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BIO 102 Final Actual Exam 2026/2027:
Questions and Verified Answers | Graded A+
- Portage Learning for Biology Success – Pass
Guaranteed - A+ Graded
SECTION 1: CELL BIOLOGY AND GENETICS (25 Questions)
Q1: In a monohybrid cross between two heterozygous pea plants (Tt × Tt) where T represents
tall and t represents short, what is the expected phenotypic ratio of offspring?
A. 1 tall : 1 short
B. 2 tall : 1 short
C. 3 tall : 1 short [CORRECT]
D. 4 tall : 0 short
Correct Answer: C
Rationale: Using a Punnett square for Tt × Tt, the genotypic ratio is 1 TT : 2 Tt : 1 tt. Since both
TT and Tt express the dominant tall phenotype, the phenotypic ratio is 3 tall (1 TT + 2 Tt) to 1
short (1 tt). This demonstrates Mendel's law of segregation and the principle of
dominant/recessive inheritance. Option A describes a test cross ratio, while B and D represent
incorrect calculations.
Q2: During which phase of mitosis do the sister chromatids separate and move toward opposite
poles of the cell?
A. Prophase
B. Metaphase
C. Anaphase [CORRECT]
D. Telophase
Correct Answer: C
Rationale: Anaphase is characterized by the separation of sister chromatids at the centromere,
with spindle fibers pulling them to opposite poles. This ensures each daughter cell receives
identical genetic material. Prophase (A) involves chromosome condensation and spindle
formation. Metaphase (B) aligns chromosomes at the equator. Telophase (D) involves nuclear
envelope reformation. This tests understanding of the sequential events in mitosis.
,2
Q3: A diploid organism has 2n = 16 chromosomes. How many chromosomes will be present in
each gamete after meiosis?
A. 4
B. 8 [CORRECT]
C. 16
D. 32
Correct Answer: B
Rationale: Meiosis reduces chromosome number by half. If 2n = 16 (diploid), then n = 8
(haploid). Gametes are haploid cells containing one set of chromosomes. The formula is: diploid
number ÷ 2 = haploid number. Option A represents an incorrect calculation. Option C is the
diploid number, and D doubles the diploid count.
Q4: In DNA replication, which enzyme synthesizes new DNA strands by adding nucleotides in
the 5' to 3' direction?
A. DNA helicase
B. DNA ligase
C. DNA polymerase [CORRECT]
D. DNA primase
Correct Answer: C
Rationale: DNA polymerase is the primary enzyme that catalyzes phosphodiester bond
formation between nucleotides, synthesizing new DNA in the 5'→3' direction only. DNA
helicase (A) unwinds the double helix. DNA ligase (B) joins Okazaki fragments on the lagging
strand. DNA primase (D) synthesizes RNA primers. This tests understanding of enzyme
functions in DNA replication.
Q5: A dihybrid cross between two heterozygous plants (RrYy × RrYy) where the genes assort
independently. What is the expected phenotypic ratio in the offspring?
A. 9:3:3:1 [CORRECT]
B. 3:1:3:1
C. 1:2:1:2:4:2:1:2:1
D. 12:3:1
Correct Answer: A
Rationale: For two heterozygous parents with independently assorting genes, the phenotypic
ratio is 9 (dominant-dominant) : 3 (dominant-recessive) : 3 (recessive-dominant) : 1 (recessive-
, 3
recessive). This is Mendel's law of independent assortment. The ratio 1:2:1:2:4:2:1:2:1 (C) is the
genotypic ratio, not phenotypic. Options B and D represent incorrect phenotypic combinations.
Q6: 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 instead of thymine. DNA uses thymine (T) to pair with adenine
(A), while RNA uses uracil (U) to pair with adenine. Both nucleic acids contain adenine (A),
guanine (G), and cytosine (C). This is a fundamental difference between DNA and RNA
structure.
Q7: In the lac operon of E. coli, what happens when lactose is present and glucose is absent?
A. The repressor binds to the operator, blocking transcription
B. The repressor is inactivated, allowing transcription [CORRECT]
C. The operon is always off regardless of lactose presence
D. RNA polymerase cannot bind to the promoter
Correct Answer: B
Rationale: When lactose is present, allolactose (an isomer of lactose) binds to the repressor
protein, causing a conformational change that prevents the repressor from binding to the
operator. This allows RNA polymerase to transcribe the structural genes (lacZ, lacY, lacA). In the
absence of lactose, the active repressor binds the operator and blocks transcription. This tests
understanding of negative gene regulation in prokaryotes.
Q8: A male with hemophilia (X-linked recessive) has children with a female carrier. What is the
probability that their son will have hemophilia?
A. 0%
B. 25%
C. 50% [CORRECT]
D. 100%
Correct Answer: C
Rationale: The male is X^hY (affected) and the female is X^HX^h (carrier). For sons: they
BIO 102 Final Actual Exam 2026/2027:
Questions and Verified Answers | Graded A+
- Portage Learning for Biology Success – Pass
Guaranteed - A+ Graded
SECTION 1: CELL BIOLOGY AND GENETICS (25 Questions)
Q1: In a monohybrid cross between two heterozygous pea plants (Tt × Tt) where T represents
tall and t represents short, what is the expected phenotypic ratio of offspring?
A. 1 tall : 1 short
B. 2 tall : 1 short
C. 3 tall : 1 short [CORRECT]
D. 4 tall : 0 short
Correct Answer: C
Rationale: Using a Punnett square for Tt × Tt, the genotypic ratio is 1 TT : 2 Tt : 1 tt. Since both
TT and Tt express the dominant tall phenotype, the phenotypic ratio is 3 tall (1 TT + 2 Tt) to 1
short (1 tt). This demonstrates Mendel's law of segregation and the principle of
dominant/recessive inheritance. Option A describes a test cross ratio, while B and D represent
incorrect calculations.
Q2: During which phase of mitosis do the sister chromatids separate and move toward opposite
poles of the cell?
A. Prophase
B. Metaphase
C. Anaphase [CORRECT]
D. Telophase
Correct Answer: C
Rationale: Anaphase is characterized by the separation of sister chromatids at the centromere,
with spindle fibers pulling them to opposite poles. This ensures each daughter cell receives
identical genetic material. Prophase (A) involves chromosome condensation and spindle
formation. Metaphase (B) aligns chromosomes at the equator. Telophase (D) involves nuclear
envelope reformation. This tests understanding of the sequential events in mitosis.
,2
Q3: A diploid organism has 2n = 16 chromosomes. How many chromosomes will be present in
each gamete after meiosis?
A. 4
B. 8 [CORRECT]
C. 16
D. 32
Correct Answer: B
Rationale: Meiosis reduces chromosome number by half. If 2n = 16 (diploid), then n = 8
(haploid). Gametes are haploid cells containing one set of chromosomes. The formula is: diploid
number ÷ 2 = haploid number. Option A represents an incorrect calculation. Option C is the
diploid number, and D doubles the diploid count.
Q4: In DNA replication, which enzyme synthesizes new DNA strands by adding nucleotides in
the 5' to 3' direction?
A. DNA helicase
B. DNA ligase
C. DNA polymerase [CORRECT]
D. DNA primase
Correct Answer: C
Rationale: DNA polymerase is the primary enzyme that catalyzes phosphodiester bond
formation between nucleotides, synthesizing new DNA in the 5'→3' direction only. DNA
helicase (A) unwinds the double helix. DNA ligase (B) joins Okazaki fragments on the lagging
strand. DNA primase (D) synthesizes RNA primers. This tests understanding of enzyme
functions in DNA replication.
Q5: A dihybrid cross between two heterozygous plants (RrYy × RrYy) where the genes assort
independently. What is the expected phenotypic ratio in the offspring?
A. 9:3:3:1 [CORRECT]
B. 3:1:3:1
C. 1:2:1:2:4:2:1:2:1
D. 12:3:1
Correct Answer: A
Rationale: For two heterozygous parents with independently assorting genes, the phenotypic
ratio is 9 (dominant-dominant) : 3 (dominant-recessive) : 3 (recessive-dominant) : 1 (recessive-
, 3
recessive). This is Mendel's law of independent assortment. The ratio 1:2:1:2:4:2:1:2:1 (C) is the
genotypic ratio, not phenotypic. Options B and D represent incorrect phenotypic combinations.
Q6: 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 instead of thymine. DNA uses thymine (T) to pair with adenine
(A), while RNA uses uracil (U) to pair with adenine. Both nucleic acids contain adenine (A),
guanine (G), and cytosine (C). This is a fundamental difference between DNA and RNA
structure.
Q7: In the lac operon of E. coli, what happens when lactose is present and glucose is absent?
A. The repressor binds to the operator, blocking transcription
B. The repressor is inactivated, allowing transcription [CORRECT]
C. The operon is always off regardless of lactose presence
D. RNA polymerase cannot bind to the promoter
Correct Answer: B
Rationale: When lactose is present, allolactose (an isomer of lactose) binds to the repressor
protein, causing a conformational change that prevents the repressor from binding to the
operator. This allows RNA polymerase to transcribe the structural genes (lacZ, lacY, lacA). In the
absence of lactose, the active repressor binds the operator and blocks transcription. This tests
understanding of negative gene regulation in prokaryotes.
Q8: A male with hemophilia (X-linked recessive) has children with a female carrier. What is the
probability that their son will have hemophilia?
A. 0%
B. 25%
C. 50% [CORRECT]
D. 100%
Correct Answer: C
Rationale: The male is X^hY (affected) and the female is X^HX^h (carrier). For sons: they
