BIOL 3301 Dauwalder | Meisel Exam 3 |
Full Questions and Answers | 2026/27
Updated | 100% Correct - UH.
1. A Drosophila embryo lacking the maternal factor Nanos is expected to show which
molecular defect in the posterior pole plasm?
A. Ectopic translation of hunchback mRNA leading to anterior-like fates
B. Failure to localize bicoid mRNA to the anterior cortex
C. Loss of germ cell formation due to defective pole cell specification
D. Premature degradation of oskar mRNA before localization
Answer: A. Ectopic translation of hunchback mRNA leading to anterior-like fates
Rationale: Nanos protein represses translation of maternal hunchback mRNA in the
posterior; without Nanos, Hunchback protein is ectopically expressed posteriorly, causing
anterior-like patterning. Bicoid localization is Nanos-independent, and pole cell formation
primarily requires Oskar and germ plasm components, not Nanos alone. Oskar mRNA
degradation is not a primary Nanos function.
2. In a yeast two-hybrid assay, a transcription factor's activation domain is fused to
Protein X and its DNA-binding domain to Protein Y. A positive signal indicates that:
A. Protein X and Protein Y directly interact in vivo
B. Protein X binds DNA upstream of the reporter gene
C. Protein Y activates transcription independently
D. Protein X and Protein Y are both nuclear-localized
Answer: A. Protein X and Protein Y directly interact in vivo
Rationale: The two-hybrid system reconstitutes a functional transcription factor only when
the bait and prey proteins physically interact, bringing the activation domain to the
reporter promoter. DNA binding is mediated by the DB-Y fusion, not X, and nuclear
localization is necessary but not sufficient for signal. Independent activation by Y would
occur without X, contradicting the assay design.
Page 1 of 75
,3. A female heterozygous for an X-linked recessive allele expresses the mutant
phenotype in patches of skin. Which molecular mechanism best explains this mosaic
expression?
A. Genomic imprinting of the maternal X allele
B. Random X-chromosome inactivation during early embryogenesis
C. Somatic recombination between X homologs
D. Transcriptional silencing by a dominant-negative mutation
Answer: B. Random X-chromosome inactivation during early embryogenesis
Rationale: Random X-inactivation in female mammals silences one X per cell, creating
mosaic expression of X-linked genes. Imprinting affects specific loci, not whole X
mosaicism, and somatic recombination is rare. A dominant-negative would affect all cells
uniformly, not produce patches.
4. ChIP-seq for RNA polymerase II shows a strong signal across the entire coding
region of Gene Z but no signal at its promoter. What is the most likely explanation?
A. Gene Z is transcribed by RNA polymerase I
B. The promoter is occupied by a paused polymerase that dissociates during crosslinking
C. Gene Z is a non-coding RNA transcribed by RNA polymerase III
D. The antibody used detects only the elongating form of RNA polymerase II
Answer: D. The antibody used detects only the elongating form of RNA polymerase II
Rationale: Antibodies specific to phosphorylated CTD (e.g., Ser2) detect elongating Pol II,
which is present in gene bodies but not at promoters. Pol I and Pol III transcribe different
RNA classes, and a paused polymerase would still crosslink at the promoter. Thus, the
antibody specificity explains the pattern.
Page 2 of 75
,5. A mutation in the insulator protein CTCF abolishes the boundary between two
topologically associating domains (TADs). What is the most likely consequence?
A. Enhancers from one TAD can inappropriately activate genes in the adjacent TAD
B. Global loss of histone acetylation across both TADs
C. Premature termination of transcription at the TAD boundary
D. Failure of cohesin loading onto chromatin
Answer: A. Enhancers from one TAD can inappropriately activate genes in the
adjacent TAD
Rationale: CTCF insulators block enhancer-promoter interactions between adjacent TADs;
losing CTCF allows ectopic enhancer activity. Histone acetylation and transcription
termination are not directly controlled by TAD boundaries, and cohesin loading occurs
independently of CTCF.
6. In a reporter assay, deleting a 200-bp region 10 kb upstream of a gene abolishes
expression, but the region lacks promoter or enhancer chromatin marks. Which
element is most likely affected?
A. A silencer
B. A locus control region (LCR)
C. A promoter-proximal element
D. A polyadenylation signal
Answer: B. A locus control region (LCR)
Rationale: LCRs are distal regulatory elements that can be located far upstream and are
required for high-level, copy-number-dependent expression, often without classic enhancer
marks. Silencers repress, promoters are proximal, and polyadenylation signals are at the 3'
end.
Page 3 of 75
, 7. A drosophila genetic screen identifies a maternal-effect mutation that causes
embryos to lack pole cells. Which gene is most likely mutated?
A. bicoid
B. nanos
C. oskar
D. torso
Answer: C. oskar
Rationale: Oskar is essential for pole plasm assembly and germ cell formation; oskar
mutants lack pole cells. Bicoid and Nanos pattern the anterior-posterior axis but are not
required for pole cell formation, and torso controls terminal structures.
8. In a mammalian cell, a transgene integrated near a heterochromatic region shows
variegated expression. Treatment with a histone deacetylase inhibitor (HDACi) is
expected to:
A. Increase the proportion of cells expressing the transgene
B. Decrease the proportion of cells expressing the transgene
C. Have no effect on transgene expression
D. Cause complete silencing of the transgene in all cells
Answer: A. Increase the proportion of cells expressing the transgene
Rationale: HDAC inhibitors promote histone acetylation and counteract heterochromatin
spreading, increasing the likelihood of transgene expression. They do not enhance
silencing and are unlikely to have no effect given the heterochromatic context.
Page 4 of 75
Full Questions and Answers | 2026/27
Updated | 100% Correct - UH.
1. A Drosophila embryo lacking the maternal factor Nanos is expected to show which
molecular defect in the posterior pole plasm?
A. Ectopic translation of hunchback mRNA leading to anterior-like fates
B. Failure to localize bicoid mRNA to the anterior cortex
C. Loss of germ cell formation due to defective pole cell specification
D. Premature degradation of oskar mRNA before localization
Answer: A. Ectopic translation of hunchback mRNA leading to anterior-like fates
Rationale: Nanos protein represses translation of maternal hunchback mRNA in the
posterior; without Nanos, Hunchback protein is ectopically expressed posteriorly, causing
anterior-like patterning. Bicoid localization is Nanos-independent, and pole cell formation
primarily requires Oskar and germ plasm components, not Nanos alone. Oskar mRNA
degradation is not a primary Nanos function.
2. In a yeast two-hybrid assay, a transcription factor's activation domain is fused to
Protein X and its DNA-binding domain to Protein Y. A positive signal indicates that:
A. Protein X and Protein Y directly interact in vivo
B. Protein X binds DNA upstream of the reporter gene
C. Protein Y activates transcription independently
D. Protein X and Protein Y are both nuclear-localized
Answer: A. Protein X and Protein Y directly interact in vivo
Rationale: The two-hybrid system reconstitutes a functional transcription factor only when
the bait and prey proteins physically interact, bringing the activation domain to the
reporter promoter. DNA binding is mediated by the DB-Y fusion, not X, and nuclear
localization is necessary but not sufficient for signal. Independent activation by Y would
occur without X, contradicting the assay design.
Page 1 of 75
,3. A female heterozygous for an X-linked recessive allele expresses the mutant
phenotype in patches of skin. Which molecular mechanism best explains this mosaic
expression?
A. Genomic imprinting of the maternal X allele
B. Random X-chromosome inactivation during early embryogenesis
C. Somatic recombination between X homologs
D. Transcriptional silencing by a dominant-negative mutation
Answer: B. Random X-chromosome inactivation during early embryogenesis
Rationale: Random X-inactivation in female mammals silences one X per cell, creating
mosaic expression of X-linked genes. Imprinting affects specific loci, not whole X
mosaicism, and somatic recombination is rare. A dominant-negative would affect all cells
uniformly, not produce patches.
4. ChIP-seq for RNA polymerase II shows a strong signal across the entire coding
region of Gene Z but no signal at its promoter. What is the most likely explanation?
A. Gene Z is transcribed by RNA polymerase I
B. The promoter is occupied by a paused polymerase that dissociates during crosslinking
C. Gene Z is a non-coding RNA transcribed by RNA polymerase III
D. The antibody used detects only the elongating form of RNA polymerase II
Answer: D. The antibody used detects only the elongating form of RNA polymerase II
Rationale: Antibodies specific to phosphorylated CTD (e.g., Ser2) detect elongating Pol II,
which is present in gene bodies but not at promoters. Pol I and Pol III transcribe different
RNA classes, and a paused polymerase would still crosslink at the promoter. Thus, the
antibody specificity explains the pattern.
Page 2 of 75
,5. A mutation in the insulator protein CTCF abolishes the boundary between two
topologically associating domains (TADs). What is the most likely consequence?
A. Enhancers from one TAD can inappropriately activate genes in the adjacent TAD
B. Global loss of histone acetylation across both TADs
C. Premature termination of transcription at the TAD boundary
D. Failure of cohesin loading onto chromatin
Answer: A. Enhancers from one TAD can inappropriately activate genes in the
adjacent TAD
Rationale: CTCF insulators block enhancer-promoter interactions between adjacent TADs;
losing CTCF allows ectopic enhancer activity. Histone acetylation and transcription
termination are not directly controlled by TAD boundaries, and cohesin loading occurs
independently of CTCF.
6. In a reporter assay, deleting a 200-bp region 10 kb upstream of a gene abolishes
expression, but the region lacks promoter or enhancer chromatin marks. Which
element is most likely affected?
A. A silencer
B. A locus control region (LCR)
C. A promoter-proximal element
D. A polyadenylation signal
Answer: B. A locus control region (LCR)
Rationale: LCRs are distal regulatory elements that can be located far upstream and are
required for high-level, copy-number-dependent expression, often without classic enhancer
marks. Silencers repress, promoters are proximal, and polyadenylation signals are at the 3'
end.
Page 3 of 75
, 7. A drosophila genetic screen identifies a maternal-effect mutation that causes
embryos to lack pole cells. Which gene is most likely mutated?
A. bicoid
B. nanos
C. oskar
D. torso
Answer: C. oskar
Rationale: Oskar is essential for pole plasm assembly and germ cell formation; oskar
mutants lack pole cells. Bicoid and Nanos pattern the anterior-posterior axis but are not
required for pole cell formation, and torso controls terminal structures.
8. In a mammalian cell, a transgene integrated near a heterochromatic region shows
variegated expression. Treatment with a histone deacetylase inhibitor (HDACi) is
expected to:
A. Increase the proportion of cells expressing the transgene
B. Decrease the proportion of cells expressing the transgene
C. Have no effect on transgene expression
D. Cause complete silencing of the transgene in all cells
Answer: A. Increase the proportion of cells expressing the transgene
Rationale: HDAC inhibitors promote histone acetylation and counteract heterochromatin
spreading, increasing the likelihood of transgene expression. They do not enhance
silencing and are unlikely to have no effect given the heterochromatic context.
Page 4 of 75
