is the
genotype of the purple-flowered parent?
A. Pp (heterozygous) [CORRECT]
B. PP (homozygous dominant)
C. pp (homozygous recessive)
D. Cannot be determined
Correct Answer: A
Rationale: A test cross uses a homozygous recessive individual (pp) to reveal the genotype of a
dominant phenotype. If the unknown parent is PP, all offspring are Pp (100% purple). If the
unknown is Pp, the cross Pp × pp produces a 1 Pp : 1 pp ratio (50% purple : 50% white). The
observed 1:1 phenotypic ratio therefore indicates the unknown parent is heterozygous (Pp).
Q5. Using the product rule, what is the probability that a heterozygous couple (Pp ×
Pp) will have three children in a row who all display the dominant phenotype?
(Assume complete dominance.)
A. (1/4)^3 = 1/64
B. (1/2)^3 = 1/8
C. (3/4)(1/4) = 3/16
D. (3/4)^3 = 27/64 [CORRECT]
Correct Answer: D
Rationale: Each child is an independent event. In a Pp × Pp cross, the probability of any one
child showing the dominant phenotype is 3/4 (PP + Pp). The product rule multiplies the
probabilities of independent events: 3/4 × 3/4 × 3/4 = 27/64 ≈ 42.2%. Option B (1/64) is the
probability of three recessive children; option C incorrectly uses 1/2; option D multiplies
dominant and recessive probabilities for one child only.
Q6. In humans, albinism is autosomal recessive (a). Two normally-pigmented
parents have an albino child. What is the probability that their next child will also be
albino?
A. 1/2 (50%)
B. 3/4 (75%)
C. 1/4 (25%) [CORRECT]
D. 100%
Correct Answer: C
Rationale: Because both parents are phenotypically normal but have an albino (aa) child, both
parents must be carriers (Aa). The cross Aa × Aa produces a 1 AA : 2 Aa : 1 aa genotypic ratio.
The probability that the next child is albino (aa) is 1/4 (25%) — each conception is an
independent event; the outcome of one child does not affect the probability for the next.
Genetics Module 2
Questions and Verified Answers
GRADE A | 100% CORRECT
Portage Learning | Genetics Course
Comprehensive 105-Question Module 2 Examination Guide
Detailed Rationales with Cross Demonstrations and Probability Calculations
Total Questions 105 MCQs (4 options, one correct)
Sections 8 genetics content domains
Cognitive Mix 30% recall / 50% application / 20% analysis
Format 65% application-based / 35% direct recall
Rationale Depth 2-4 sentences with genetic reasoning, cross demonstration, and error analysis
Portage Learning BIOD 210 | Genetics Module 2 | Z.ai Academic Series Confidential - Educational Use
,BIOD 210 Module 2 - Genetics - Questions and Verified Answers Latest 2026/2027 | Grade A | 100% Correct
Examination Overview
Purpose: This 105-question examination guide is designed to prepare Portage Learning
students for the BIOD 210 Genetics Module 2 assessment. The questions cover the eight core
content domains: Mendelian genetics and monohybrid crosses; dihybrid crosses and
independent assortment; non-Mendelian inheritance patterns; sex determination and
sex-linked inheritance; pedigree analysis and inheritance pattern recognition; linkage,
crossing over, and recombination; extranuclear inheritance and epigenetics; and
multifactorial traits, population genetics basics, and applied genetics. Each question includes
a detailed rationale with genetic reasoning, cross demonstration, and error analysis to support
Grade A preparation.
How to use this document: Attempt all 105 questions across the 8 sections. Verify each answer
against the rationale. Pay special attention to "What is the probability that..." probability
calculation questions, "What is the most likely mode of inheritance?" pedigree interpretation
questions, blood group (multiple allele) application questions, and recombination frequency
calculations — these are highest-yield on the BIOD 210 Module 2 assessment. The rationales
are designed to double as a genetics study guide with cross demonstrations and common error
analysis.
Section Title Q Range # Qs
Section 1 Mendelian Genetics and Monohybrid Crosses Q1-Q16 16
Section 2 Dihybrid Crosses and Independent Assortment Q17-Q28 12
Section 3 Non-Mendelian Inheritance Patterns Q29-Q42 14
Section 4 Sex Determination and Sex-Linked Inheritance Q43-Q55 13
Section 5 Pedigree Analysis and Inheritance Pattern Recognition Q56-Q69 14
Section 6 Linkage, Crossing Over, and Recombination Q70-Q80 11
Section 7 Extranuclear Inheritance and Epigenetics Q81-Q89 9
Section 8 Multifactorial Traits, Population Genetics Basics, and Applied Genetics
Q90-Q105 16
TOTAL 105
BIOD 210 Module 2 Page 2 105 Questions
, Section 1: Mendelian Genetics and Monohybrid Crosses
Alleles, Genotypes, Phenotypes, Punnett Squares, & Probability Rules
Q1. In pea plants, purple flower color (P) is dominant over white (p). If two
heterozygous plants (Pp × Pp) are crossed, what is the expected phenotypic ratio of
the offspring?
A. 1:1 purple:white
B. 1:2:1 (PP:Pp:pp)
C. 3:1 purple:white [CORRECT]
D. 9:3:3:1
Correct Answer: C
Rationale: Pp × Pp yields a 1 PP : 2 Pp : 1 pp genotypic ratio. Because P is completely
dominant, both PP and Pp plants are purple (3/4) while only pp is white (1/4). The phenotypic
ratio is therefore 3 purple : 1 white. The 1:2:1 ratio is the genotypic ratio (incorrect as a
phenotypic answer), 1:1 occurs in a testcross (Pp × pp), and 9:3:3:1 is the dihybrid phenotypic
ratio — not applicable to a single-gene cross.
Q2. A homozygous dominant (TT) tall pea plant is crossed with a homozygous
recessive (tt) short plant. What proportion of the F1 generation will be tall?
A. 25%
B. 100% [CORRECT]
C. 50%
D. 75%
Correct Answer: B
Rationale: TT × tt produces F1 offspring that are all Tt (heterozygous). Because T (tall) is
dominant over t (short), every F1 plant expresses the tall phenotype. The Punnett square yields
four Tt combinations out of four — 100% tall. The 25%, 50%, and 75% values would apply to
other cross types (e.g., a Tt × Tt cross gives 75% tall).
Q3. Mendel's Law of Segregation states that:
A. The two alleles for a trait separate (segregate) during gamete formation, with each
gamete receiving only one allele [CORRECT]
B. Alleles of different genes assort independently during gamete formation
C. Dominant alleles always mask recessive alleles in the phenotype
D. Genes located on the same chromosome are always inherited together
Correct Answer: A
Rationale: The Law of Segregation: diploid individuals carry two alleles per gene, and these
alleles separate during meiosis so that each gamete receives only one allele. This is the
molecular basis of the 3:1 monohybrid ratio. The first option describes the Law of Independent
Assortment; the third describes dominance (a separate principle); the fourth describes
complete linkage (a violation of independent assortment).
, Q4. A plant with purple flowers of unknown genotype is test-crossed with a
white-flowered plant. If the offspring are 50% purple and 50% white, what is the
genotype of the purple-flowered parent?
A. Pp (heterozygous) [CORRECT]
B. PP (homozygous dominant)
C. pp (homozygous recessive)
D. Cannot be determined
Correct Answer: A
Rationale: A test cross uses a homozygous recessive individual (pp) to reveal the genotype of a
dominant phenotype. If the unknown parent is PP, all offspring are Pp (100% purple). If the
unknown is Pp, the cross Pp × pp produces a 1 Pp : 1 pp ratio (50% purple : 50% white). The
observed 1:1 phenotypic ratio therefore indicates the unknown parent is heterozygous (Pp).
Q5. Using the product rule, what is the probability that a heterozygous couple (Pp ×
Pp) will have three children in a row who all display the dominant phenotype?
(Assume complete dominance.)
A. (1/4)^3 = 1/64
B. (1/2)^3 = 1/8
C. (3/4)(1/4) = 3/16
D. (3/4)^3 = 27/64 [CORRECT]
Correct Answer: D
Rationale: Each child is an independent event. In a Pp × Pp cross, the probability of any one
child showing the dominant phenotype is 3/4 (PP + Pp). The product rule multiplies the
probabilities of independent events: 3/4 × 3/4 × 3/4 = 27/64 ≈ 42.2%. Option B (1/64) is the
probability of three recessive children; option C incorrectly uses 1/2; option D multiplies
dominant and recessive probabilities for one child only.
Q6. In humans, albinism is autosomal recessive (a). Two normally-pigmented
parents have an albino child. What is the probability that their next child will also be
albino?
A. 1/2 (50%)
B. 3/4 (75%)
C. 1/4 (25%) [CORRECT]
D. 100%
Correct Answer: C
Rationale: Because both parents are phenotypically normal but have an albino (aa) child, both
parents must be carriers (Aa). The cross Aa × Aa produces a 1 AA : 2 Aa : 1 aa genotypic ratio.
The probability that the next child is albino (aa) is 1/4 (25%) — each conception is an
independent event; the outcome of one child does not affect the probability for the next.
