LIFESCI 7A MIDTERM 2 EXAM |
QUESTIONS AND ANSWERS | 2026/27
UPDATED | 100% CORRECT - UCLA.
73 Questions with Answers and Detailed Rationales
100 PERCENT GUARANTEED PASS
INSTANT DOWNLOAD ANSWERS INCLUDED
IMPORTANCE OF THIS DOCUMENT
This comprehensive examination preparation guide has been meticulously developed to help you succeed in the
LIFESCI 7A MIDTERM 2 EXAM | QUESTIONS AND ANSWERS | 2026/27 UPDATED | 100% CORRECT -
UCLA.. It contains 73 carefully selected questions that reflect the most current exam content and testing
strategies. Each question is accompanied by a correct answer and a detailed rationale that explains the
underlying pathophysiology, pharmacology, or clinical reasoning.
Self-Assessment – Test your knowledge and Exam Preparation – Familiarize yourself with the
identify areas requiring further question format and content
study areas
Concept Reinforcement – Deepen your Confidence Building – Develop test-taking
understanding through strategies and reduce
evidence-based exam anxiety
rationales
Time Management – Practice answering
questions under simulated
exam conditions
Review Summary 73 Questions
Foundations - Application - Lifesci 7a 2 AND 2026/27 Updated 100 Correct - UCLA LIFE Sciences /
Molecular Biology Undergraduate YEAR 2/3 Introductory LIFE Sciences FOR Majors
All answers with rationales
,Table of Contents
Content Area Questions Key Topics
Central Dogma DNA 1-13 Likely, Polymerase, Eukaryotic, Mutation, Subunit
Replication Transcription
AND Translation
GENE Regulation IN 14-26 Protein, Mutant, Potential, Genetic, Phenotype
Prokaryotes AND Eukaryotes
Mutation AND DNA Repair 27-39 Likely, Transcription, Protein, Factor, Studying
Mechanisms
CELL Cycle AND Mitosis 40-52 Effect, Concentration, Plasma, Explains, Reaction
Meiosis AND Sexual 53-65 Agonist, Therapeutic, Receptor, Concentration, Effect
Reproduction
Mendelian Genetics AND 66-73 Operon, Presence, Strand, Synthesized, Researcher
Inheritance Patterns
TOTAL 73 All questions include answers and detailed rationales
,Section A - Central Dogma DNA Replication Transcription
AND Translation
Q1.
A bacterial strain carries a mutation in the gene encoding the subunit of RNA polymerase
that eliminates its interaction with the C-terminal domain of the subunit. Which regulatory
mechanism is most directly impaired?
A. Promoter recognition by sigma factors B. Activation by CAP-cAMP at
catabolite-sensitive promoters
C. Rho-dependent transcription termination D. Stringent response mediated by ppGpp
Correct: B - Activation by CAP-cAMP at catabolite-sensitive promoters
Rationale:The ±-CTD is required for interaction with activator proteins like CAP. Loss of this
interaction prevents CAP-dependent activation, while sigma factors bind to the core enzyme
independently. Rho termination and ppGpp effects do not require -CTD.
Why the other answers are wrong:
A. Sigma factors bind to the core enzyme via conserved regions, not through -CTD.
C. Rho termination relies on Rho and RNA sequences, not -CTD.
D. ppGpp acts by altering RNA polymerase conformation, not via -CTD.
Reference: Alberts et al., Molecular Biology of the Cell, 7th ed., Ch. 6
Q2.
In a eukaryotic cell, a specific mRNA has a very short half-life due to AU-rich elements
(AREs) in its 3' UTR. Which mechanism is LEAST likely to increase the stability of this
mRNA?
A. Binding of HuR protein to the ARE B. Deadenylation by CCR4-NOT complex
C. Recruitment of the RNA-induced D. Loss of function mutation in the gene
silencing complex (RISC) by a microRNA encoding TTP (tristetraprolin)
Correct: B - Deadenylation by CCR4-NOT complex
Rationale:Deadenylation promotes mRNA decay, so its enhancement would decrease
stability. HuR binding stabilizes ARE-mRNAs, TTP promotes degradation (loss increases
stability), and RISC would typically degrade mRNA, but if it targets a different mRNA, it may
not affect this one; however, RISC recruitment would generally decrease stability, so it's least
likely to increase stability, but the question asks LEAST likely to increase, and deadenylation
clearly promotes decay.
Why the other answers are wrong:
A. HuR protects ARE-mRNAs from degradation, increasing stability.
Page 3
, Section A - Central Dogma DNA Replication Transcription AND Translation
C. RISC generally suppresses translation or degrades mRNA, not increase stability.
D. TTP destabilizes ARE-mRNAs; its loss would increase stability.
Reference: Alberts et al., Molecular Biology of the Cell, 7th ed., Ch. 7
Q3.
A researcher performs a chromatin immunoprecipitation (ChIP) experiment targeting
H3K27me3 in a gene promoter and observes high enrichment. However, the gene is
actively transcribed. Which explanation best reconciles this observation?
A. H3K27me3 is present only on the inactive B. The antibody cross-reacts with H3K27ac,
X chromosome, and the gene is on the which marks active enhancers.
active X.
C. The gene is subject to bivalent domains, D. The gene is silenced by Polycomb in
having both H3K4me3 and H3K27me3 in somatic cells, but transcription is from an
embryonic stem cells. alternative promoter lacking the mark.
Correct: C - The gene is subject to bivalent domains, having both H3K4me3 and
H3K27me3 in embryonic stem cells.
Rationale:Bivalent domains contain both H3K4me3 (activation) and H3K27me3 (repression)
and are poised for transcription. This allows gene to be transcribed despite the repressive
mark. The other options do not explain active transcription at the same promoter.
Why the other answers are wrong:
A. X-inactivation silencing would prevent transcription, not allow it.
B. Cross-reactivity would not be specific to active transcription.
D. If the gene is silenced, transcription would not occur; alternative promoter would not have
H3K27me3.
Reference: Alberts et al., Molecular Biology of the Cell, 7th ed., Ch. 7
Q4.
In a classic Meselson-Stahl experiment, E. coli was grown for many generations in 15N
medium, then shifted to 14N medium. DNA was extracted after one and two generations
and centrifuged to equilibrium in CsCl. If DNA replication were dispersive rather than
semiconservative, what would be the predicted banding pattern after the first and second
generations?
A. One hybrid band after the first generation B. One hybrid band after the first generation
and one light band after the second and two bands (hybrid and light) after the
second
C. One band of intermediate density after D. Two bands (heavy and light) after the first
the first generation and one band of generation and one light band after the
intermediate density after the second second
Correct: C - One band of intermediate density after the first generation and one band of
Page 4
QUESTIONS AND ANSWERS | 2026/27
UPDATED | 100% CORRECT - UCLA.
73 Questions with Answers and Detailed Rationales
100 PERCENT GUARANTEED PASS
INSTANT DOWNLOAD ANSWERS INCLUDED
IMPORTANCE OF THIS DOCUMENT
This comprehensive examination preparation guide has been meticulously developed to help you succeed in the
LIFESCI 7A MIDTERM 2 EXAM | QUESTIONS AND ANSWERS | 2026/27 UPDATED | 100% CORRECT -
UCLA.. It contains 73 carefully selected questions that reflect the most current exam content and testing
strategies. Each question is accompanied by a correct answer and a detailed rationale that explains the
underlying pathophysiology, pharmacology, or clinical reasoning.
Self-Assessment – Test your knowledge and Exam Preparation – Familiarize yourself with the
identify areas requiring further question format and content
study areas
Concept Reinforcement – Deepen your Confidence Building – Develop test-taking
understanding through strategies and reduce
evidence-based exam anxiety
rationales
Time Management – Practice answering
questions under simulated
exam conditions
Review Summary 73 Questions
Foundations - Application - Lifesci 7a 2 AND 2026/27 Updated 100 Correct - UCLA LIFE Sciences /
Molecular Biology Undergraduate YEAR 2/3 Introductory LIFE Sciences FOR Majors
All answers with rationales
,Table of Contents
Content Area Questions Key Topics
Central Dogma DNA 1-13 Likely, Polymerase, Eukaryotic, Mutation, Subunit
Replication Transcription
AND Translation
GENE Regulation IN 14-26 Protein, Mutant, Potential, Genetic, Phenotype
Prokaryotes AND Eukaryotes
Mutation AND DNA Repair 27-39 Likely, Transcription, Protein, Factor, Studying
Mechanisms
CELL Cycle AND Mitosis 40-52 Effect, Concentration, Plasma, Explains, Reaction
Meiosis AND Sexual 53-65 Agonist, Therapeutic, Receptor, Concentration, Effect
Reproduction
Mendelian Genetics AND 66-73 Operon, Presence, Strand, Synthesized, Researcher
Inheritance Patterns
TOTAL 73 All questions include answers and detailed rationales
,Section A - Central Dogma DNA Replication Transcription
AND Translation
Q1.
A bacterial strain carries a mutation in the gene encoding the subunit of RNA polymerase
that eliminates its interaction with the C-terminal domain of the subunit. Which regulatory
mechanism is most directly impaired?
A. Promoter recognition by sigma factors B. Activation by CAP-cAMP at
catabolite-sensitive promoters
C. Rho-dependent transcription termination D. Stringent response mediated by ppGpp
Correct: B - Activation by CAP-cAMP at catabolite-sensitive promoters
Rationale:The ±-CTD is required for interaction with activator proteins like CAP. Loss of this
interaction prevents CAP-dependent activation, while sigma factors bind to the core enzyme
independently. Rho termination and ppGpp effects do not require -CTD.
Why the other answers are wrong:
A. Sigma factors bind to the core enzyme via conserved regions, not through -CTD.
C. Rho termination relies on Rho and RNA sequences, not -CTD.
D. ppGpp acts by altering RNA polymerase conformation, not via -CTD.
Reference: Alberts et al., Molecular Biology of the Cell, 7th ed., Ch. 6
Q2.
In a eukaryotic cell, a specific mRNA has a very short half-life due to AU-rich elements
(AREs) in its 3' UTR. Which mechanism is LEAST likely to increase the stability of this
mRNA?
A. Binding of HuR protein to the ARE B. Deadenylation by CCR4-NOT complex
C. Recruitment of the RNA-induced D. Loss of function mutation in the gene
silencing complex (RISC) by a microRNA encoding TTP (tristetraprolin)
Correct: B - Deadenylation by CCR4-NOT complex
Rationale:Deadenylation promotes mRNA decay, so its enhancement would decrease
stability. HuR binding stabilizes ARE-mRNAs, TTP promotes degradation (loss increases
stability), and RISC would typically degrade mRNA, but if it targets a different mRNA, it may
not affect this one; however, RISC recruitment would generally decrease stability, so it's least
likely to increase stability, but the question asks LEAST likely to increase, and deadenylation
clearly promotes decay.
Why the other answers are wrong:
A. HuR protects ARE-mRNAs from degradation, increasing stability.
Page 3
, Section A - Central Dogma DNA Replication Transcription AND Translation
C. RISC generally suppresses translation or degrades mRNA, not increase stability.
D. TTP destabilizes ARE-mRNAs; its loss would increase stability.
Reference: Alberts et al., Molecular Biology of the Cell, 7th ed., Ch. 7
Q3.
A researcher performs a chromatin immunoprecipitation (ChIP) experiment targeting
H3K27me3 in a gene promoter and observes high enrichment. However, the gene is
actively transcribed. Which explanation best reconciles this observation?
A. H3K27me3 is present only on the inactive B. The antibody cross-reacts with H3K27ac,
X chromosome, and the gene is on the which marks active enhancers.
active X.
C. The gene is subject to bivalent domains, D. The gene is silenced by Polycomb in
having both H3K4me3 and H3K27me3 in somatic cells, but transcription is from an
embryonic stem cells. alternative promoter lacking the mark.
Correct: C - The gene is subject to bivalent domains, having both H3K4me3 and
H3K27me3 in embryonic stem cells.
Rationale:Bivalent domains contain both H3K4me3 (activation) and H3K27me3 (repression)
and are poised for transcription. This allows gene to be transcribed despite the repressive
mark. The other options do not explain active transcription at the same promoter.
Why the other answers are wrong:
A. X-inactivation silencing would prevent transcription, not allow it.
B. Cross-reactivity would not be specific to active transcription.
D. If the gene is silenced, transcription would not occur; alternative promoter would not have
H3K27me3.
Reference: Alberts et al., Molecular Biology of the Cell, 7th ed., Ch. 7
Q4.
In a classic Meselson-Stahl experiment, E. coli was grown for many generations in 15N
medium, then shifted to 14N medium. DNA was extracted after one and two generations
and centrifuged to equilibrium in CsCl. If DNA replication were dispersive rather than
semiconservative, what would be the predicted banding pattern after the first and second
generations?
A. One hybrid band after the first generation B. One hybrid band after the first generation
and one light band after the second and two bands (hybrid and light) after the
second
C. One band of intermediate density after D. Two bands (heavy and light) after the first
the first generation and one band of generation and one light band after the
intermediate density after the second second
Correct: C - One band of intermediate density after the first generation and one band of
Page 4
