BIOD 210 Module 1-7 Exam Genetics - Complete Question &
Answer Bank (2026/2027 Update) | Portage Learning | 100%
Verified Pass Guide - 174 Questions and Answers Already
Graded A+ Premium Exam Tested And Verified
Subject Area BIOD 210 Module 1-7 Exam Genetics - Complete Question & Answer Bank
(2026/2027 Update) | Portage Learning | 100% Verified Pass Guide
Description Comprehensive examination on BIOD 210 Module 1-7 Exam Genetics -
Complete Question & Answer Bank (2026/2027 Update) | Portage Learning |
100% Verified Pass Guide.
Expected Grade A+
Total Questions 174
Duration 3 hours
Learning Outcomes 1. Demonstrate mastery of core concepts
Accreditation Aligned with US university standards.
Page 1
,1. A mutation in the 3'->5' exonuclease domain of DNA polymerase III results
in a 100-fold increase in replication errors. Which repair pathway is most
directly compromised?
A. Mismatch repair
B. Base excision repair
C. Proofreading
D. Nucleotide excision repair
Answer: C. Proofreading
The 3'->5' exonuclease activity is responsible for proofreading during DNA
replication. Loss of this activity increases misincorporation rates, which are later
corrected by mismatch repair, but the immediate defect is proofreading. Base
excision repair deals with damaged bases, and nucleotide excision repair removes
bulky lesions.
2. A pre-mRNA from a gene with four exons undergoes alternative splicing to
produce two isoforms: one including exon 3 (isoform A) and one skipping exon
3 (isoform B). A mutation in the branch point sequence upstream of exon 3
leads to preferential inclusion of exon 3. What is the most likely effect on
splicing regulation?
A. Loss of U2AF binding reduces recognition of the 3' splice site, favoring exon
skipping
B. Disruption of the spliceosome assembly at the intron upstream of exon 3, causing
inclusion of exon 3 due to exon definition
C. Strengthening of the 5' splice site of exon 3 by the mutation, promoting inclusion
D. The mutation creates a new silencer element that represses exon 3 skipping
Answer: B. Disruption of the spliceosome assembly at the intron upstream of
exon 3, causing inclusion of exon 3 due to exon definition
The branch point is critical for lariat formation and spliceosome assembly. A
mutation in the branch point upstream of exon 3 impairs intron removal. In exon
definition, failure to splice the upstream intron can lead to inclusion of the
downstream exon because the splicing machinery recognizes the exon as a unit.
Option A is incorrect because branch point mutations do not directly affect U2AF
binding; U2AF binds the polypyrimidine tract. Option C is incorrect because
branch point mutations do not affect the 5' splice site. Option D is unlikely;
creating a silencer would promote skipping, not inclusion.
Page 2
,3. A tRNA with the anticodon 5'-IIU-3' (I = inosine) is expected to recognize
which of the following sets of codons?
A. 5'-AAU-3' and 5'-AAC-3'
B. 5'-UUA-3', 5'-UUG-3', and 5'-CUA-3'
C. 5'-AAU-3', 5'-AAC-3', and 5'-AAA-3'
D. 5'-UAA-3' and 5'-UAG-3'
Answer: C. 5'-AAU-3', 5'-AAC-3', and 5'-AAA-3'
Inosine can pair with U, C, or A by wobble rules. The anticodon IIU (reading
5'->3' anticodon) pairs with 3'->5' codon. Actually, anticodon is 5'-IIU-3'; the
corresponding codon is 5'-AAU-3' (since IIU pairs with AAU? Let's follow
standard: anticodon IIU (5'->3') binds to codon 3'->5' complementary: anticodon
bases: I I U; codon bases: C C A? No, careful: anticodon is written 5' to 3',
complementary antiparallel to codon. So anticodon 5'-IIU-3' pairs with codon
3'-YYA-5', which in 5'->3' is 5'-AYY-3'? That's confusing. Better: tRNA
anticodon I I U will bind to codons where the first base of codon (5' end) pairs
with I. Inosine can pair with A, U, or C. Second base of codon pairs with I (again
inosine can pair with A, U, C). Third base of codon pairs with U (so
complementary: A). So codons: first base: A, U, or C; second base: A, U, or C;
third base: A. That gives 3x3=9 possible codons, but only options include three
codons. Option C: AAU (first A, second A, third U) - third U pairs with U? No,
anticodon third base U pairs with A on codon. So codon third base must be A.
Thus AAU would pair? Actually, anticodon I I U: first I pairs with A, U, C;
second I pairs with A, U, C; third U pairs with A. So codon third base must be A.
So codons like AAU, AAC, AAA all have third base U, C, A respectively. But
AAU has third U, which does not pair with U; it should pair with A. So incorrect.
Let's recalc strictly: The standard wobble rules: I pairs with U, C, A. So
anticodon 5'-IIU-3' binds to codon 3'-UUI-5'? That would be 5'-IUU-3'? This is
messy. Actually, the correct pairing: anticodon 5'-IIU-3' corresponds to codon
3'-YYA-5' (where Y is a base complementary to I). Since I pairs with three bases,
Y can be A, G, or U? No, I pairs with U, C, A in anticodon, so in codon it would
be A, G, U? No, complementarity: I (anticodon) pairs with U, C, A (codon third
position). So anticodon first base I pairs with codon third base U, C, or A. So
codon third base can be U, C, or A. Anticodon second base I pairs with codon
second base U, C, or A. Anticodon third base U pairs with codon first base A. So
codons: first base A, second base any of U, C, A, third base any of U, C, A. That
gives 9 codons: A U U, A U C, A U A, A C U, A C C, A C A, A A U, A A C, A A
A. Option C lists AAU, AAC, AAA (third base U, C, A? Actually AAU: A A U,
that is first A, second A, third U - fits if second base A is allowed. Yes, second
base can be A. So AAU (A A U), AAC (A A C), AAA (A A A) all fit. Option A:
AAU and AAC but missing AAA. Option B: UUA, UUG, CUA - first base not A.
Option D: UAA, UAG - first base not A. So correct is C.
Page 3
, 4. In the trp operon, a mutation that introduces a stop codon early in the leader
peptide sequence (trpL) is most likely to cause:
A. Constitutive attenuation, leading to increased transcription when tryptophan is
abundant
B. Decreased transcription when tryptophan is scarce due to failure of ribosome
stalling
C. No effect on attenuation if the stop codon is upstream of region 1
D. Increased transcription when tryptophan is scarce because ribosome translates
quickly through the leader
Answer: A. Constitutive attenuation, leading to increased transcription when
tryptophan is abundant
In attenuation, ribosome stalling at trp codons in the leader peptide allows
formation of antiterminator structures. A premature stop codon causes the
ribosome to dissociate early, preventing it from covering region 1, so region 2 can
pair with region 3 to form the antiterminator regardless of tryptophan levels,
leading to constitutive readthrough (transcription) even when tryptophan is high.
Option B is opposite; C is false because a stop upstream of region 1 still affects
ribosome position; D is false because quick translation prevents stalling.
5. Which of the following experimental observations would best demonstrate
that a DNA sequence acts as an insulator?
A. The sequence enhances transcription of a reporter gene when placed upstream of
the promoter
B. The sequence blocks the ability of an enhancer to activate a promoter when inserted
between them
C. The sequence binds to the nuclear matrix and is associated with heterochromatin
D. The sequence is methylated in cells where the downstream gene is silenced
Answer: B. The sequence blocks the ability of an enhancer to activate a
promoter when inserted between them
Insulators are defined by their ability to block enhancer-promoter
communication when positioned between them, without affecting the enhancer or
promoter directly. Option A describes an enhancer; option C describes a
scaffold/matrix attachment region (SAR/MAR) that may organize chromatin but
not necessarily block enhancers; option D describes a consequence of silencing
but not a direct assay for insulation.
Page 4
Answer Bank (2026/2027 Update) | Portage Learning | 100%
Verified Pass Guide - 174 Questions and Answers Already
Graded A+ Premium Exam Tested And Verified
Subject Area BIOD 210 Module 1-7 Exam Genetics - Complete Question & Answer Bank
(2026/2027 Update) | Portage Learning | 100% Verified Pass Guide
Description Comprehensive examination on BIOD 210 Module 1-7 Exam Genetics -
Complete Question & Answer Bank (2026/2027 Update) | Portage Learning |
100% Verified Pass Guide.
Expected Grade A+
Total Questions 174
Duration 3 hours
Learning Outcomes 1. Demonstrate mastery of core concepts
Accreditation Aligned with US university standards.
Page 1
,1. A mutation in the 3'->5' exonuclease domain of DNA polymerase III results
in a 100-fold increase in replication errors. Which repair pathway is most
directly compromised?
A. Mismatch repair
B. Base excision repair
C. Proofreading
D. Nucleotide excision repair
Answer: C. Proofreading
The 3'->5' exonuclease activity is responsible for proofreading during DNA
replication. Loss of this activity increases misincorporation rates, which are later
corrected by mismatch repair, but the immediate defect is proofreading. Base
excision repair deals with damaged bases, and nucleotide excision repair removes
bulky lesions.
2. A pre-mRNA from a gene with four exons undergoes alternative splicing to
produce two isoforms: one including exon 3 (isoform A) and one skipping exon
3 (isoform B). A mutation in the branch point sequence upstream of exon 3
leads to preferential inclusion of exon 3. What is the most likely effect on
splicing regulation?
A. Loss of U2AF binding reduces recognition of the 3' splice site, favoring exon
skipping
B. Disruption of the spliceosome assembly at the intron upstream of exon 3, causing
inclusion of exon 3 due to exon definition
C. Strengthening of the 5' splice site of exon 3 by the mutation, promoting inclusion
D. The mutation creates a new silencer element that represses exon 3 skipping
Answer: B. Disruption of the spliceosome assembly at the intron upstream of
exon 3, causing inclusion of exon 3 due to exon definition
The branch point is critical for lariat formation and spliceosome assembly. A
mutation in the branch point upstream of exon 3 impairs intron removal. In exon
definition, failure to splice the upstream intron can lead to inclusion of the
downstream exon because the splicing machinery recognizes the exon as a unit.
Option A is incorrect because branch point mutations do not directly affect U2AF
binding; U2AF binds the polypyrimidine tract. Option C is incorrect because
branch point mutations do not affect the 5' splice site. Option D is unlikely;
creating a silencer would promote skipping, not inclusion.
Page 2
,3. A tRNA with the anticodon 5'-IIU-3' (I = inosine) is expected to recognize
which of the following sets of codons?
A. 5'-AAU-3' and 5'-AAC-3'
B. 5'-UUA-3', 5'-UUG-3', and 5'-CUA-3'
C. 5'-AAU-3', 5'-AAC-3', and 5'-AAA-3'
D. 5'-UAA-3' and 5'-UAG-3'
Answer: C. 5'-AAU-3', 5'-AAC-3', and 5'-AAA-3'
Inosine can pair with U, C, or A by wobble rules. The anticodon IIU (reading
5'->3' anticodon) pairs with 3'->5' codon. Actually, anticodon is 5'-IIU-3'; the
corresponding codon is 5'-AAU-3' (since IIU pairs with AAU? Let's follow
standard: anticodon IIU (5'->3') binds to codon 3'->5' complementary: anticodon
bases: I I U; codon bases: C C A? No, careful: anticodon is written 5' to 3',
complementary antiparallel to codon. So anticodon 5'-IIU-3' pairs with codon
3'-YYA-5', which in 5'->3' is 5'-AYY-3'? That's confusing. Better: tRNA
anticodon I I U will bind to codons where the first base of codon (5' end) pairs
with I. Inosine can pair with A, U, or C. Second base of codon pairs with I (again
inosine can pair with A, U, C). Third base of codon pairs with U (so
complementary: A). So codons: first base: A, U, or C; second base: A, U, or C;
third base: A. That gives 3x3=9 possible codons, but only options include three
codons. Option C: AAU (first A, second A, third U) - third U pairs with U? No,
anticodon third base U pairs with A on codon. So codon third base must be A.
Thus AAU would pair? Actually, anticodon I I U: first I pairs with A, U, C;
second I pairs with A, U, C; third U pairs with A. So codon third base must be A.
So codons like AAU, AAC, AAA all have third base U, C, A respectively. But
AAU has third U, which does not pair with U; it should pair with A. So incorrect.
Let's recalc strictly: The standard wobble rules: I pairs with U, C, A. So
anticodon 5'-IIU-3' binds to codon 3'-UUI-5'? That would be 5'-IUU-3'? This is
messy. Actually, the correct pairing: anticodon 5'-IIU-3' corresponds to codon
3'-YYA-5' (where Y is a base complementary to I). Since I pairs with three bases,
Y can be A, G, or U? No, I pairs with U, C, A in anticodon, so in codon it would
be A, G, U? No, complementarity: I (anticodon) pairs with U, C, A (codon third
position). So anticodon first base I pairs with codon third base U, C, or A. So
codon third base can be U, C, or A. Anticodon second base I pairs with codon
second base U, C, or A. Anticodon third base U pairs with codon first base A. So
codons: first base A, second base any of U, C, A, third base any of U, C, A. That
gives 9 codons: A U U, A U C, A U A, A C U, A C C, A C A, A A U, A A C, A A
A. Option C lists AAU, AAC, AAA (third base U, C, A? Actually AAU: A A U,
that is first A, second A, third U - fits if second base A is allowed. Yes, second
base can be A. So AAU (A A U), AAC (A A C), AAA (A A A) all fit. Option A:
AAU and AAC but missing AAA. Option B: UUA, UUG, CUA - first base not A.
Option D: UAA, UAG - first base not A. So correct is C.
Page 3
, 4. In the trp operon, a mutation that introduces a stop codon early in the leader
peptide sequence (trpL) is most likely to cause:
A. Constitutive attenuation, leading to increased transcription when tryptophan is
abundant
B. Decreased transcription when tryptophan is scarce due to failure of ribosome
stalling
C. No effect on attenuation if the stop codon is upstream of region 1
D. Increased transcription when tryptophan is scarce because ribosome translates
quickly through the leader
Answer: A. Constitutive attenuation, leading to increased transcription when
tryptophan is abundant
In attenuation, ribosome stalling at trp codons in the leader peptide allows
formation of antiterminator structures. A premature stop codon causes the
ribosome to dissociate early, preventing it from covering region 1, so region 2 can
pair with region 3 to form the antiterminator regardless of tryptophan levels,
leading to constitutive readthrough (transcription) even when tryptophan is high.
Option B is opposite; C is false because a stop upstream of region 1 still affects
ribosome position; D is false because quick translation prevents stalling.
5. Which of the following experimental observations would best demonstrate
that a DNA sequence acts as an insulator?
A. The sequence enhances transcription of a reporter gene when placed upstream of
the promoter
B. The sequence blocks the ability of an enhancer to activate a promoter when inserted
between them
C. The sequence binds to the nuclear matrix and is associated with heterochromatin
D. The sequence is methylated in cells where the downstream gene is silenced
Answer: B. The sequence blocks the ability of an enhancer to activate a
promoter when inserted between them
Insulators are defined by their ability to block enhancer-promoter
communication when positioned between them, without affecting the enhancer or
promoter directly. Option A describes an enhancer; option C describes a
scaffold/matrix attachment region (SAR/MAR) that may organize chromatin but
not necessarily block enhancers; option D describes a consequence of silencing
but not a direct assay for insulation.
Page 4
