exam 2 Flashcards
What feature of DNA allows it to store information?
a. the 5’ to 3’ direction of the strands
b. the base sequence
c. the relative locations of the major and minor groove
d. the spiral structure
b. the base sequence
Chargaff’s results suggest that (in DNA),
a. the amount of G = C and A = T.
b. the amount of G = A = C = T.
c. the amount of G = T and A = C.
d. all of the above.
a. the amount of G = C and A = T.
Which of the following could be the components of a single nucleotide found within an RNA strand?
a. deoxyribose, thymine, and phosphate
b. deoxyribose, guanine, and three phosphates
c. ribose, thymine, and phosphate
d. ribose, uracil, and phosphate
d. ribose, uracil, and phosphate
Which component(s) of a nucleotide is/are found along the DNA backbone?
a. bases only
b. deoxyribose and bases
c. phosphate only
d. deoxyribose and phosphate
d. deoxyribose and phosphate
Which of the following is/are ways that a retroelement (retrotransposon) can proliferate?
a. Using reverse transcriptase and integrase
b. Using transposase
c. Target-site primed reverse transcription
d. Both a and c
d. Both a and c
Which of the following is found in a eukaryotic chromosome but not in a bacterial chromosome?
a. Many genes
b. One centromere
c. Some repetitive sequences
d. One or more origins of replication
b. One centromere
A region in a cell nucleus that is occupied by a single chromosome is a
a. nucleosome.
b. radial loop domain.
c. chromosome territory.
d. euchromatic region.
c. chromosome territory.
With regard to chromosome structure, the loop extrusion model proposes that __________ promote loop formation and ___________ stablize the loop.
a. CTCFs, histones
b. SMC proteins, histones
c. CTCFs, SMC proteins
d. SMC proteins, CTCFs
d. SMC proteins, CTCFs
The reason for using chromosome conformation capture (3C) methods is to
a. determine if a gene is transcriptionally active.
b. distinguish interphase chromosomes from metaphase chromosomes.
c. detect changes in chromosome structure.
d. identify DNA segments that are close to together in three-dimensional space.
d. identify DNA segments that are close to together in three-dimensional space.
Protein complexes known as ______________ are responsible for forming ___________ in ___________.
a. condensins, euchromatin, interphase chromosomes
b. cohesins, radial loop arrays, interphase chromosomes
c. condensins, radial loop arrays, metaphase chromosomes
d. cohesins, radial loop arrays, metaphase chromosomes
c. condensins, radial loop arrays, metaphase chromosomes
From a structural point of view, DNA can be replicated to produce two double helices with the identical base sequences because
a. of the AT/GC rule.
b. the strands are antiparallel.
c. of base stacking.
d. DNA polymerase can make DNA only in the 5’ to 3’ direction.
a. of the AT/GC rule.
What are the expected results for the Meselson and Stahl experiment after 4 generations (i.e, 4 rounds of DNA replication in the presence of light nitrogen)? Note: during generation zero, the DNA is all heavy, and subsequent generations only make light DNA.
a. 1/4 half heavy, 3/4 light
b. 1/8 half heavy, 7/8 light
c. 1/16 half-heavy, 15/16 light
d. 1/32 half-heavy, 31/32 light
b. 1/8 half heavy, 7/8 light
At the E. coli origin of replication, the DNA begins to unwind at the
a. DnaA boxes
b. AT-rich region
c. GATC methylation sites
d. None of the above
b. AT-rich region
How many replication forks are formed at an origin of replication?
a. One
b. Two
c. four
d. many
b. two
Primase is needed during DNA replication because DNA polymerase is not able to
a. synthesize DNA in the 5’ to 3’ direction.
b. synthesize DNA in the 3’ to 5’ direction.
c. begin synthesis on a bare template strand.
d. proofread the DNA.
c. begin synthesis on a bare template strand.
How is a new DNA strand made in the leading strand?
a. DNA is made in the direction toward the replication fork and is made as one continuous strand.
b. DNA is made in the direction away from the replication fork and is made as one continuous strand.
c. DNA is made in the direction toward the replication fork and is made as Okazaki fragments.
d. DNA is made in the direction away from the replication fork and is made as Okazaki fragments.
a. DNA is made in the direction toward the replication fork and is made as one continuous strand.
Which of the following is not found in a replisome?
a. two DNA polymerases
b. helicase
c. primase
d. All of the above are found in a replisome.
d. All of the above are found in a replisome.
The proofreading function of DNA polymerase involves the recognition of a ________ and the removal of a short segment of DNA in the __________ direction.
a. missing base, 5’ to 3’
b. base mismatch, 5’ to 3’
c. missing base, 3’ to 5’
d. base mismatch, 3’ to 5’
d. base mismatch, 3’ to 5’
Below is a DNA molecule, with the ends labeled A, B, C and D. Which ends would get shorter during DNA replication if telomerase did not function?
A 5’———————————————–3’ B
C 3’———————————————–5’ D
a. A and C
b. A and D
c. B and D
d. B and C
d. B and C
A small effector molecule that enhances transcription binds to a regulatory protein and causes it to not bind to the DNA. The regulatory protein
a. is a repressor.
b. is an activator.
c. could be a repressor or an activator.
d. s neither a repressor or an activator.
a. is a repressor.
During diauxic growth involving glucose and lactose, bacterial cells
a. use up lactose first, then they metabolize glucose.
b. use up glucose first, then they metabolize lactose.
c. metabolize both sugars at maximal rates at the same time.
d. None of the above
b. use up glucose first, then they metabolize lactose.
Let’s suppose you have isolated a mutant strain of E. coli in which the lac operon is constitutively expressed. You create a merozygote in which the mutant strain also contains an F’ factor with a normal lac operon and a normal lacI gene. You then compare the mutant strain and the merozygote with regard to their β-galactosidase activities in the presence and absence of lactose. You obtain the following results:
Strain Addition of lactose (no glucose) Amount β-Galactosidase
Normal Yes 100
Mutant No 100
Mutant Yes 100
Merozygote No 1
Merozygote Yes 200
Which of the following is consistent with these data?
a. The mutation is in the lacI gene and results in a lac repressor that cannot bind to the operator.
b. The mutation is in the lac operator and prevents the lac repressor from binding to the operator.
c. The mutation is in the lacI gene and results in a lac repressor that cannot bind allolactose.
d. All of the above.
a. The mutation is in the lacI gene and results in a lac repressor that cannot bind to the operator.
For a riboswitch that controls transcription, the binding of a small molecule such as TPP controls whether the RNA
a. has an antiterminator or terminator stem-loop.
b. has a Shine-Dalgarno antisequestor or the Shine-Dalgarno sequence in a stem-loop.
c. is degraded from its 5’ end.
d. has both a and b.
a. has an antiterminator or terminator stem-loop.
What is the role of mediator with regard to eukaryotic gene regulation?
a. Mediator controls the switch to the elongation phase of transcription.
b. Mediator controls whether or not RNA polymerase can bind to the core promoter.
c. Mediator causes histones to be displaced from the core promoter.
d. Mediator controls whether or not TFIID can bind to the core promoter.
a. Mediator controls the switch to the elongation phase of transcription.
During transcription, the removal of histone proteins from the DNA is important for
a. the phosphorylation of CTD.
b. the formation of the preinitiation complex.
c. transcriptional elongation.
d. both B and C.
d. both B and C.
The following list describes events that are required for transcription of a eukaryotic gene. Put them in the correct order.
Elongation
Binding of activators
Chromatin remodeling and histone modification
Formation of the pre-initiation complex
a. 2, 3, 4, 1
b. 1, 4, 2, 3
c. 4, 3, 2, 1
d. 2, 1, 4, 3
a. 2, 3, 4, 1
During the phenomenon of proximal promoter pausing,
a. the CTD (carboxyl terminal domain) of RNA polymerase gets phoshorylated.
b. RNA polymerase stops transcribing the DNA due to the binding of NELF (negative elongation factor) and DSIF (DRB-sensitivity-inducing factor).
c. TFIIH converts the preinitiation complex into an open complex.
d. TFIIB, TFIIE, TFIIH, and mediator are released from RNA polymerase.
b. RNA polymerase stops transcribing the DNA due to the binding of NELF (negative elongation factor) and DSIF (DRB-sensitivity-inducing factor).
Which of the following does not happen for the glucocorticoid receptor to activate a gene?
a. Protein-protein interaction (e.g., dimer formation)
b. Covalent modification of the glucocorticoid receptor
c. The binding of a small effector molecule to the glucocorticoid receptor subunits
d. All of the above happen.
b. Covalent modification of the glucocorticoid receptor
Which of the following phenomena would not be studied using ChIp-Seq?
a. To see if a regulatory transcription factor binds to mediator and/or TFIID
b. To determine the sites where a particular histone variant are bound in a genome.
c. To analyze the binding of regulatory transcription factor to an enhancer on a genome-wide scale.
d. To determine if a mutation in a regulatory transcription factor alters its ability to bind to different sites in a genome.
a. To see if a regulatory transcription factor binds to mediator and/or TFIID
Which of the following is not a function of heterochromatin formation?
a. Gene silencing
b. Gene activation
c. Prevention of transposable element movement
d. Prevention of viral proliferation
b. Gene activation
Due to imprinting, the Igf2 gene is activated because ______________ of the ICR and DMR _________ occur during _____________.
a. covalent histone modification, does, oogenesis
d. DNA methylation, does not, spermatogenesis
c. DNA methylation, does, spermatogenesis
d. DNA methylation, does not, oogenesis
c. DNA methylation, does, spermatogenesis
Which of the following is NOT involved in regulating gene transcription in eukaryotes?
a. RISC (RNA-induced silencing complex)
b. HOTAIR
c. Pioneer factors
d. All of the above are involved in transcriptional regulation.
a. RISC (RNA-induced silencing complex)
The main role of the polycomb group complexes is to
a. cause X-chromosome inactivation.
b. cause gene repression in specific cell types.
c. activate genes in response to certain environmental agents.
d. remove histone proteins during the elongation phase of transcription.
b. cause gene repression in specific cell types.
The Avy allele of the Agouti gene involves the insertion of a transposable element upstream from the normal Agouti promoter. The transposable element carries a promoter that causes the over expression of the Agouti gene. Mice carrying this allele tend to have coat colors that are more yellow than mice that don’t have this transposable element. If pregnant female mice are fed a diet that contains chemicals that increase DNA methylation, how would you expect that this diet would affect the coat color of offspring carrying the Avy allele?
a. Their fur would be more yellow because the Agouti gene would tend to be over expressed.
b. Their fur would be more yellow because the Agouti gene would tend to be under expressed.
c. Their fur would be less yellow (more dark brown) because the Agouti gene would tend to be over expressed.
d. Their fur would be less yellow (more dark brown) because the Agouti gene would tend to be under expressed.
d. Their fur would be less yellow (more dark brown) because the Agouti gene would tend to be under expressed.
Let’s suppose a mutation in the HOTAIR gene prevents HOTAIR from fully repressing other genes, such as HoxD. Which of the following types of mutation could prevent HOTAIR from fully repressing the genes it is supposed to repress?
a. A mutation that prevents HOTAIR from binding to the PRC2 complex
b. A mutation that prevents HOTAIR from binding to the LSD1 complex
c. A mutation that alters the site in HOTAIR that binds to GA-rich regions
d. All of the above
d. All of the above
Which of the following is not one of the steps in RNA interference (RNAi) that occurs with miRNA (microRNA)?
a. Transcription to produce Pri-miRNA
b. Cleavage of pre-miRNA by dicer
c. Binding of miRNA to form a P-body
d. Binding of RISC to a complementary mRNA
c. Binding of miRNA to form a P-body
Which genes of the CRISPR-Cas system are not needed for the expression and interference phases?
a. Cas1, Cas2
b. Cas9
c. tracr
d. Crispr
a. Cas1, Cas2
