BIOD210 GENETICS | PORTAGE LEARNING | COMPLETE EXAM
PACKAGE 2026/2027 Exams | Actual Questions & Verified Answers |
Comprehensive Genetics Assessment | Pass Guarantee
1. A pea plant heterozygous for stem length (Tt) is self-fertilized. What proportion
of the offspring will be homozygous?
A. 1/4
B. 1/2
C. 3/4
D. All
Correct Answer: B
Rationale: Punnett square of Tt × Tt → 1 TT : 2 Tt : 1 tt; 2/4 = ½ are homozygous (TT +
tt).
Distractor: A (1/4) counts only TT, ignoring tt.
2. In peas, round (R) is dominant to wrinkled (r). A round-seeded plant crossed with
a wrinkled-seeded plant produces 50 % round and 50 % wrinkled progeny. What is
the genotype of the round parent?
A. RR
B. Rr
C. rr
D. Cannot determine
Correct Answer: B
Rationale: 1:1 ratio is characteristic of a monohybrid test-cross; round parent must be
heterozygous (Rr).
Distractor: A (RR) would give 100 % round.
, 3. Which of the following represents a dihybrid genotype?
A. AABB
B. AaBB
C. AaBb
D. aaBb
Correct Answer: C
Rationale: Dihybrid = heterozygous at two loci (AaBb).
Distractor: B (AaBB) is monohybrid heterozygous.
4. The probability of drawing two aces in a row without replacement from a 52-card
deck is approximately:
A. 1/221
B. 1/52
C. 4/52
D. 1/13
Correct Answer: A
Rationale: (4/52) × (3/51) = 12/2652 = 1/221.
Distractor: B ignores the second draw’s dependence.
5. A couple both heterozygous for an autosomal recessive condition have four
children. What is the probability that exactly three are unaffected carriers?
A. 1/256
B. 3/64
C. 27/64
D. 1/4
Correct Answer: C
Rationale: Each child: ½ carrier (unaffected), ¼ affected, ¼ homozygous normal. Use
binomial: C(4,3) × (½)³ × (½)¹ = 4 × 1/16 = ¼, but only carriers count → ½ chance per
child to be carrier; C(4,3) × (½)³ × (½)¹ = 4/16 = ¼; however, exactly three carriers among
PACKAGE 2026/2027 Exams | Actual Questions & Verified Answers |
Comprehensive Genetics Assessment | Pass Guarantee
1. A pea plant heterozygous for stem length (Tt) is self-fertilized. What proportion
of the offspring will be homozygous?
A. 1/4
B. 1/2
C. 3/4
D. All
Correct Answer: B
Rationale: Punnett square of Tt × Tt → 1 TT : 2 Tt : 1 tt; 2/4 = ½ are homozygous (TT +
tt).
Distractor: A (1/4) counts only TT, ignoring tt.
2. In peas, round (R) is dominant to wrinkled (r). A round-seeded plant crossed with
a wrinkled-seeded plant produces 50 % round and 50 % wrinkled progeny. What is
the genotype of the round parent?
A. RR
B. Rr
C. rr
D. Cannot determine
Correct Answer: B
Rationale: 1:1 ratio is characteristic of a monohybrid test-cross; round parent must be
heterozygous (Rr).
Distractor: A (RR) would give 100 % round.
, 3. Which of the following represents a dihybrid genotype?
A. AABB
B. AaBB
C. AaBb
D. aaBb
Correct Answer: C
Rationale: Dihybrid = heterozygous at two loci (AaBb).
Distractor: B (AaBB) is monohybrid heterozygous.
4. The probability of drawing two aces in a row without replacement from a 52-card
deck is approximately:
A. 1/221
B. 1/52
C. 4/52
D. 1/13
Correct Answer: A
Rationale: (4/52) × (3/51) = 12/2652 = 1/221.
Distractor: B ignores the second draw’s dependence.
5. A couple both heterozygous for an autosomal recessive condition have four
children. What is the probability that exactly three are unaffected carriers?
A. 1/256
B. 3/64
C. 27/64
D. 1/4
Correct Answer: C
Rationale: Each child: ½ carrier (unaffected), ¼ affected, ¼ homozygous normal. Use
binomial: C(4,3) × (½)³ × (½)¹ = 4 × 1/16 = ¼, but only carriers count → ½ chance per
child to be carrier; C(4,3) × (½)³ × (½)¹ = 4/16 = ¼; however, exactly three carriers among
