BIOC 385 - EXAM 4 || with 100% Error-free Solutions.
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Module
BIOC 385
Institution
BIOC 385
What would be the effect of a mutation in the HIV-1 reverse transcriptase gene that decreased the rate of mismatches? correct answers Any decrease in the error rate of HIVRT would result in an overall decrease in the rate of viral mutations. This would probably not be an advantage for the virus, as...
BIOC 385 - EXAM 4 || with 100% Error-free Solutions.
What would be the effect of a mutation in the HIV-1 reverse transcriptase gene that decreased
the rate of mismatches? correct answers Any decrease in the error rate of HIVRT would result in
an overall decrease in the rate of viral mutations. This would probably not be an advantage for
the virus, as the relatively high error rate often helps to thwart the use of antiviral drugs.
Approximately how many ATP are required by helicase for the replication of the E. coli
chromosome? correct answers Because helicase requires 1 ATP per two bases, it requires 2.3 x
10^6 molecules of ATP to replicate 4.6 x 10^6 bases
What must the rate of nucleotide incorporation by primase be such that it completes a primer in
exactly the same time that an okazaki fragment is complete? correct answers The average
Okazaki fragment is 2,000 bases and synthesized in 2 seconds (1,000bases/second). A typical
primer is about 10-12 bases; thus primase would need to rate a 5-6 nucleotides/second to avoid
being the rate-limiting step of replications
An Ames test of a suspected mutagen was examined both before and after incubation with rat
liver extract, giving the following results. What can you conclude about the suspected mutagen?
Control: 2 colonies
Suspected mutagen: 16 colonies
Suspected mutagen after incubation with rat liver extract: 30+ colonies correct answers The
results suggest the suspected mutagen is mutagenic without liver metabolism, and even more
mutagenic after metabolism by liver enzymes.
If the sequence 5'-AACGC-3' were damaged by reactive oxygen species, what would be the most
prevalent product, and what would be the result of replication (show both strands after
replication)? correct answers The guanine would become 8-hydroxyguanine and tautomer, 8-
oxoguanine. These base pair with adenine, resulting in a G-A mismatch that is not likely to be
repaired by DNA repair mechanisms. After a second round of replication, the result is a G->C
substitution in one of the daughter strands.
First round of replication: 5′-AACGC-3′
5′-AACGC-3′
(8-hydroxyguanine = G): 3′-TTG*A*G-5′
3′-TTGCG-5′
Second round of replication: 5′-AAC*C*C-3′
5′-AACGC-3′
3′-TTG*A*G-5′
3′-TTGCG-5′
5′-AACGC-3′
,5′-AACGC-3′
3′-TTG*A*G-5′
3′-TTGCG-5′
From the following sequence, locate the sites of potential photoproduct formation. Indicate what
photoproducts will most likely form and what the potential effect on the DNA would be if left
unrepaired
5'-ACGTCAGTTAACGTACTGACGT correct answers 5′-
ACG*TC*AG*TT*ACGTA*CT*GACGT. The TC photoproduct is likely to be a (6-4)
photoproduct, which can stall replication forks if not repaired. The TT and CT photoproducts
would most likely be pyrimidine dimers and, if unrepaired, would most likely result in adenine
on the daughter strand. The TT photoproduct would not result in a mutation, but CT would result
in a mismatch, with high potential for mutation, depending upon when repair occurs. If the
daughter strand were used to correct the lesion, then the C would be converted to a T.
What would happen if a mutation in the lambda phage Xis gene occurred such that the resulting
protein was not functional? correct answers Because Xis is required for lambda phage DNA
excision, a mutation rendering Xis nonfunctional would prevent excision and prevent initiation
of a lytic cycle
Describe the typical major elements of a replication fork correct answers Ahead of the
replication fork, topoisomerase adds negative supercoils to relieve the positive supercoiling
caused by separation of strands. Helicase and primase are coupled at the front of the replication
fork to separate the strands and synthesize RNA primers, respectively. Single-stranded binding
proteins maintain single-stranded DNA and protect it from damage. DNA polymerase III,
tethered to the helicase, consists of a core enzyme and beta-clamp , which increases processivity
of the polymerase. DNA polymerase I and ligase follow the replication fork to remove RNA
primers and seal nicks in the newly formed DNA.
What are the major events that occur at oriC to allow initiation of DNA synthesis? correct
answers Binding of DnaA beings DNA melting and guides helicase to single-stranded DNA
(with the help of DnaC). Gyrase allows further opening of the double helix by relieving torsion
created by the single-strand region, and single-stranded DNA binding proteins maintain the
single-stranded state. DNA Pol III binds to release DnaA and start the replication process
What is unique about the structure of telomerase that allows it to extend the ends of
chromosomes? correct answers Human telomerase contains a 451-nucleotide RNA with a
tandem RNA at the 3' end that is complementary to the chromosome end. After association of the
RNA with the end of the chromosome, telomerase extends the chromosome using the RNA as
template. When the end of the template is reached, telomerase releases from the chromosome.
Describe the Ames test and explain it importance correct answers the Ames test uses a mutant
strain of Salmonella that cannot produce histidine to test the mutagenicity of various molecules.
The test molecule is incubated with Salmonella and liver extract to simulate mammalian
, metabolism. If colonies are produced on minimal-histidine media, then a back mutation has
occurred in the histidine synthesis gene so the pathway is functional. Substances that are
mutagenic to bacteria can be identified, which are potentially carcinogenic to humanss
What happens during cytosine deamination, and how is it repaired? correct answers Cytosine can
be spontaneously deaminated, producing uracil. Uracil is removed by uracil DNA glycosylase to
produce an abasic site that is repaired by base excision repair, where a nick is created in the
DNA by AP endonuclease. The nucleotide is replaced by short-path repair or a short section of
nucleotides is replaced by long-path repair
HW 1
What is the function of the MutS-MutL-MutH protein complex?
a) DNA methylation
b) DNA adenylation
c) base excision
d) none
e) resolution of Holliday junctions correct answers None of the answers are correct
HW 1
How can DNA strands be synthesized using bidirectional replication?
a) none
b) they are synthesized through Okazaki fragments
c) they are synthesized using RNA polymerase
d) one strand is synthesized 5'-3' whereas the other is 3'-5'
e) They are no synthesized simultaneously correct answers They are synthesized through
Okazaki fragments
HW 1
What accounts for the high fidelity of DNA polymerization? Choose the ONE best answer
-Nucleotide excision repair corrects all DNA replication errors, so DNA polymerase does not
need a proofreading activity.
-The high rate of accuracy in DNA polymerization comes from both nucleotide proofreading and
3' to 5' exonuclease activity of DNA Pol I.
-Active site geometry ensures that only correct sized nucleotide base pairs are formed, and 3' to
5' proofreading removes incorrect nucleotides.
-The 5' to 3' exonuclease proofreading activity removes nucleotides, which provides a
mechanism to correct errors at the fork.
-Active site geometry but not nucleotide proofreading is important to the high fidelity of DNA
polymerization.
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