BIO 251 Final Exam 2026: Complete Study Guide & Test Bank -
Real Exam Questions with Correct Detailed Answers (Most
Recent) - 150 Questions
Section 1: Cell Biology and Genetics (Questions 1-15)
1 A researcher treats cultured human fibroblasts with a specific inhibitor that blocks the interaction between the
anaphase-promoting complex/cyclosome (APC/C) and its activator Cdc20. Which of the following immediate
consequences is most likely to be observed in the treated cells?
A) Premature sister chromatid separation due to loss of securin degradation
B) Arrest in metaphase due to stabilization of securin and cyclin B
C) Exit from mitosis without chromosome segregation due to cyclin B degradation
D) Activation of the spindle assembly checkpoint, causing anaphase onset delay
Answer: B
Rationale: APC/C-Cdc20 targets securin and cyclin B for degradation to trigger anaphase and mitotic exit.
Inhibiting their interaction stabilizes both proteins, leading to metaphase arrest. Option A is incorrect because
securin degradation is blocked, not promoted. Option C is incorrect because cyclin B is not degraded. Option D is
incorrect because the spindle checkpoint is not directly activated by APC/C inhibition.
2 In a study of a rare genetic disorder, a family pedigree reveals that affected individuals inherit the disease from
their mothers, but never from their fathers. Sequencing of the mitochondrial genome shows a heteroplasmic
mutation in the MT-ND5 gene. Which of the following best explains the observed inheritance pattern?
A) The mutation is in a nuclear gene that escapes X-inactivation and is imprinted in paternal germline
B) The mutation is in a mitochondrial gene, and sperm mitochondria are selectively eliminated after fertilization
C) The mutation is in a Y-linked gene that is only expressed in males
D) The mutation is in an autosomal gene with incomplete penetrance and sex-limited expression
Answer: B
Rationale: Mitochondrial DNA is maternally inherited because sperm mitochondria are degraded post-fertilization.
A heteroplasmic mitochondrial mutation thus only passes from mother to all offspring. Option A describes
X-linked or imprinted inheritance, not mitochondrial. Option C is Y-linked, which would show male-to-male
transmission. Option D is autosomal, which would show transmission from both parents.
3 A laboratory studies the role of autophagy in cancer. They engineer cells to express a mutant form of ATG5 that
cannot be cleaved by calpain, preventing the conversion of ATG5 to a truncated form required for
autophagosome elongation. Compared to wild-type cells, which of the following is most likely to be observed in
the mutant cells under nutrient starvation?
A) Increased LC3-II levels and accumulation of autophagic vesicles
B) Decreased LC3-II levels and impaired autophagic flux
C) Normal autophagic flux but reduced mitophagy
D) Enhanced autophagosome-lysosome fusion and increased degradation
Answer: B
Rationale: Calpain cleavage of ATG5 produces a truncated form essential for autophagosome elongation. Without
this cleavage, LC3-II conjugation is impaired, leading to decreased LC3-II and blocked autophagic flux. Option A
is incorrect because LC3-II would not accumulate; it would be lower. Option C is incorrect because general
autophagy is affected. Option D is incorrect because fusion is not enhanced.
,4 A researcher performs chromatin immunoprecipitation followed by sequencing (ChIP-seq) using an antibody
against histone H3 trimethylated at lysine 4 (H3K4me3) in embryonic stem cells. Which of the following
genomic features is most likely to be enriched in the immunoprecipitated DNA?
A) Heterochromatic regions and centromeres
B) Promoters of actively transcribed genes
C) Enhancers marked by H3K27ac
D) Telomeric repeats and subtelomeric regions
Answer: B
Rationale: H3K4me3 is a hallmark of active promoters, particularly at transcription start sites. It correlates with
gene activation. Option A is incorrect because heterochromatin is marked by H3K9me3 or H3K27me3. Option C is
incorrect because enhancers are typically marked by H3K4me1 and H3K27ac. Option D is incorrect because
telomeres are associated with H3K9me3 and other repressive marks.
5 In a genetic screen for modifiers of the Ras-MAPK pathway in Drosophila, you isolate a mutation that
suppresses the rough eye phenotype caused by an activated Ras allele. The mutation maps to a gene encoding a
protein with a conserved PH domain. Which of the following is the most likely function of the wild-type
protein?
A) Binding to phosphoinositides at the plasma membrane to recruit downstream effectors
B) Acting as a GTPase-activating protein (GAP) for Ras
C) Functioning as a guanine nucleotide exchange factor (GEF) for Ras
D) Serving as a scaffold protein for the MAPK cascade
Answer: A
Rationale: PH domains bind phosphoinositides, often targeting proteins to membranes. In the Ras-MAPK pathway,
membrane recruitment of effectors like Raf is critical. A mutation in a PH domain protein that suppresses activated
Ras likely disrupts downstream signaling. Option B (GAP) would suppress by inactivating Ras, but activated Ras is
resistant to GAPs. Option C (GEF) would activate Ras, not suppress. Option D (scaffold) could suppress but is less
directly linked to PH domain function.
6 A patient with a family history of early-onset breast cancer undergoes genetic testing and is found to have a
heterozygous nonsense mutation in the BRCA1 gene. Which of the following best describes the molecular
consequence of this mutation at the cellular level?
A) Production of a truncated BRCA1 protein that retains partial function due to alternative translation initiation
B) Loss of heterozygosity in tumor cells leading to complete absence of functional BRCA1
C) Haploinsufficiency causing reduced homologous recombination repair efficiency in all cells
D) Dominant-negative effect of the mutant protein that inhibits wild-type BRCA1 function
Answer: B
Rationale: BRCA1 is a tumor suppressor that follows the two-hit hypothesis. A heterozygous germline mutation
predisposes to cancer, but tumorigenesis typically requires loss of the wild-type allele (loss of heterozygosity) in
somatic cells. Option A is incorrect because nonsense mutations usually cause nonsense-mediated decay or
truncation with loss of function. Option C is incorrect because haploinsufficiency is not the primary mechanism;
LOH is. Option D is incorrect because BRCA1 mutations are recessive at the cellular level.
7 A researcher investigates the role of microRNAs (miRNAs) in regulating the expression of a target gene X.
They transfect cells with a synthetic miRNA mimic that perfectly base-pairs with the 3' UTR of gene X mRNA.
Which of the following is the most likely direct effect on gene X expression?
A) Increased translation due to stabilization of the mRNA
B) mRNA cleavage and degradation via the RNA-induced silencing complex (RISC)
,C) Translational repression without mRNA degradation
D) Enhanced nuclear export of the mRNA
Answer: B
Rationale: Perfect complementarity between miRNA and target mRNA typically leads to Ago2-mediated
endonucleolytic cleavage and mRNA degradation. In contrast, imperfect pairing often causes translational
repression. Option A is incorrect because miRNAs generally repress expression. Option C is more common for
imperfect matches. Option D is not a known miRNA effect.
8 In a genetic mapping experiment, you cross two strains of yeast that differ in two linked genes: HIS4 and LEU2.
You obtain 1000 tetrads and classify them as parental ditype (PD), nonparental ditype (NPD), and tetratype (T).
The numbers are: PD= 450, NPD= 50, T= 500. What is the map distance between HIS4 and LEU2?
A) 10 cM
B) 20 cM
C) 30 cM
D) 40 cM
Answer: C
Rationale: Map distance in cM = (NPD + 1/2 T) / total tetrads × 100. Here, NPD=50, T=500, total=1000. So
distance = (50 + 250)/1000 × 100 = 300/1000 × 100 = 30 cM. Option A (10 cM) is too low, option B (20 cM) is
incorrect, option D (40 cM) is too high.
9 A researcher is studying a signaling pathway where ligand binding to a receptor tyrosine kinase (RTK) leads to
activation of PI3K and subsequent production of PIP3. Which of the following experimental observations would
most strongly indicate that the pathway is negatively regulated by the phosphatase PTEN?
A) PTEN knockout cells show decreased AKT phosphorylation compared to wild-type
B) Overexpression of PTEN increases PIP3 levels and enhances cell proliferation
C) PTEN knockdown results in increased PIP3 levels and AKT activation
D) PTEN localizes to the nucleus and dephosphorylates AKT directly
Answer: C
Rationale: PTEN dephosphorylates PIP3 to PIP2, negatively regulating PI3K signaling. Knockdown of PTEN
should increase PIP3 and activate AKT. Option A is incorrect because PTEN loss would increase AKT
phosphorylation, not decrease. Option B is incorrect because PTEN overexpression would decrease PIP3. Option D
is incorrect because PTEN acts on PIP3, not AKT.
10 Consider a hypothetical eukaryote with a genome of 3 × 10^9 base pairs. If replication occurs at a rate of 50
nucleotides per second per replication fork, and there are 10,000 origins of replication that fire simultaneously,
how long (in minutes) would it take to replicate the entire genome, assuming bidirectional replication from
each origin?
A) 50 minutes
B) 100 minutes
C) 250 minutes
D) 500 minutes
Answer: A
Rationale: Each origin initiates two forks, so total forks = 2 × 10,000 = 20,000. Each fork replicates at 50 nt/s, so
total rate = 20,000 × 50 = 1,000,000 nt/s. Genome size = 3 × 10^9 nt. Time = 3 × 10^ × 10^6 = 3000 s = 50
minutes. Option B (100 min) is double, option C (250 min) is five times, option D (500 min) is ten times the
correct answer.
, 11 In a study of cell cycle regulation, a novel compound is found to inhibit the activity of the anaphase-promoting
complex/cyclosome (APC/C). Which of the following molecular events would be most directly blocked by this
inhibition?
A) Phosphorylation of cohesin by Plk1
B) Ubiquitination of securin and cyclin B
C) Cleavage of cohesin by separase
D) Activation of Cdk1 by cyclin B binding
Answer: B
Rationale: APC/C is an E3 ubiquitin ligase that targets securin and cyclin B for degradation, allowing separase
activation and mitotic exit. Inhibiting APC/C prevents ubiquitination, blocking anaphase onset and cyclin B
degradation. Phosphorylation of cohesin by Plk1 is not directly APC/C-dependent, and separase cleavage requires
securin degradation (APC/C-mediated). Cdk1 activation by cyclin B occurs earlier in mitosis.
12 A researcher introduces a point mutation in the gene encoding the catalytic subunit of DNA polymerase III in
E. coli. The mutation eliminates the 3'->5' exonuclease activity but leaves the 5'->3' polymerase activity intact.
Which of the following outcomes is most likely to be observed in a culture of these mutant cells?
A) Increased frequency of frameshift mutations due to slippage
B) Accumulation of Okazaki fragments with RNA primers still attached
C) Elevated rate of base substitution mutations during replication
D) Stalled replication forks due to inability to remove mismatches
Answer: C
Rationale: The 3'!’5' exonuclease activity of DNA polymerase III is the proofreading function that removes
mismatched bases. Without it, misincorporated nucleotides persist, leading to a higher rate of base substitution
mutations. Frameshift mutations are more associated with slippage, not proofreading loss. Okazaki fragment
processing involves RNase H and DNA ligase, not the proofreading subunit. Stalled forks from mismatches are less
likely; the polymerase can continue with errors.
13 A researcher is studying a eukaryotic cell line that has a mutation in the gene encoding the large ribosomal
subunit protein L10. The mutation disrupts the interaction between L10 and the 5.8S rRNA. Which of the
following processes would be most directly impaired?
A) Initiation of translation at the start codon
B) Peptidyl transferase activity of the ribosome
C) Assembly of the 60S ribosomal subunit
D) Binding of tRNA to the A site
Answer: C
Rationale: Ribosomal protein L10 is a component of the 60S large subunit and is involved in its assembly,
particularly in stabilizing 5.8S rRNA incorporation. Disruption of L10-5.8S interaction impairs 60S subunit
biogenesis, reducing the pool of functional ribosomes. Initiation, peptidyl transferase (catalyzed by 23S/28S
rRNA), and A-site binding are not directly dependent on L10.
14 In a genetic screen, you identify a mutant yeast strain that fails to repress transcription of the GAL genes in the
absence of galactose. The mutation is in a gene encoding a protein that binds to the UASg element. Which of
the following proteins is most likely mutated?
A) Gal3p
B) Gal80p
C) Gal4p
D) Mig1p
Real Exam Questions with Correct Detailed Answers (Most
Recent) - 150 Questions
Section 1: Cell Biology and Genetics (Questions 1-15)
1 A researcher treats cultured human fibroblasts with a specific inhibitor that blocks the interaction between the
anaphase-promoting complex/cyclosome (APC/C) and its activator Cdc20. Which of the following immediate
consequences is most likely to be observed in the treated cells?
A) Premature sister chromatid separation due to loss of securin degradation
B) Arrest in metaphase due to stabilization of securin and cyclin B
C) Exit from mitosis without chromosome segregation due to cyclin B degradation
D) Activation of the spindle assembly checkpoint, causing anaphase onset delay
Answer: B
Rationale: APC/C-Cdc20 targets securin and cyclin B for degradation to trigger anaphase and mitotic exit.
Inhibiting their interaction stabilizes both proteins, leading to metaphase arrest. Option A is incorrect because
securin degradation is blocked, not promoted. Option C is incorrect because cyclin B is not degraded. Option D is
incorrect because the spindle checkpoint is not directly activated by APC/C inhibition.
2 In a study of a rare genetic disorder, a family pedigree reveals that affected individuals inherit the disease from
their mothers, but never from their fathers. Sequencing of the mitochondrial genome shows a heteroplasmic
mutation in the MT-ND5 gene. Which of the following best explains the observed inheritance pattern?
A) The mutation is in a nuclear gene that escapes X-inactivation and is imprinted in paternal germline
B) The mutation is in a mitochondrial gene, and sperm mitochondria are selectively eliminated after fertilization
C) The mutation is in a Y-linked gene that is only expressed in males
D) The mutation is in an autosomal gene with incomplete penetrance and sex-limited expression
Answer: B
Rationale: Mitochondrial DNA is maternally inherited because sperm mitochondria are degraded post-fertilization.
A heteroplasmic mitochondrial mutation thus only passes from mother to all offspring. Option A describes
X-linked or imprinted inheritance, not mitochondrial. Option C is Y-linked, which would show male-to-male
transmission. Option D is autosomal, which would show transmission from both parents.
3 A laboratory studies the role of autophagy in cancer. They engineer cells to express a mutant form of ATG5 that
cannot be cleaved by calpain, preventing the conversion of ATG5 to a truncated form required for
autophagosome elongation. Compared to wild-type cells, which of the following is most likely to be observed in
the mutant cells under nutrient starvation?
A) Increased LC3-II levels and accumulation of autophagic vesicles
B) Decreased LC3-II levels and impaired autophagic flux
C) Normal autophagic flux but reduced mitophagy
D) Enhanced autophagosome-lysosome fusion and increased degradation
Answer: B
Rationale: Calpain cleavage of ATG5 produces a truncated form essential for autophagosome elongation. Without
this cleavage, LC3-II conjugation is impaired, leading to decreased LC3-II and blocked autophagic flux. Option A
is incorrect because LC3-II would not accumulate; it would be lower. Option C is incorrect because general
autophagy is affected. Option D is incorrect because fusion is not enhanced.
,4 A researcher performs chromatin immunoprecipitation followed by sequencing (ChIP-seq) using an antibody
against histone H3 trimethylated at lysine 4 (H3K4me3) in embryonic stem cells. Which of the following
genomic features is most likely to be enriched in the immunoprecipitated DNA?
A) Heterochromatic regions and centromeres
B) Promoters of actively transcribed genes
C) Enhancers marked by H3K27ac
D) Telomeric repeats and subtelomeric regions
Answer: B
Rationale: H3K4me3 is a hallmark of active promoters, particularly at transcription start sites. It correlates with
gene activation. Option A is incorrect because heterochromatin is marked by H3K9me3 or H3K27me3. Option C is
incorrect because enhancers are typically marked by H3K4me1 and H3K27ac. Option D is incorrect because
telomeres are associated with H3K9me3 and other repressive marks.
5 In a genetic screen for modifiers of the Ras-MAPK pathway in Drosophila, you isolate a mutation that
suppresses the rough eye phenotype caused by an activated Ras allele. The mutation maps to a gene encoding a
protein with a conserved PH domain. Which of the following is the most likely function of the wild-type
protein?
A) Binding to phosphoinositides at the plasma membrane to recruit downstream effectors
B) Acting as a GTPase-activating protein (GAP) for Ras
C) Functioning as a guanine nucleotide exchange factor (GEF) for Ras
D) Serving as a scaffold protein for the MAPK cascade
Answer: A
Rationale: PH domains bind phosphoinositides, often targeting proteins to membranes. In the Ras-MAPK pathway,
membrane recruitment of effectors like Raf is critical. A mutation in a PH domain protein that suppresses activated
Ras likely disrupts downstream signaling. Option B (GAP) would suppress by inactivating Ras, but activated Ras is
resistant to GAPs. Option C (GEF) would activate Ras, not suppress. Option D (scaffold) could suppress but is less
directly linked to PH domain function.
6 A patient with a family history of early-onset breast cancer undergoes genetic testing and is found to have a
heterozygous nonsense mutation in the BRCA1 gene. Which of the following best describes the molecular
consequence of this mutation at the cellular level?
A) Production of a truncated BRCA1 protein that retains partial function due to alternative translation initiation
B) Loss of heterozygosity in tumor cells leading to complete absence of functional BRCA1
C) Haploinsufficiency causing reduced homologous recombination repair efficiency in all cells
D) Dominant-negative effect of the mutant protein that inhibits wild-type BRCA1 function
Answer: B
Rationale: BRCA1 is a tumor suppressor that follows the two-hit hypothesis. A heterozygous germline mutation
predisposes to cancer, but tumorigenesis typically requires loss of the wild-type allele (loss of heterozygosity) in
somatic cells. Option A is incorrect because nonsense mutations usually cause nonsense-mediated decay or
truncation with loss of function. Option C is incorrect because haploinsufficiency is not the primary mechanism;
LOH is. Option D is incorrect because BRCA1 mutations are recessive at the cellular level.
7 A researcher investigates the role of microRNAs (miRNAs) in regulating the expression of a target gene X.
They transfect cells with a synthetic miRNA mimic that perfectly base-pairs with the 3' UTR of gene X mRNA.
Which of the following is the most likely direct effect on gene X expression?
A) Increased translation due to stabilization of the mRNA
B) mRNA cleavage and degradation via the RNA-induced silencing complex (RISC)
,C) Translational repression without mRNA degradation
D) Enhanced nuclear export of the mRNA
Answer: B
Rationale: Perfect complementarity between miRNA and target mRNA typically leads to Ago2-mediated
endonucleolytic cleavage and mRNA degradation. In contrast, imperfect pairing often causes translational
repression. Option A is incorrect because miRNAs generally repress expression. Option C is more common for
imperfect matches. Option D is not a known miRNA effect.
8 In a genetic mapping experiment, you cross two strains of yeast that differ in two linked genes: HIS4 and LEU2.
You obtain 1000 tetrads and classify them as parental ditype (PD), nonparental ditype (NPD), and tetratype (T).
The numbers are: PD= 450, NPD= 50, T= 500. What is the map distance between HIS4 and LEU2?
A) 10 cM
B) 20 cM
C) 30 cM
D) 40 cM
Answer: C
Rationale: Map distance in cM = (NPD + 1/2 T) / total tetrads × 100. Here, NPD=50, T=500, total=1000. So
distance = (50 + 250)/1000 × 100 = 300/1000 × 100 = 30 cM. Option A (10 cM) is too low, option B (20 cM) is
incorrect, option D (40 cM) is too high.
9 A researcher is studying a signaling pathway where ligand binding to a receptor tyrosine kinase (RTK) leads to
activation of PI3K and subsequent production of PIP3. Which of the following experimental observations would
most strongly indicate that the pathway is negatively regulated by the phosphatase PTEN?
A) PTEN knockout cells show decreased AKT phosphorylation compared to wild-type
B) Overexpression of PTEN increases PIP3 levels and enhances cell proliferation
C) PTEN knockdown results in increased PIP3 levels and AKT activation
D) PTEN localizes to the nucleus and dephosphorylates AKT directly
Answer: C
Rationale: PTEN dephosphorylates PIP3 to PIP2, negatively regulating PI3K signaling. Knockdown of PTEN
should increase PIP3 and activate AKT. Option A is incorrect because PTEN loss would increase AKT
phosphorylation, not decrease. Option B is incorrect because PTEN overexpression would decrease PIP3. Option D
is incorrect because PTEN acts on PIP3, not AKT.
10 Consider a hypothetical eukaryote with a genome of 3 × 10^9 base pairs. If replication occurs at a rate of 50
nucleotides per second per replication fork, and there are 10,000 origins of replication that fire simultaneously,
how long (in minutes) would it take to replicate the entire genome, assuming bidirectional replication from
each origin?
A) 50 minutes
B) 100 minutes
C) 250 minutes
D) 500 minutes
Answer: A
Rationale: Each origin initiates two forks, so total forks = 2 × 10,000 = 20,000. Each fork replicates at 50 nt/s, so
total rate = 20,000 × 50 = 1,000,000 nt/s. Genome size = 3 × 10^9 nt. Time = 3 × 10^ × 10^6 = 3000 s = 50
minutes. Option B (100 min) is double, option C (250 min) is five times, option D (500 min) is ten times the
correct answer.
, 11 In a study of cell cycle regulation, a novel compound is found to inhibit the activity of the anaphase-promoting
complex/cyclosome (APC/C). Which of the following molecular events would be most directly blocked by this
inhibition?
A) Phosphorylation of cohesin by Plk1
B) Ubiquitination of securin and cyclin B
C) Cleavage of cohesin by separase
D) Activation of Cdk1 by cyclin B binding
Answer: B
Rationale: APC/C is an E3 ubiquitin ligase that targets securin and cyclin B for degradation, allowing separase
activation and mitotic exit. Inhibiting APC/C prevents ubiquitination, blocking anaphase onset and cyclin B
degradation. Phosphorylation of cohesin by Plk1 is not directly APC/C-dependent, and separase cleavage requires
securin degradation (APC/C-mediated). Cdk1 activation by cyclin B occurs earlier in mitosis.
12 A researcher introduces a point mutation in the gene encoding the catalytic subunit of DNA polymerase III in
E. coli. The mutation eliminates the 3'->5' exonuclease activity but leaves the 5'->3' polymerase activity intact.
Which of the following outcomes is most likely to be observed in a culture of these mutant cells?
A) Increased frequency of frameshift mutations due to slippage
B) Accumulation of Okazaki fragments with RNA primers still attached
C) Elevated rate of base substitution mutations during replication
D) Stalled replication forks due to inability to remove mismatches
Answer: C
Rationale: The 3'!’5' exonuclease activity of DNA polymerase III is the proofreading function that removes
mismatched bases. Without it, misincorporated nucleotides persist, leading to a higher rate of base substitution
mutations. Frameshift mutations are more associated with slippage, not proofreading loss. Okazaki fragment
processing involves RNase H and DNA ligase, not the proofreading subunit. Stalled forks from mismatches are less
likely; the polymerase can continue with errors.
13 A researcher is studying a eukaryotic cell line that has a mutation in the gene encoding the large ribosomal
subunit protein L10. The mutation disrupts the interaction between L10 and the 5.8S rRNA. Which of the
following processes would be most directly impaired?
A) Initiation of translation at the start codon
B) Peptidyl transferase activity of the ribosome
C) Assembly of the 60S ribosomal subunit
D) Binding of tRNA to the A site
Answer: C
Rationale: Ribosomal protein L10 is a component of the 60S large subunit and is involved in its assembly,
particularly in stabilizing 5.8S rRNA incorporation. Disruption of L10-5.8S interaction impairs 60S subunit
biogenesis, reducing the pool of functional ribosomes. Initiation, peptidyl transferase (catalyzed by 23S/28S
rRNA), and A-site binding are not directly dependent on L10.
14 In a genetic screen, you identify a mutant yeast strain that fails to repress transcription of the GAL genes in the
absence of galactose. The mutation is in a gene encoding a protein that binds to the UASg element. Which of
the following proteins is most likely mutated?
A) Gal3p
B) Gal80p
C) Gal4p
D) Mig1p
