BIOD210 GENETICS | PORTAGE LEARNING | COMPLETE EXAM
PACKAGE 2026/2027 Exams | Actual Questions & Verified Answers |
Comprehensive Genetics Assessment | Pass Guarantee
BIOD210 Exam 1: Mendelian Foundations
1. In garden peas, the trait for tall stems (T) is dominant to short (t). If a
heterozygous tall plant is crossed with a short plant, what proportion of the
offspring will be short?
A. 1/4
B. 1/2
C. 3/4
D. 0
Correct Answer: B
Rationale: Cross = Tt × tt → 1/2 Tt (tall) : 1/2 tt (short).
Why others wrong: A assumes Tt self-cross; C reverses ratio; D ignores tt class.
2. Mendel’s principle of segregation states that:
A. Genes for different traits assort independently.
B. Alleles separate during gamete formation.
C. Dominant alleles always mask recessive.
D. Genetic material replicates before division.
Correct Answer: B
Rationale: Segregation = alleles of a locus separate into different gametes.
Why others wrong: A = independent assortment; C = dominance definition; D =
basic mitosis.
3. A monohybrid cross produces a 3:1 phenotypic ratio in F₂. This ratio reflects:
A. Complete dominance
B. Codominance
C. Epistasis
D. Polygenic inheritance
Correct Answer: A
, Rationale: 3 dominant : 1 recessive hallmark of complete dominance.
Why others wrong: Codominance shows both phenotypes together; epistasis &
polygenic modify ratios.
4. In humans, unattached earlobes (E) are dominant. Two parents with unattached
earlobes have a child with attached earlobes. What are the parental genotypes?
A. EE × Ee
B. Ee × Ee
C. EE × EE
D. Ee × ee
Correct Answer: B
Rationale: Attached child = ee → each parent must carry e.
Why others wrong: A & C cannot produce ee; D would give 1/2 attached, but both
parents show unattached.
5. (Problem) In pea plants, yellow seed (Y) is dominant to green (y) and round (R)
dominant to wrinkled (r). A plant heterozygous for both traits is self-crossed.
What proportion of offspring will be green and wrinkled?
A. 1/16
B. 3/16
C. 9/16
D. 1/4
Correct Answer: A
Work: YyRr × YyRr → 1/4 yy × 1/4 rr = 1/16 yyrr.
Why others wrong: B = single recessive; C = dominant both; D = 1/4 only one trait
recessive.
6. The Punnett square visually represents:
A. Gene linkage
B. Random fertilization following gamete formation
C. Epistatic interactions
D. Chromosome mapping
Correct Answer: B
Rationale: Gametes placed on axes show all possible zygote combinations.
Why others wrong: Linkage & mapping require locus distance; epistasis needs
modifier genes.
PACKAGE 2026/2027 Exams | Actual Questions & Verified Answers |
Comprehensive Genetics Assessment | Pass Guarantee
BIOD210 Exam 1: Mendelian Foundations
1. In garden peas, the trait for tall stems (T) is dominant to short (t). If a
heterozygous tall plant is crossed with a short plant, what proportion of the
offspring will be short?
A. 1/4
B. 1/2
C. 3/4
D. 0
Correct Answer: B
Rationale: Cross = Tt × tt → 1/2 Tt (tall) : 1/2 tt (short).
Why others wrong: A assumes Tt self-cross; C reverses ratio; D ignores tt class.
2. Mendel’s principle of segregation states that:
A. Genes for different traits assort independently.
B. Alleles separate during gamete formation.
C. Dominant alleles always mask recessive.
D. Genetic material replicates before division.
Correct Answer: B
Rationale: Segregation = alleles of a locus separate into different gametes.
Why others wrong: A = independent assortment; C = dominance definition; D =
basic mitosis.
3. A monohybrid cross produces a 3:1 phenotypic ratio in F₂. This ratio reflects:
A. Complete dominance
B. Codominance
C. Epistasis
D. Polygenic inheritance
Correct Answer: A
, Rationale: 3 dominant : 1 recessive hallmark of complete dominance.
Why others wrong: Codominance shows both phenotypes together; epistasis &
polygenic modify ratios.
4. In humans, unattached earlobes (E) are dominant. Two parents with unattached
earlobes have a child with attached earlobes. What are the parental genotypes?
A. EE × Ee
B. Ee × Ee
C. EE × EE
D. Ee × ee
Correct Answer: B
Rationale: Attached child = ee → each parent must carry e.
Why others wrong: A & C cannot produce ee; D would give 1/2 attached, but both
parents show unattached.
5. (Problem) In pea plants, yellow seed (Y) is dominant to green (y) and round (R)
dominant to wrinkled (r). A plant heterozygous for both traits is self-crossed.
What proportion of offspring will be green and wrinkled?
A. 1/16
B. 3/16
C. 9/16
D. 1/4
Correct Answer: A
Work: YyRr × YyRr → 1/4 yy × 1/4 rr = 1/16 yyrr.
Why others wrong: B = single recessive; C = dominant both; D = 1/4 only one trait
recessive.
6. The Punnett square visually represents:
A. Gene linkage
B. Random fertilization following gamete formation
C. Epistatic interactions
D. Chromosome mapping
Correct Answer: B
Rationale: Gametes placed on axes show all possible zygote combinations.
Why others wrong: Linkage & mapping require locus distance; epistasis needs
modifier genes.
