quiz questions exam 3

Question 1

Mutations in the coding sequence of the CFTR gene can lead to cystic fibrosis. One of the most common CFTR mutations found in people with cystic fibrosis is detailed below with spaces between nucleotide triplets indicating the open reading frame and starting at nucleotides that encode amino acid 506 in the protein:
Normal CFTR sequence ATC ATC TTT GGT GTT
Disease CFTR sequence ATC ATC GGT GTT
Which statement must be true of this mutation?

It occurred during mitosis.

It is a deletion mutation that results in the eventual production of a protein one amino acid shorter than the normal protein.

It is an insertion mutation.

It is a deletion mutation that is repaired in all cells except lung cells.

Answer 1

It is a deletion mutation that results in the eventual production of a protein one amino acid shorter than the normal protein

Question 2

The E. coli origin, oriC, contains:

four 9-base-pair consensus sequences, three AT-rich direct repeats, and eleven GATC sites.

eleven GATC sites.

four 9-base-pair consensus sequences and three AT-rich direct repeats.

three AT-rich direct repeats.

Answer 2

four 9-base-pair consensus sequences, three AT-rich direct repeats, and eleven GATC sites.

Question 3

Which of the following is required for formation of the open complex of replication initiation?

binding of DnaB helicase to single strands of the replication bubble

binding of DnaC to the four 9-mer consensus sequences, which destabilizes the AT-rich repeat sequences

binding of DnaA to the four 9-mer consensus sequences, which destabilizes the AT-rich repeat sequences

DnaA/DnaC/ATP complex loads DnaA onto the origin and ATP hydrolysis allows release of the DnaC protein

Answer 3

binding of DnaA to the four 9-mer consensus sequences, which destabilizes the AT-rich repeat sequences

Question 4

Which of the following enzymes or enzyme subunits does not rely on ATP hydrolysis for its correct function?

DnaB helicase
primase
DnaA
DnaC

Answer 4

primase

Question 5

Which of the following is not true about the termination of replication in bacteria?

Termination occurs at sequences called Ter sites.

There are two clusters of Ter sites oriented in opposite directions.

Tus-Ter complex activity is polar; the replication forks can pass through the first set but are blocked by the second.

Replication forks can proceed past both Tus-Ter clusters, which causes problems when it runs into RNA polymerase head first.

Answer 5

Replication forks can proceed past both Tus-Ter clusters, which causes problems when it runs into RNA polymerase head first.

Question 6

What structures solve the end replication problem for linear chromosomes?

telomerases
centromeres
telomeres
origin of replication

Answer 6

telomeres

Question 7

What is required for a DNA lesion to result in a mutation?

A base is altered and no longer hydrogen bonds properly.

Replication occurs before repair of a lesion producing a base pair that is different from the original base pair.

An incorrect base is inserted into a nucleotide sequence.

A break occurs in one of the DNA strands.

Answer 7

Replication occurs before repair of a lesion producing a base pair that is different from the original base pair.

Question 8

The base-pairing atoms of thymine are not directly involved in the cyclobutane ring of a pyrimidine dimer, formed by UV irradiation. Why, then, does a pyrimidine dimer stall the replicative DNA polymerase?

The base pairing of the pyrimidines in the dimer to their respective purines is stronger than usual, therefore the strands remain annealed and the DNA polymerase stalls.

The cross-linked pyrimidine dimer causes a distortion in the DNA resulting in a break in the DNA, therefore the DNA polymerase stalls.

The cross-linked pyrimidine dimer causes a distortion in the DNA that prevents base pairing in the active site of the DNA polymerase, therefore it stalls.

The cross-linked pyrimidine dimer binds to the DNA polymerase complex, causing it to stall.

Answer 8

The cross-linked pyrimidine dimer causes a distortion in the DNA that prevents base pairing in the active site of the DNA polymerase, therefore it stalls.

Question 9

XPF and XPG together have a unique activity among the different DNA repair mechanisms; they make up an excinuclease. This means:

these enzymes make four total incisions—one on each side of the site needing repair, and on both strands.

these enzymes make two total incisions—one on each strand on opposite sides of the DNA needing repair.

these enzymes make two total incisions—one on each side of the DNA needing repair, on the same strand.

None of the above.

Answer 9

these enzymes make two total incisions—one on each side of the DNA needing repair, on the same strand.

Question 10

Humans have at least 10 trans-lesion synthesis (TLS) polymerases, even though these polymerases all lack proofreading activity and are thus highly prone to errors. This could be due to any of the following, EXCEPT:

it is vital that cells keep growing, especially once they have committed to division and have started replicating their DNA.

some DNA has to divide faster than other DNA and this DNA tends to be less sensitive to mutation.

given the need for DNA replication to complete once it has begun, many TLS polymerases have evolved to deal with the numerous types of lesions that might arise.

lack of proofreading is the “biological price” a species must pay to overcome replication blocks—allowing a few mutant cells to survive.

Answer 10

some DNA has to divide faster than other DNA and this DNA tends to be less sensitive to mutation.

Question 11

Which of the following is not part of the mismatch repair (MMR) mechanism?

MutS and MutL form a complex at the site of the mismatch and while scanning in both directions along the DNA form a loop.

Mismatched bases are recognized by MutS.

When the MutS/MutL complex finds a GATC site, MutH binds to the complex and cleaves the methylated strand.

Helicase and exonuclease activities unwind and degrade the strand containing the mismatched base.

Answer 11

When the MutS/MutL complex finds a GATC site, MutH binds to the complex and cleaves the methylated strand.

Question 12

Which repair mechanism requires light energy?

nucleotide excision repair (NER)

base excision repair (BER)

mismatch repair (MMR)

photoreactivation

Answer 12

photoreactivation

Question 13

Which of the following is not true about base excision repair (BER) in eukaryotes?

Since eukaryotes lack a polymerase with 5’3’ exonuclease activity, the abasic strand can’ t be degraded from a nick.

Long-patch repair eliminates the strand containing the abasic site by displacement to create a “flap” that is removed by the flap endonuclease.

Short-patch repair only removes the 5’-deoxyribose phosphate and replaces it with a nucleotide.

BER in eukaryotes cannot be used to repair abasic sites created by hydrolysis.

Answer 13

BER in eukaryotes cannot be used to repair abasic sites created by hydrolysis.

Question 14

Which of the following damaged templates encountered by a replication fork is repaired by gap repair?

A replication fork encounters a lesion and translesion synthesis occurs.

The replication fork encounters a lesion and stalls.

The replication fork encounters a lesion and bypasses it, restarting synthesis on the other side of the lesion.

The replication fork encounters a lesion at which repair has been initiated and the fork collapses.

Answer 14

The replication fork encounters a lesion and bypasses it, restarting synthesis on the other side of the lesion.

Question 15

Which of the following is a benefit of recombination during meiosis in eukaryotes?

Branched intermediates aid in segregation by keeping the homologous chromosomes together.

Recombination introduces mutations that lead to increased variation in the population.

Recombination does not occur during meiosis.

None of the choices given is correct.

Answer 15

Branched intermediates aid in segregation by keeping the homologous chromosomes together.

Question 16

If a replication fork stalls at a lesion the usual first response is:

strand invasion.
branch migration.
fork regression.
initiation of gap repair.

Answer 16

fork regression

Question 17

The RecBCD enzyme:

has both polymerase and helicase activities.

prepares the 5’-ending single-strand by degrading the 3’-ending strand at the site of a double-stranded break.

binds tightly to chi sites through interactions with the RecB subunit.

directly loads the RecA protein onto single-stranded DNA overhangs.

Answer 17

directly loads the RecA protein onto single-stranded DNA overhangs.

Question 18

Which of the following is not true of RecA binding and function?

Single-stranded binding proteins (SSB) bound to single-strand DNA prevent RecA binding.

RecB recruits RecA to the 3’ end of the single-strand overhang.

RecFOR can recruit RecA to single-strand gaps.

RecA can displace SSB with the help of RecBCD but not RecFOR complexes.

Answer 18

RecA can displace SSB with the help of RecBCD but not RecFOR complexes.

Question 19

The gene for β-galactosidase has 3,075 bp. How long would it take for the E. coli RNA polymerase to transcribe this gene, assuming initiation has occurred upstream prior to its encounter with the gene?

~30-60 seconds
~0.5-2 seconds
~2.5-5 minutes
~1-2 minutes

Answer 19

~30-60 seconds

Question 20

What are the DNA sequence requirements for homologous genetic recombination?

specific DNA sequences that are recognized, bound, and recombined by recombinases

specific DNA sequences that are recognized by editing polymerases

two DNA molecules that have identical or very similar sequences over a significant region; any sequence is permitted

it occurs at almost any sequence

Answer 20

two DNA molecules that have identical or very similar sequences over a significant region; any sequence is permitted

Question 21

Many retroviruses have sequences complementary to the 3′ end of one of the tRNA molecules prominent in host cells infected by the virus. What is the purpose of this complementarity?

The tRNA accelerates viral protein synthesis at the expense of host protein synthesis.

The retroviral sequences bind host tRNA and interfere with host protein synthesis such that viral sequences are preferentially translated.

The retroviral sequences hybridize to host mRNA and interfere with host protein synthesis such that viral sequences are preferentially translated.

The tRNA provides the primer for DNA synthesis by reverse transcriptase.

Answer 21

The tRNA provides the primer for DNA synthesis by reverse transcriptase.

Question 22

Many TP (non-LTR) retrotransposons encode a reverse transcriptase that also has endonuclease activity. What is the primary function of the endonuclease activity?

The endonuclease cleaves the target DNA to facilitate recombination with the circular cDNA copy of the retroviral genome.

The endonuclease activity cleaves a phosphodiester bond in the target, exposing the required primer terminus. The reverse transcriptase then uses this 3′ end termini to prime viral DNA synthesis.

The endonuclease activity cleaves a phosphodiester bond in the target, exposing the required primer terminus. The reverse transcriptase then uses this 5′ end termini to prime viral DNA synthesis.

The endonuclease cleaves the target DNA to facilitate recombination with the linear cDNA copy of the retroviral genome.

Answer 22

The endonuclease activity cleaves a phosphodiester bond in the target, exposing the required primer terminus. The reverse transcriptase then uses this 3′ end termini to prime viral DNA synthesis.

Question 23

Which of the following will not result from site-specific recombination?

insertion
deletion
translocation
inversion

Answer 23

translocation

Question 24

The target of a site-specific recombinase consists of:

two short, asymmetric core sequences; two inverted-repeat binding sites; and one cleavage site.

one short, palindromic core sequence; two inverted-repeat binding sites; and one cleavage site.

one short, asymmetric core sequence; two inverted-repeat binding sites; and one cleavage site.

one short, asymmetric core sequence; two direct-repeat binding sites; and one cleavage site.

Answer 24

one short, asymmetric core sequence; two inverted-repeat binding sites; and one cleavage site.

Question 25

Which of the following characteristics of DNA-dependent DNA synthesis is not the same for DNA-dependent RNA synthesis?

Initiation involves the recognition of a specific DNA sequence.

Synthesis is catalyzed in the 5’ to 3’ direction.

Synthesis requires a template.

Initiation of synthesis requires a primer.

Answer 25

Initiation of synthesis requires a primer.

Question 26

Which of the following is not usually involved in the processing of primary mRNA transcripts?

addition of multiple adenosines to the 3’ end

covalent joining of exons

methylation of nucleotides at the 5’ end

insertion of intron sequences

Answer 26

insertion of intron sequences

Question 27

Polyadenylation of mRNA requires all of the following except:

Pol II extends the transcript beyond the site where the poly(A) tail is added.

The site of poly(A) addition is marked by specific sequence elements in the transcript.

The transcript is cleaved by an endonuclease that associates with the C-terminal domain of Pol II.

Polyadenylate polymerase uses a poly(U) RNA template to synthesize the poly(A) tail.

Answer 27

Polyadenylate polymerase uses a poly(U) RNA template to synthesize the poly(A) tail.

Question 28

An RNA processing event that frequently leads to different protein products encoded by a single gene is:

transcription.
alternative splicing.
alternative transcription.
polyadenylation.

Answer 28

alternative splicing

Question 29

Differential RNA processing may result in:

a shift in the ratio of mRNA produced from two adjacent genes.

inversion of certain exons in the mature mRNA.

the production of the same protein from two different genes.

the production of two proteins with different activities from a single gene.

Answer 29

the production of two proteins with different activities from a single gene.

Question 30

The nucleophile of the first step of spliceosome-mediated pre-mRNA splicing is the:

2’ hydroxyl of the branch point adenosine.

3’ hydroxyl of a free guanine nucleotide.

3’ hydroxyl of the 5’ splice site exon.

2’ hydroxyl of the 3’ splice site intron.

Answer 30

2’ hydroxyl of the branch point adenosine.

Question 31

What is the function of the sigma subunit of E. coli RNA polymerase in transcription?

It contains the polymerase active site.

It contains the separate exonuclease site required for proofreading.

It directs the polymerase to bind a specific type of promoter sequence.

It signals the polymerase to terminate transcription at specific sites.

Answer 31

It directs the polymerase to bind a specific type of promoter sequence.

Question 32

“DNA footprinting” is a technique that can be used to identify:

a region of DNA that has been damaged by mutation.

the position of a particular gene of a chromosome.

the position of internally double-stranded regions in a single-stranded DNA molecule.

the specific binding site of a repressor, polymerase, or other protein on the DNA.

Answer 32

the specific binding site of a repressor, polymerase, or other protein on the DNA.

Question 33

Which of the following is not true of the interaction of TBP with DNA?

TBP is able to interact with DNA nonspecifically with respect to sequence.

TBP binds the TATA box through minor groove DNA contacts.

A helix-turn-helix motif in TBP makes the DNA contacts.

The DNA bound by TBP is bent.

Answer 33

A helix-turn-helix motif in TBP makes the DNA contacts.

Question 34

In which of the following steps of transcription is the C-terminal domain of RNA polymerase II first phosphorylated?

assembly of the preinitiation complex

initiation

elongation

termination complex formation

Answer 34

initiation