BIOA01H3 Lecture Notes - Lecture 2: Dna Replication, Telomerase, Okazaki Fragments

41 views5 pages

viridianworm191

13 Nov 2017

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

3363410481 and 37710 others unlocked

BIOA01H3 Full Course Notes

Verified Note

18 documents

Document Summary

Replication origins: cells can do proofreading for a couple of hours because it begins replication simultaneously in many spots in our genome. Linear molecules often need to have multiple replication origins in order for us to complete active rounds of dna replication. Not only prokaryotes, but also organelles within eukaryotic cell (for instance, mitochondria and chloroplast have their own. Dna) also have dna that resembles this type of circular molecule, they also replicate dna in this fashion. Linear dna after each round of replication (end replication problem) we"re looking at the template strands and it"s the very end of the replication fork, where we"re going to arrive at the very ends of that template dna strand. The newly synthesized dna on the bottom is going to be synthesized continuously, there"s the leading strand, and the leading strand will be synthesized all the way to the very end to that template. Even if we"re to imagine that when we remove that.

Related Questions

Multiple choice.

1.Origins of replication...

A. occur at many sites on bacterial and human chromosomes

B. contain one replication fork each where the daughter stands are being sythesized

C. initiate replication in a uni-directional manner

D. result in bubbles opened by helicases

E. occur at one site per chromosome

2.Telomerase...

A. is important for the replication of circular chromosomes

B. is important for the replication of linear chromosomes

C. carries a template in the form of a DNA molecule

D. sythesizes RNA primers

E. both the 1st and 2nd option are correct

3.What are topoisomerases?

A.Enzymes that lay down an RNA primer during DNA synthesis.

B.The ends of the chromosomes.

C. Proteins that unwind DNA.

D. A strain of mice.

4.Which enzyme produces the first piece of nucleic acid synthesized in DNA replication?

A. Primase

B. DNA polymerase

C.ORC

D. DNA ligase

5. Which of the following does not contribute to the mutagenesis of our DNA?

		A. alterations to bases (nucleotides) that change their hydrogen bonding characteristics
		B. passing on of mismatched DNA into daughter cells
		C. inefficiency of MutS to recognize errors in base pairing
		D. DNA polymerase slipping during synthesis and creating insertions or deletions
		E. DNA polymerase erroneously synthesizing DNA in the reverse direction during S phase

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.

10.

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.

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 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.

QUESTION 1

A mutation caused by exposure to gamma rays is called a spontaneous mutation

	a.	True
	b.	False
	c.	Sometimes, depending on the type of base change
	d.	Gamma rays do not cause mutations

QUESTION 2

A recessive mutation affecting an essential biochemical pathway can be detected________

	a.	when the organism is heterozygous for that mutation.
	b.	when the organism is homozygous for that mutation.
	c.	in either homozygous or heterozygous conditions.
	d.	only by sequencing (recessive mutations never show a phenotype).

QUESTION 3

During DNA replication, the newly synthesized chain grows by adding

	a.	nitrogenous bases
	b.	the sugar-phosphate backbone
	c.	nucleosides
	d.	dNTPs

QUESTION 4

During replication, the DNA strand 5â- AAGTCTAGCCTAG -3â will serve as a template for the polymerization of:

	a.	5â- CTAGGCTAGACTT -3â
	b.	5â - TTCAGATCGGATC -3â
	c.	5â - GATCCGATCTGAA -3â
	d.	3â - AAGTCTAGCCTAG -5â

QUESTIOn 5

Given the following DNA sequences, select which statement(s) is(are) correct.

Sequence 1:

5â-TGGACGCTAA-3â

3â-ACCTGCGATT-5â

Sequence 2:

5â-AATCGCAGGT-3â

3â-TTAGCGTCCA-5â

Sequence 3:

5â-ACCTGCGATT-3â

3â-TGGACGCTAA-5â

	a.	Sequences 1 and 2 are the same
	b.	Sequences 1 and 3 are the same
	c.	Sequences 2 and 3 are the same
	d.	Sequences 1, 2 and 3 are all different

QUESTION 6

Repetitive sequences in the genome are hotspots for:

	a.	Deamination
	b.	Depurination
	c.	Thymine dimer formation
	d.	Replication errors

QUESTION 7

Select which statement(s) about the Ames test is(are) correct

	a.	it has been designed to understand mutation repair systems in Salmonella typhimurium
	b.	it uses mammalian liver extract
	c.	it allows to study whether chemical compounds or their enzymatic breakdown products are mutagenic
	d.	it is based on whether a chemical compound causes reversion mutations from his+ to his-
	e.	two of the above are correct
	f.	three of the above are correct

QUESTION 8

Select which statement(s) is(are) correct.

	a.	All DNA strands have a direction, and it is specified by the carbons in the sugar backbone.
	b.	All DNA strands have a direction, and it is specified by hydrogen bonds between nucleotides.
	c.	In order to form proper base pairs in a double stranded DNA molecule, the two strands must run in opposite directions.
	d.	A and C are correct.
	e.	B and C are correct.

QUESTION 9

Studies of gene mutation frequencies have shown that:

	a.	mutations are rare, and genomes are generally stable.
	b.	mutation frequencies differ among organisms and also between genes, suggesting certain genes are more susceptible to mutation.
	c.	mutation frequencies are consistent between organisms, and each region of DNA is equally susceptible to random mutations.
	d.	Both A and B are correct.
	e.	Both A and C are correct.

QUESTION 10

The compound 5-bromodeoxyuridine (BrdU) is a derivative of uracil, and if BrdU becomes incorporated during DNA replication, it pairs with adenine. This compound is best classified as which type of mutagen?

	a.	base analog
	b.	base inducer
	c.	intercalating agent
	d.	oxidative agent
	e.	alkylating agent

QUESTION 11

The rate of mutation of the fruit fly is higher than the rate of mutation of Algae

True

False

QUESTION 12

Thymine dimers are most commonly caused by which of the following?

	a.	X-rays
	b.	Alkylating agents
	c.	U.V. irradiation
	d.	DNA intercalating agents

QUESTION 13

What chemical group is found in the 3â end of a DNA strand?

	a.	an alcohol
	b.	a hydroxyl
	c.	a methyl
	d.	a phosphate

QUESTION 14

What chemical group is found in the 5â end of a DNA strand?

	a.	an alcohol
	b.	a hydroxyl
	c.	a methyl
	d.	a phosphate

QUESTION 15

Which is the correct order of molecules binding to DNA during DNA replication?

	a.	Helicase, SSB, primase, DNA pol III, DNA pol I, ligase
	b.	SSB, DNA pol I, ligase, helicase, DNA pol I, primase
	c.	primase, helicase, DNA pol III, ligase, DNA pol I, SSB
	d.	SSB, helicase, primase, DNA pol III, DNA pol I, ligase

QUESTION 16

Which statement(s) is(are) correct about strand slippage?

	a.	It can cause disorders such as Huntingon disease.
	b.	It is a process that causes mutations altering the number of DNA repeats.
	c.	It is a process that incorporates nucleotide base analogs and trinucleotide repeats.
	d.	a and b are correct.
	e.	b and c are correct.
	f.	a, b and c are correct.

help needed asap.

chocolatesea-lion434

BIOA01H3 Lecture Notes - Lecture 2: Dna Replication, Telomerase, Okazaki Fragments

BIOA01H3 Full Course Notes

Document Summary

Get access

Related Documents

BIOA01H3 Lecture Notes - Lecture 6: Spindle Apparatus, Sister Chromatids, Nuclear Membrane

BIOA01H3 Midterm: Module 2 Review

BIOA01H3 Chapter Notes - Chapter 8: Nucleolus, Telophase, Telomere

Related Questions