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BCEM 393 Lecture Notes - Lecture 14: Helicase, Dna Supercoil, Dnaa

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rosefrog223
21 Jun 2018
School
University of Calgary
Department
Biochemistry
Course
BCEM 393
Professor
Robert Allan Edwards

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DNA Replication
SEMICONSERVATIVE REPLICATION
-As DNA is replicated, one of the strands
of each daughter DNA molecule would
be newly synthesized, whereas the oth-
er would be passed unchanged from
the parent DNA molecule
-The three theoretical mechanisms of
DNA replication are shown on the right.
The first mechanism is semiconserva-
tive replication, second is conservative
replication, and third is the dispersive
model
MESELSON AND STAHL DEMONSTRATED
THAT DNA REPLICATION IS
SEMICONSERVATIVE
-In their experiment, they:
-Grew bacteria in 15N (heavy) labelled
medium
-Transferred to regular 14N medium
(light)
-Centrifuged and used densities of
DNA present
-In later generations, the absence of
15N DNA indicated that the parental
DNA was not preserved
-In generation 1, the single band be-
tween 14N and 15N that all of the
daughter DNA derived some of their
atoms from the parent DNA
-In generation 2, there were equal
amounts of two bands on DNA
OVERVIEW OF THE DNA SYNTHESIS REACTION
-(DNA)n + dNTP —> (DNA)n+1 + PPi
-The reaction requires a DNA template
-Each of the deoxynucleotides (dATP, dGTP, dCTP, and dTTP) plus Mg2+
-DNA polymerase (primarily DNA polymerase III in prokaryotes)
-A priming sequence (with a free 3’-OH group)
-Elongation of the DNA strand proceeds in the 5’-to-3’ direction
-The final product is a complete copy of the template
HIGH FIDELITY OF DNA REPLICATION — DNTP BINDING CHANGES ENZYME CONFORMATION
BY INDUCED FIT
-High fidelity means low error rates
-DNA polymerases close down around the incoming dNTP
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-The binding triggers a conformational change
-The finger domain rotates to form a tight pocket into which only a properly shaped base pair
will readily fit
-Importance —> Contributes to DNA polymerases low error rate
-Discrimination between
corrupt and incorrect nu-
cleotides relies not just
on complementary base
pairing (hydrogen bond-
ing), but also on base
pairing geometry
-The active site of DNa
polymerase I accommo-
dates only base pairs with this geometry
-DNA polymerases insert one incorrect nucleotide for every 10^4 to 10^5 correct ones
HIGH FIDELITY OF DNA REPLICATION
-The geometry of incorrectly paired bases can exclude them from the active site, as occurs on
DNA polymerase
-DNA polymerases insert one incorrect nucleotide for every 10^4 to 10^5 correct ones, but,
replication proceeds with one mistake for every 10^9 to 10^10 nucleotides added
-Proofreading Enzymatic Activity
-3’-5’ endonuclease activity removes mismatched nucleotides from the 3’ end of DNA by hy-
drolysis
-Increases fidelity by ~1000-fold
HOW DOES DSDNA ACT AS A TEMPLATE?
-Helicases are enzymes that move along the DNA and separate the strands (breaks hydrogen
bonds between base pairs), using chemical energy from ATP
-Mechanism:
-One of the strands of the double helix passes through the hole in the centre of the helicase,
bound to the loops of two adjacent subunits
-2 of the subunits do not contain
bound nucleotides
-On the binding of ATP to these 2
subunits, and the release of ADP +
Pi from 2 other subunits, the heli-
case hexamer undergoes a confor-
mational change, pulling the DNA
through the helicase
-The helicase acts as a wedge to
force separation of the two strands
of DNA
TOPOISOMERASES PREPARE THE DOUBLE HELIX FOR UNWINDING
-The local helicase-mediated unwinding in one region leads to stressful overwinding in sur-
rounding regions
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Document Summary

As dna is replicated, one of the strands of each daughter dna molecule would be newly synthesized, whereas the oth- er would be passed unchanged from the parent dna molecule. The first mechanism is semiconserva- tive replication, second is conservative replication, and third is the dispersive model. Grew bacteria in 15n (heavy) labelled medium. In generation 1, the single band be- tween 14n and 15n that all of the daughter dna derived some of their atoms from the parent dna. In generation 2, there were equal amounts of two bands on dna. (dna)n + dntp > (dna)n+1 + ppi. Each of the deoxynucleotides (datp, dgtp, dctp, and dttp) plus mg2+ Dna polymerase (primarily dna polymerase iii in prokaryotes) A priming sequence (with a free 3"-oh group) Elongation of the dna strand proceeds in the 5"-to-3" direction. The final product is a complete copy of the template. High fidelity of dna replication dntp binding changes enzyme conformation.

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Related Questions

The DNA phage QX174 was found to have a ratio of bases of 25A:33T:24G:18C. This departs from the usual A/T = 1 and G/C = 1. How can you explain this observation?





b. Phage P1 has double stranded DNA with 91,500 base pairs (91.5 Kb)
i. Approximately how many double-helical turns does this DNA contain?




ii. How many amino acids can be coded by this DNA? What is the molecular weight of the expected polypeptide in kilodaltons? (One amino acid has a molecular mass of 100 daltons)




iii. What is the molecular weight of this DNA? (Assume that the average base pair weighs 660 Daltons).



iv. If P1 phage original DNA is labelled with 15N and transferred to 14N medium to grow for six generations, What % of the DNA strands will be labelled with 15N if DNA replication is semiconservative.



c. In DNA denaturation and renaturation, what is a Cot curve and what factors influence the renaturation efficiency of DNA? Draw a DNA melting curve, label the axes, and point out the Tm. What are the factors that would influence the T m of your hypothetical DNA

1. In addition to identifying the genetic material, the experiments of Avery, MacLeod, and McCarty with different strains of Streptococcus pneumoniae demonstrated that
DNA is a double helix.
DNA may be taken up by bacterial cells and be active.
DNA is present in bacteriophage T2.
eukaryotic cells may be genetically transformed.
DNA binds to the cell wall of the bacterium.
2. In order to show that DNA in cell extracts is responsible for genetic transformation in Streptococcus pneumoniae, important corroborating evidence should indicate that _______ also destroy transforming activity.
temperatures high enough to kill cells
enzymes that hydrolyze proteins
enzymes that hydrolyze DNA
enzymes that hydrolyze polysaccharides
enzymes that hydrolyze RNA
3. Based on what you have learned about the experiments conducted by Griffith and Avery and colleagues with bacteria, which of the following would result in transformation of living R cells?
Pure RNA from S cells
Cell-membrane extracts from S cells
A mixture of pure RNA and protein from S cells
Cell-free extracts from S cells
Pure protein from S cells
4. A-T base pairs in a DNA double helix
form three hydrogen bonds with each other.
are more heat-stable than G-C base pairs.
are of a substantially different length than G-C base pairs.
are not accessible to DNA binding proteins.
are chemically distinct from G-C base pairs.
5. If 23 percent of the bases in a sample of double-stranded DNA are adenine, what percentage of the bases are uracil?
27
50
0
73
23
6. The uniform diameter of the DNA structure provides evidence for
antiparallel DNA strands.
a right-handed helix.
a double-stranded helix.
3′ end phosphorylation.
purine–pyrimidine pairing.
7. If a sequence of one strand of DNA is 5′-TGACTATC-3′, what is the complementary strand?
5′-ACTGATAG-3′
3′-TGACTATC-5′
3′-ACTGATAG-5′
3′-ACTGATAG-3′
5′-TGACTATC-3′
8. What structural aspect of the DNA facilitates dissociation of the two DNA strands for replication?
3′ end phosphorylation
Purine–pyrimidine pairing
Antiparallel DNA strands
Covalent bonds between sugars and phosphate groups
Hydrogen bonding between paired bases
9. If the Meselson–Stahl density gradient experiment had resulted in two bands of DNA molecules after only one round of replication, one containing only 15N and the second only 14N, this result would have indicated that replication was
unidirectional.
dispersive.
conservative.
bidirectional.

semiconservative.
10. The nucleoside analogue acyclovir, which is used to treat herpes simplex virus (HSV) infections, lacks a 3′ hydroxyl group (–OH). Predict what will happen if the host cell DNA polymerase incorporates a molecule of acyclovir into an elongating strand of HSV DNA.
The replication complex will no longer bind to origins of replication.
The DNA polymerase will only incorporate acyclovir molecules.
Acyclovir will bind single-strand binding proteins.
DNA polymerase will not be able to link a successive nucleotide.
Acyclovir will block the activity of the DNA helicase.
11. Which of the following does not demonstrate the stability of the DNA double helix?
Single-strand binding proteins are required to keep the strands apart.
Paired bases form hydrogen bonds.
High temperatures are required to denature it.
DNA synthesis requires energy.
DNA helicases require ATP.
12. What effect would a primase inhibitor have on DNA replication?
DNA polymerase would not be able to add bases to the DNA.
Primase would be blocked from joining the DNA replication complex.
Primase would not be able to provide primers for DNA polymerases.
DNA polymerase would not be able to use DNA as a template.
There would be no effect on DNA replication.
13. To replicate their DNA in a timely manner, most eukaryotic chromosomes
normally have uncoiled DNA.
have multiple origins of replication.
contain linear molecules of single-stranded DNA.
have two replication forks that move in the same direction.
are composed entirely of DNA.
14. Which statement about DNA replication is false?
Okazaki fragments are synthesized as parts of the leading strand.
Error rates for DNA replication are reduced by proofreading of the DNA polymerase.
The sliding clamp increases the rate of DNA synthesis.
Replication forks represent areas of active DNA synthesis on the chromosomes.
Ligases and polymerases function in the vicinity of replication forks.
15. In many eukaryotes, there are repetitive sequences called telomeres at the ends of chromosomes. After successive rounds of DNA replication, the _______ strand becomes shorter. In some cells, an enzyme called _______ repairs the shortened strand.
leading; telomerase
lagging; telomerase
lagging; DNA polymerase I
leading; DNA polymerase I
leading; replicase
16. A researcher studies normal human fibroblast cells. They can be maintained in culture but die off after about 30 cell generations. Unexpectedly, a colony of cells continues to survive and divide past 30 generations. Which scenario is most likely true for these cells?
DNA polymerase is constantly active.
Primase is able to extend the telomeres.
The mismatch repair system is extra efficient.
The telomerase gene is being expressed.
The cells do not need the ends of their chromosomes.
17. If DNA polymerase III introduces an incorrect nucleotide, what is the first corrective action made by the DNA repair system?
The replication complex excises the incorrect nucleotide.
The excision repair proteins remove the incorrect nucleotide.
The mismatch repair system scans for base-pairing mismatches.
There is no correction, because nucleotide errors by DNA polymerases are not repaired.
DNA ligase connects the correct nucleotide.
18. Choose the correct order of the following four events in the excision repair of DNA:
(1) Base-paired DNA is made complementary to the template.
(2) Damaged bases are recognized.
(3) DNA ligase seals the new strand to existing DNA.
(4) Part of a single strand is excised.
1, 2, 3, 4
2, 1, 3, 4
3, 4, 2, 1
2, 4, 1, 3
4, 2, 3, 1
19. Six complete cycles of PCR should result in a _______-fold increase in the amount of DNA.
64
32
12
128
6
20. When double-stranded DNA is heated to temperatures above 90°C, it denatures. Denaturation is a process that
breaks covalent bonds.
hydrolyzes DNA.
separates double-stranded DNA into two single strands.
causes new hydrogen bonds to form.
irreversibly destroys the double helical structure.

You decide to identify the CFTR mutation by analyzing the genomic DNA of your patients compared to healthy individuals. You specifically are looking to see whether a specific 3' gene truncation has occurred in the patients. You will determine this using hybridization techniques with samples from healthy and CF patients. Which of the following will allow you to accomplish this?

Using an RNA probe complementary to the region not removed by the truncation.

Using an RNA probe complementary to the region removed by the truncation.

Using an DNA probe complementary to the region not removed by the truncation.

Using an DNA probe complementary to the region removed by the truncation.

To conduct the hybridization experiment, you are trying to decide between using a DNA or RNA probe. Which would be ideal to use and why?

As both are composed of nucleic acids, using either would result in identical results.

An RNA probe because RNA has uracil bases.

An RNA probe because it could also be used in a translation experiment.

A DNA probe because it is more stable than RNA.

A DNA probe because RNA cannot bind to DNA.

Imagine Hershey/Chase had used an RNA virus (genome composed of RNA) instead of a DNA virus in their experiment. Would radioactivity still have been found in the pellet?

No, because only DNA can be labeled with radioactivity.

No, because the RNA genome would not enter the bacteria upon infection.

No, because while DNA and RNA nucleotides are similar, they are not identical.

Yes, because DNA and RNA nucleotides are similar.

Yes, because genome in any form (DNA, RNA, protein) would be labeled similarly.

The human genome consists mostly of non-coding DNA. Which of the following are benefits of this?

Random DNA mutations generally won't affect RNA and protein function.

It is faster to duplicate the genome when these are present.

The existence of introns can lead to multiple variations of proteins encoded by a single gene.

It is unlikely transposons would exist in the genome if there was too much protein coding DNA.

You accidentally add a mutant dNTP (which has an H instead of an OH connected to the 3’ carbon) to a reaction where DNA is being replicated. Which of the following is true of this mutant dNTP?

It can be incorporated into DNA strand but cannot form a phosphodiester bond with an incoming wild type dNTP

It can be incorporated into a DNA strand but cannot base pair with a complementary nucleotide

It can be incorporated into a DNA strand and can form a phosphodiester bond with an incoming dNTP, but only if it is another mutant dNTP

It cannot be incorporated into a DNA strand.

Andrew Murray's sister, Andrea, is adding to her brother's work on chromosomes. She is using cells that are unable to synthesize adenine (∆ade) and histidine (∆his). The plasmid she is currently working with consists of an origin of replication and the Ade gene.

Following her transformation of the plasmid into her yeast, what media will the cells be plated on to select for cells that have picked up the plasmid?

Media containing histidine

Media containing adenine

Media lacking adenine

Media lacking histidine

Andrew Murray's sister, Andrea, is adding to her brother's work on chromosomes. She is using cells that are unable to synthesize adenine (∆ade) and histidine (∆his). The plasmid she is currently working with consists of an origin of replication and the Ade gene.

She starts by attempting to add the centromere DNA into a plasmid containing the origin of replication. Unfortunately, when adding the centromere sequence, the origin of replication is removed, thus leaving a plasmid with only a centromere and selection marker. Following plasmid transformation, what growth result will she see on her plates?

No colonies

Little colonies

Big colonies

Andrew Murray's sister, Andrea, is adding to her brother's work on chromosomes. She is using cells that are unable to synthesize adenine (∆ade) and histidine (∆his). The plasmid she is currently working with consists of an origin of replication and the Ade gene.

She fixes the mistakes from the previous experiment and now has a complete plasmid (selection marker, origin of replication, centromere). She then inserts telomere sequences into the plasmid. How will this impact her transformation?

It will not impact her transformation

Her transformation will no longer work because plasmids don’t require telomeres

She will now see much larger colonies

She will now see fewer but larger colonies

She will now see smaller colonies

In the Meselson/Stahl experiment, E. coli were first grown in media containing heavy nitrogen, 15N, and then transferred to light nitrogen, 14N, at the beginning of the experiment.

Imagine that their data showed that replication occurs in a conservative manner instead of semi-conservative. What fraction of the DNA helices will consist of mixed DNA after 4 rounds of replication in this case?

None

More than 75%

25-50%

51-75%

Less than 25%