Practice Problems from Review & In Class Flashcards
The following are three hypothesized mechanisms for DNA replication. Mechanism A represents _____ replication
(one other DNA splits into two different colored DNA strands)
a) semiconserved
b) conservative
c) dispersive
d) discontinuous
b) conservative
What accounts for the high fidelity of DNA polymerization?
a) the 5’ to 3’ exonuclease activity is a proofreading function that removes incorrect
nucleotides at the fork
b) The 3’ to 5’ exonuclease activity is a proofreading function that removes incorrect
nucleotides at the fork
c) DNA polymerase fidelity is ensured by the G-C and A-T base pairs forming
between two DNA strand
d) Mg 2+ in the active site provides proofreading functions through ionic bonds with
only correct nucleotides
e) DNA repair takes care of all errors so no need for proofreading activity in DNA
polymerase III enzyme
b) The 3’ to 5’ exonuclease activity is a proofreading function that removes incorrect
nucleotides at the fork
Match the bacterial DNA replication component on the left with its primary function to the right
a) clamp protein
b) topoisomerase
c) DNA helicase
d) single-strand binding protein
e) primase
f) DNA A protein
A) Unwind the DNA at Ori C
B) unwind the DNA double helix
C) Synthesize an RNA primer
D) Maximize processivity of DNA Pol
E) Remove supercoils ahead of fork
F) Prevent DNA strand reannealing
a) E D B F C A
b) D E B F C A
c) D E B F C A
d) A E B F C D
e) A F B E C A
c) D E B F C A
Tus-Ter terminates DNA synthesis by
a) breaking the single strand DNA
b) forcing the primase to dissociate
c) blocking the opening of helicase
d) methylating the DNA strand
c) blocking the opening of helicase
An Ames test of a compound suspected to be a mutagen was examined before and after incubation with rat liver extract. Based on the results shown, what do you conclude
a) The Ames test does not work considering the liver is worse
b) the control plate shows low mutation rate, which is surpirsing
c) all three bacterial plates must contain a large amount of histidine to prevent colonies from forming
d) compound is mutagenic without liver enzymes and this activity is further enhanced by liver enzymes
e) compound is not mutagenic at all because colonies grow meaning that their DNA is undamaged
d) compound is mutagenic without liver enzymes and this activity is further enhanced by liver enzymes
The DNA alkylation of guanine is possible because the _______ of guanine is _______________ and reacts rapidly with __________ alkylating agents
a) N-7; nucleophilic; electrophilic
b) N-5; nucleophilic; electrophilic
c) N-7; electrophilic; nucleophilic
d) N-5; electrophilic; nucleophilic
a) N-7; nucleophilic; electrophilic
What is the function of
1) the resolvase enzyme in DNA recombination?
2) the MutS-MutL-MutH protein complex
3) the BRACA1/BRACA2 proteins?
A methylation of DNA daughter strands
B repair of DNA double strand breaks
C Repair of Holliday junctions in recombination
D Base excision repair of DNA mutations
E mismatch repair of DNA mutations
a) 1) B 2) A 3) C
b) 1) C 2) E 3) B
c) 1) A 2) E 3) B
d) C 2) B 3) E
e) 1) A 2) D 3) E
b) 1) C 2) E 3) B
What is the recombination product if resolvase enzyme cleaves and rejoins as a) site 1 (half way on the resolvase, so red stays with red and green stays with greens), b) site 2? ( halfway in the middle, so red would break in half (joining with blue) and orange would break (joining with green)
A a) no change in sequence, b) sequence is a mixture
B a) a sequence is a mixture, b) sequence has reverse polarity (3’ to 5’)
C a) sequence is a mixture b) no change in sequence
D a) sequence has reverse polarity (3’ to 5’) b) DNA sequence is degraded
A a) no change in sequence, b) sequence is a mixture
the sequence is 4 parts of red, blue, orange, and green
Answer the following two questions regarding the structure and function of RNA
a) Which of the RNAs listed below
are noncoding RNA considering
that mRNA is the only type of
coding RNA?
- ribosomal RNA
- snoRNA
- RNase P RNA
- miRNA
- All of these
b) Which two noncoding RNA
molecules listed below are
recycled intact after protein
synthesis?
- rRNA and miRNA
- siRNA and rRNA
- rRNA and tRNA
- rRNA and lncRNA
- snRNA and mRNA
A. 2 and 9
B. 5 and 6
C. 5 and 8
D. 4 and 10
E. 3 and 7
C. 5 and 8
The most significant influence on why mRNA is processed differently in prokaryotes than eukaryotes is the fact that…
A. eukaryotes separate transcription and translation with a nucleus.
B. prokaryotes are often polycistronic.
C. eukaryotes are multicellular organisms.
D. prokaryotes do not add a poly(A) tai
A. eukaryotes separate transcription and translation with a nucleus.
The processes of prokaryotic (PRK) and eukaryotic (EUK) transcriptional initiation share some
similarities, but there are also some differences. Which statement below correctly describes a similarity
and a difference between PRK and EUK transcriptional initiation?
A. Similarity; a transcription factor binds to a conserved A-T rich sequence in both promoters
Difference; PRK RNA pol has proof-reading function, the EUK RNA pol does not have proofreading
B. Similarity; the sigma factor and TBP protein both dissociate from DNA after RNA pol initiates
Difference; PRK RNA pol transcribes coding strand, EUK RNA pol transcribes template strand
C. Similarity; a transcription factor binds to a conserved A-T rich sequence in both promoters
Difference; PRK sigma factor binds to a G-C rich sequence, EUK TBP binds to A-T rich sequence
D. Similarity; a transcription factor binds to a conserved A-T rich sequence in both promoters
Difference; PRK sigma factor dissociates from DNA after RNA pol, EUK TBP remains bound to DNA
E. Similarity; the PRK and EUK RNA polymerases both have a phosphorylated C-terminal domains
Difference; PRK sigma factor binds DNA 5’ of start site, EUK TBP binds DNA 3’ of start site
Review Question (M11Q1)
D. Similarity; a transcription factor binds to a conserved A-T rich sequence in both promoters
Eukaryotic promoters are much more complex than prokaryotic promoters
because they…
A. require three RNA polymerases.
B. are largely conserved across the genome.
C. can stimulate DNA polymerases by direct binding.
D. control translational as well as transcriptional events.
Review Question (M11Q2)
A. require three RNA polymerases.
Put the 10 steps shown below in the correct order describing
the spliceosome reaction cycle.
___ The U1-U2 snRNP complex binds to the GU sequence at the 5’ end of the intron.
___ The U2-U5-U6 snRNP complex is bound to the lariat RNA structure and dissociates from mRNA.
___ The U2 snRNP translocates to the adenine residue at the intronic branch site.
___ The hydroxyl group at the 5’ end of the intron attacks the phosphate at the 3’ end of the intron.
___ The U2 snRNP is released from the complex and U5 and U6 snRNPs associate with U4 snRNP.
___ The hydroxyl group of the branch point adenine attacks the phosphate at 5’ end of the intron.
___ RNA Pol II finishes transcribing across an entire intron and is in the middle of a 3’ exon.
___ The U4-U5-U6 tri-snRNP complex binds to U2 at the branch point along with the U1 snRNP.
___ The U1 and U4 snRNPs dissociate from the precatalytic complex leaving U2-U5-U6 behind.
___ The U2, U5, and U6 snRNPs dissociate from lariat RNA structure and intronic RNA is released.
A. 2, 9, 3, 7, 10, 6, 1, 4, 5, 8
B. 2, 8, 3, 7, 10, 5, 1, 4, 6, 9
C. 2, 8, 4, 7, 9, 6, 1, 3, 5, 10
D. 2, 8, 3, 7, 10, 6, 1, 4, 5, 9
E. 2, 7, 3, 8, 10, 6, 1, 4, 5, 9
D. 2, 8, 3, 7, 10, 6, 1, 4, 5, 9
All RNA splicing events are a series of ____________ reactions where the intron
is removed and the exonic sequences are rejoined to create a mature transcript.
A. transesterification
B. oxidation–reduction
C. β-elimination
D. nucleophilic addition
A. transesterification
What is the correct order of RNA processing events in eukaryotes beginning with
phosphorylation of the RNA Pol II CTD?
A. 5’ capping → splicing → 3’ processing → termination
B. 5’ capping → 3’ processing → splicing → termination
C. 3’ capping → splicing → 5’ processing → termination
D. splicing → 5’ capping → 3’ processing → termination
A. 5’ capping → splicing → 3’ processing → termination
What is the type of molecular defect that has been linked to the disease retinitis pigmentosa?
A. Mutation in the mRNA sequence causing an alternate splicing reaction to occur.
B. Mutation in the mRNA pathway as a result of poly(A) tail deadenylation.
C. Mutation in a splicing protein causing defects in the spliceosome machinery.
D. Mutation causing mRNA redirection during nuclear export owing to loss of m 7 G cap.
E. Mutation disrupting RNA decay by the miRNA directed Dicer complex in retinal cells
C. Mutation in a splicing protein causing defects in the spliceosome machinery.
Nirenberg and colleagues determined a portion of the genetic code using filter binding
assays that included ribosomes, trinucleotides, and tRNA molecules charged with
radioactively-labeled amino acids. Choose the answer below that correctly answers each
of the following questions.
1) If trinucleotide sequence is UAG, which amino acid is bound to filter paper?
2) If 5’ position of trinucleotide is G, which amino acid is bound to the filter?
3) True/False: There are 61 tRNAs to recognize all but three of 64 codons.
A. 1) Gln, 2) none 3) False
B. 1) none, 2) Glu, 3) False
C. 1) Gln, 2) none 3) True
D. 1) Trp, 2) Gln, 3) False
E. 1) none, 2) none, 3) True
B. 1) none, 2) Glu, 3) False
If a tRNA anticodon is 5’-IGC-3’, how many possible mRNA codons can this tRNA bind to?
A. Zero, this is a stop codon
B. One
C. Three
D. Four
E. Five
C. Three
Number the steps from start to finish for prokaryotic mRNA translation (ae).
(a) ____ Translational elongation proceeds in the 3’ direction along the mRNA strand
(b) ____ Binding of 50S subunit, GTP hydrolysis, release of IFs, 70S ribosome forms
(c) ____ Initiation factors (IFs) bind to the 30S ribosome along with GTP
(d) ____ Stop codon enters A site, binding of RF protein, disassembly of ribosome.
(e) ____ mRNA Shine-Dalgarno sequence base-pairs with 16S rRNA in 30S subunit
A. 3, 4, 1, 5, 2
B. 4, 1, 3, 5, 2
C. 4, 3, 1, 5, 2
D. 4, 3, 5, 1, 2
E. 4, 3, 1, 2, 5
C. 4, 3, 1, 5, 2
How does the antibiotic puromycin block protein synthesis?
A. It binds to the 50S subunit and prevents peptide bond formation.
B. It terminates protein synthesis by accepting the polypeptide chain.
C. It binds to the 50S subunit and prevents translocation.
D. Alters the structure of the 30S subunit causing translation errors.
B. It terminates protein synthesis by accepting the polypeptide chain.
Which sequence is MOST likely to be modified with the fatty acid myristoylate in
the ER? (The … represents additional amino acids C-terminal to those shown)
A. Cys-Leu-Ala-Ile-Ser-Phe…
B. Leu-Gly-Phe-Ala-Ser-Phe…
C. Gly-Leu-Phe-Ala-Ile-Ser…
D. Phe-Leu-Cys-Gly-Leu-Ile…
C. Gly-Leu-Phe-Ala-Ile-Ser…
The Drosophila Xmt protein is localized to the endoplasmic reticulum (ER), whereas the Drosophila protein
Kfr is cytosolic. Two altered proteins were constructed by gene cloning, 1) del-Xmt, which lacks a short Xmt
N-terminal sequence, and 2) N-Xmt-Kfr, which contains this short N-terminal Xmt sequence linked to Kfr.
Based on what is known about signal sequences in protein targeting, choose True (T) or False (F).
T F a. Data likely showed del-Xmt protein in the ER, whereas the natural Kfr protein was in the cytosol.
T F b. Data likely showed N-Xmt-Kfr protein in the cytosol, whereas the natural Xmt protein was in the ER.
T F c. If N-Xmt-Kfr and del-Xmt proteins were found in the cytosol, then it would indicate that the
N-terminal Xmt sequence was “necessary, but not sufficient” for localization to the ER.
T F d. If the del-Xmt protein was found in the cytosol, but the N-Xmt-Kfr protein was in the ER, then it
would indicate the N-terminal Xmt sequence was “necessary and sufficient” for ER localization.
A. T, F, F, T
B. F, F, T, T
C. F, T, F, T
D. F, T, T, T
E. F, F, F, T
B. F, F, T, T
In which of the following types of prokaryotic gene regulation will the affinity of a protein for its operator sequence decrease when a ligand is added, which coincides with increased rates of transcription relative to the level of transcription in the absence of ligand?
A. Ligand-regulated activation
B. Ligand-regulated de-repression
C. Ligand-regulated repression
D. Ligand-regulated deactivation
E. This outcome is not possible.
B. Ligand-regulated de-repression
In the presence of glucose, CRP does not bind to the lac operon because cAMP
levels are __(1)__. In the presence of lactose alone, lac repressor does not bind
DNA because __(2)__ binds to the repressor, and under these conditions, the
observed level of lac operon expression is very __(3)__. When both glucose and
lactose are present, a very __(4)__ level of lac operon expression is observed
A. (1) low, (2) allolactose, (3) high, (4) low
B. (1) high, (2) allolactose, (3) high, (4) low
C. (1) low, (2) glucose, (3) high, (4) low
D. (1) low, (2) allolactose, (3) low, (4) high
E. (1) high, (2) allolactose, (3) low, (4) high
A. (1) low, (2) allolactose, (3) high, (4) low
