COMSAE PHASE 1 MOLECULAR BIOLOGY
PRACTICE EXAM WITH ACTUAL
QUESTIONS AND VERIFIED ANSWERS,
PLUS EXPLAINED RATIONALES/EXPERT
VERIFIED FOR GUARANTEED 100% PASS
2026/LATEST UPDATE/INSTANT
DOWNLOAD PDF
1.
A researcher isolates a newly discovered eukaryotic gene and determines
that its coding region contains several introns. Which molecular event is
primarily responsible for producing mature mRNA from the initial RNA
transcript?
A. DNA replication
B. RNA splicing
C. Translation initiation
D. Polyadenylation alone
Answer: B. RNA splicing
Rationale: Introns are removed from the primary RNA transcript by
the spliceosome, while the remaining exons are joined to form mature
mRNA. Although 5′ capping and 3′ polyadenylation also contribute to
mRNA maturation, they do not remove introns.
2.
A mutation changes the sequence of a promoter so that RNA polymerase
II can no longer efficiently bind. Which molecular process will be most
directly impaired?
1
,A. Translation of mature mRNA
B. DNA replication
C. Transcription initiation
D. Protein folding
Answer: C. Transcription initiation
Rationale: A promoter is a regulatory DNA sequence required for
recruitment and positioning of RNA polymerase and transcription
factors. A promoter mutation can markedly reduce or abolish
transcription initiation, resulting in decreased production of the
corresponding RNA and protein.
3.
A patient has a mutation in a gene that changes a codon from UAU to
UAA. What type of mutation has occurred?
A. Missense mutation
B. Silent mutation
C. Nonsense mutation
D. Frameshift mutation
Answer: C. Nonsense mutation
Rationale: UAU encodes tyrosine, whereas UAA is a stop codon.
Conversion of a codon specifying an amino acid into a premature
termination codon is a nonsense mutation and can produce a
truncated, usually dysfunctional protein.
4.
During DNA replication, one strand is synthesized continuously while
the other is synthesized discontinuously. Which enzyme joins the short
DNA fragments generated on the discontinuously synthesized strand?
2
,A. DNA ligase
B. DNA helicase
C. Primase
D. Topoisomerase
Answer: A. DNA ligase
Rationale: The lagging strand is synthesized as Okazaki fragments.
DNA ligase seals the phosphodiester backbone between adjacent
fragments after the RNA primers have been removed and replaced
with DNA.
5.
A mutation abolishes the activity of DNA helicase. Which step of DNA
replication would be directly disrupted?
A. Addition of RNA primers
B. Separation of parental DNA strands
C. Formation of peptide bonds
D. Removal of introns
Answer: B. Separation of parental DNA strands
Rationale: Helicase unwinds the parental double-stranded DNA by
disrupting hydrogen bonds between complementary bases. Without
helicase activity, the replication fork cannot effectively progress.
6.
A scientist performs PCR to amplify a specific segment of DNA. Which
component determines the boundaries of the DNA region that will be
amplified?
3
, A. Ribosomes
B. Primers
C. Introns
D. RNA polymerase
Answer: B. Primers
Rationale: PCR primers are short oligonucleotides complementary to
sequences flanking the target region. DNA polymerase extends these
primers, thereby defining the boundaries of the amplified fragment.
Primers
DNA polymerase
DNA building blocks
Setup
PCR ingredients are combined (25 °C).
Give feedback
7.
A researcher performs PCR but accidentally omits the DNA primers.
What is the most likely result?
A. Exponential amplification of the target sequence
B. Translation of the target sequence into protein
C. Failure of specific DNA amplification
D. Conversion of DNA into RNA
Answer: C. Failure of specific DNA amplification
Rationale: DNA polymerase requires a preexisting 3′-OH group from
which to extend DNA. PCR primers provide this starting point.
Without primers, exponential amplification of the target DNA cannot
occur.
4
PRACTICE EXAM WITH ACTUAL
QUESTIONS AND VERIFIED ANSWERS,
PLUS EXPLAINED RATIONALES/EXPERT
VERIFIED FOR GUARANTEED 100% PASS
2026/LATEST UPDATE/INSTANT
DOWNLOAD PDF
1.
A researcher isolates a newly discovered eukaryotic gene and determines
that its coding region contains several introns. Which molecular event is
primarily responsible for producing mature mRNA from the initial RNA
transcript?
A. DNA replication
B. RNA splicing
C. Translation initiation
D. Polyadenylation alone
Answer: B. RNA splicing
Rationale: Introns are removed from the primary RNA transcript by
the spliceosome, while the remaining exons are joined to form mature
mRNA. Although 5′ capping and 3′ polyadenylation also contribute to
mRNA maturation, they do not remove introns.
2.
A mutation changes the sequence of a promoter so that RNA polymerase
II can no longer efficiently bind. Which molecular process will be most
directly impaired?
1
,A. Translation of mature mRNA
B. DNA replication
C. Transcription initiation
D. Protein folding
Answer: C. Transcription initiation
Rationale: A promoter is a regulatory DNA sequence required for
recruitment and positioning of RNA polymerase and transcription
factors. A promoter mutation can markedly reduce or abolish
transcription initiation, resulting in decreased production of the
corresponding RNA and protein.
3.
A patient has a mutation in a gene that changes a codon from UAU to
UAA. What type of mutation has occurred?
A. Missense mutation
B. Silent mutation
C. Nonsense mutation
D. Frameshift mutation
Answer: C. Nonsense mutation
Rationale: UAU encodes tyrosine, whereas UAA is a stop codon.
Conversion of a codon specifying an amino acid into a premature
termination codon is a nonsense mutation and can produce a
truncated, usually dysfunctional protein.
4.
During DNA replication, one strand is synthesized continuously while
the other is synthesized discontinuously. Which enzyme joins the short
DNA fragments generated on the discontinuously synthesized strand?
2
,A. DNA ligase
B. DNA helicase
C. Primase
D. Topoisomerase
Answer: A. DNA ligase
Rationale: The lagging strand is synthesized as Okazaki fragments.
DNA ligase seals the phosphodiester backbone between adjacent
fragments after the RNA primers have been removed and replaced
with DNA.
5.
A mutation abolishes the activity of DNA helicase. Which step of DNA
replication would be directly disrupted?
A. Addition of RNA primers
B. Separation of parental DNA strands
C. Formation of peptide bonds
D. Removal of introns
Answer: B. Separation of parental DNA strands
Rationale: Helicase unwinds the parental double-stranded DNA by
disrupting hydrogen bonds between complementary bases. Without
helicase activity, the replication fork cannot effectively progress.
6.
A scientist performs PCR to amplify a specific segment of DNA. Which
component determines the boundaries of the DNA region that will be
amplified?
3
, A. Ribosomes
B. Primers
C. Introns
D. RNA polymerase
Answer: B. Primers
Rationale: PCR primers are short oligonucleotides complementary to
sequences flanking the target region. DNA polymerase extends these
primers, thereby defining the boundaries of the amplified fragment.
Primers
DNA polymerase
DNA building blocks
Setup
PCR ingredients are combined (25 °C).
Give feedback
7.
A researcher performs PCR but accidentally omits the DNA primers.
What is the most likely result?
A. Exponential amplification of the target sequence
B. Translation of the target sequence into protein
C. Failure of specific DNA amplification
D. Conversion of DNA into RNA
Answer: C. Failure of specific DNA amplification
Rationale: DNA polymerase requires a preexisting 3′-OH group from
which to extend DNA. PCR primers provide this starting point.
Without primers, exponential amplification of the target DNA cannot
occur.
4
