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BI346 Lecture Notes - Lecture 10: Origin Recognition Complex, Dna, Helicase

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ultramarinedinosaur805

5 Jul 2016

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First step to opening up region to allow helicases to bind: initiation proteins bind to origins. Initiator protein belong to aaa+ family, which means they are atpases associated with atp binding proteins. In ecoli, the initiator protein is dnaa. In eukaryotes is orc origin recognition complex. In eukaryotes its more complex, they have 6 subunits. Orc 1 is required for orc to bind to dna. The origin has an at rich sequence near by because of the 2 bonds holding it together instead of the 3 hydrogen bonds in c-g bonds, so its easier to split them apart. Only when they are bound, they can multerimize and form a spiral. As it binds to dnaa it distorts and makes it easier for at rich element to split and we get strand separation in at rich area (unwinds) Need to add the helicase inside the ring, but there e. coli is linear dna no end.

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Related Questions

Molecular Biology Help!

Histone chaperones play a role in replication by:

Histone chaperones play a role in replication by:

preventing specific interactions between the positively charged histones and negatively charged DNA.

preventing nonspecific interactions between the positively charged histones and negatively charged DNA.

facilitating specific interactions between the positively charged histones and negatively charged DNA.

facilitating nonspecific interactions between the positively charged histones and negatively charged DNA.

none of the above.

Histone chaperones play a role in replication by:

Histone chaperones play a role in replication by:

preventing specific interactions between the positively charged histones and negatively charged DNA.

preventing nonspecific interactions between the positively charged histones and negatively charged DNA.

facilitating specific interactions between the positively charged histones and negatively charged DNA.

facilitating nonspecific interactions between the positively charged histones and negatively charged DNA.

none of the above.

Histone chaperones play a role in replication by:

Histone chaperones play a role in replication by:

preventing specific interactions between the positively charged histones and negatively charged DNA.

preventing nonspecific interactions between the positively charged histones and negatively charged DNA.

facilitating specific interactions between the positively charged histones and negatively charged DNA.

facilitating nonspecific interactions between the positively charged histones and negatively charged DNA.

none of the above.

Histone chaperones play a role in replication by:

Histone chaperones play a role in replication by:

preventing specific interactions between the positively charged histones and negatively charged DNA.

preventing nonspecific interactions between the positively charged histones and negatively charged DNA.

facilitating specific interactions between the positively charged histones and negatively charged DNA.

facilitating nonspecific interactions between the positively charged histones and negatively charged DNA.

none of the above.

Histone chaperones play a role in replication by:

Histone chaperones play a role in replication by:

preventing specific interactions between the positively charged histones and negatively charged DNA.

preventing nonspecific interactions between the positively charged histones and negatively charged DNA.

facilitating specific interactions between the positively charged histones and negatively charged DNA.

facilitating nonspecific interactions between the positively charged histones and negatively charged DNA.

none of the above.

________________ may play a role in aging and cancer.

________________ may play a role in aging and cancer.

DNA polymerase III

restriction enzymes

none of the above

Telomeres consist of :

Telomeres consist of :

long random sequences at the ends of the linear chromosomes

short random sequences at the ends of the linear chromosomes

long TG-rich repeats at the ends of the linear chromosomes

long AT-rich repeats at the ends of the linear chromosomes

long AC-rich repeats at the ends of the linear chromosomes

Telomerase is a _____________________.

Telomerase is a _____________________.

restriction enzyme

reverse transcriptase

none of the above

Replication of the eukaryotic nuclear genome requires which of the following?

Replication of the eukaryotic nuclear genome requires which of the following?

DNA polymerase Î±

DNA polymerase Î³

DNA polymerase Î´

DNA polymerase Îµ

all but one of the above

Which of the following is the eukaryotic DNA replication initiator?

Which of the following is the eukaryotic DNA replication initiator?

replication protein A

replication protein C

The yeast ARS element has which of the following functions and properties:

The yeast ARS element has which of the following functions and properties:

they are AT-rich and serve as replication termination sites

they are GC-rich and serve as replication termination sites

they have random sequence and serve as replication termination sites

they are AT-rich and serve as replication initiation sites

they are GC-rich and serve as replication initiation sites

Replication factor C plays which of the following roles in eukaryotic DNA replication?

Replication factor C plays which of the following roles in eukaryotic DNA replication?

it is the sliding clamp

it is the clamp loader

it is the DNA ligase

it is a single-stranded DNA binding protein

it is a cofactor required for activating the DNA polymerase

Which of the following is the replicative DNA polymerase for both the leading and lagging strands of SV40?

Which of the following is the replicative DNA polymerase for both the leading and lagging strands of SV40?

DNA polymerase Î±

DNA polymerase Î²

DNA polymerase Î³

DNA polymerase Î´

DNA polymerase Îµ

SV40 T antigen recruits which of the following to the proto-replication bubble?

SV40 T antigen recruits which of the following to the proto-replication bubble?

DNA polymerase Î±

DNA polymerase Î²

DNA polymerase Î³

DNA polymerase Î´

DNA polymerase Îµ

The SV40 large tumor (T) antigen has which of the following functions:

The SV40 large tumor (T) antigen has which of the following functions:

helicase activity only

DNA primase and DNA helicase activity

origin binding and DNA primase activity

origin binding and DNA helicase activity

origin binding and DNA helicase activity

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.

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

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

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.

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

The uniform diameter of the DNA structure provides evidence for

	antiparallel DNA strands.
	a right-handed helix.
	a double-stranded helix.
	3â² end phosphorylation.
	purineâpyrimidine pairing.

If a sequence of one strand of DNA is 5â²-TGACTATC-3â², what is the complementary strand?

	5â²-ACTGATAG-3â²
	3â²-TGACTATC-5â²
	3â²-ACTGATAG-5â²
	3â²-ACTGATAG-3â²
	5â²-TGACTATC-3â²

What structural aspect of the DNA facilitates dissociation of the two DNA strands for replication?

	3â² end phosphorylation
	Purineâpyrimidine pairing
	Antiparallel DNA strands
	Covalent bonds between sugars and phosphate groups
	Hydrogen bonding between paired bases

If the MeselsonâStahl 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.

The nucleoside analogue acyclovir, which is used to treat herpes simplex virus (HSV) infections, lacks a 3â² hydroxyl group (âOH). 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.

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.

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.

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.

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.

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

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.

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.

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

Six complete cycles of PCR should result in a _______-fold increase in the amount of DNA.

	64
	32
	12
	128
	6

When double-stranded DNA is heated to temperatures above 90Â°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.