BIOL 2000 Lecture Notes - Telomerase, Telomere, Chromosome

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3 Apr 2013

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Timing of replication in eukaryotic cell cycle is once. Timing control through cdc6 and cdt1: also ensres if only happens once, only in g1. G1 phase: where pre-replication complex is formed, ready to go for synthesis, orc+ cdc6 + cdt1 + helicase. Helicase is required for replication to occur for lining up . The other proteins inhibit helicase from proceeding. Cdt1 inhibited: replication forks are made from helicase. All the components required for dna replication: main functions. Progressively unwind the helix and keep strands separated. Catalyze main dna synthesis (phosphodiester bond formation) on both leading and lagging strands. Need to synthesize both strands at the same time. Main polymerase components: where most of dna synthesis is occurring. To read both strands at the same time. Synthesis must move in opposite direction of the replication fork. Pulls the dna strand through the polymerase in the bottom, out the top for lagging strand. Every okazaki fragment requires a primer: exam question.

Related Questions

QUESTION 1. Choose the true statement about prokaryotic and eukaryotic DNA replication.

A.Eukaryotic DNA replication involves less protein factors than does prokaryotic DNA replication.

B. Prokaryotic DNA polymerases replicate DNA 5' to 3', while eukaryotic DNA polymerases replicate DNA 3' to 5'.

C. Eukaryotic DNA replication begins at multiple origins of replication, while prokaryotic DNA replication begins at a single origin of replication.

D. Prokaryotic genomes take longer to replicate than eukaryotic genomes.

QUESTION 2. Bacterial cells use DNA replication to:

A. make back-up copies of their genome in case of a mutation.

B. manifest multiple phenotypes at once.

C. copy their genetic material prior to binary fission.

D. build functional proteins.

QUESTION 3. Choose the true statement about the genetic code.

A. A cell's mRNA code is read four nucleotides at a time to specify a single amino acid.

B. The genetic code encodes 64 amino acids, one for each of the 64 corresponding codons.

C. The genetic code is redundant, with multiple codons specifying a single amino acid.

D. The genetic code varies widely among organisms, with each having a different set of codons for a corresponding amino acid.

QUESTION 4. Of the following enzymes, which are correctly matched with their function? To be marked correct, you'll need to select all correctly matched statements, as there may be more than one correct answer.

A. Helicase, relieves torsion stress as DNA unwinds

B. Ligase, covalently bonds adjacent Okazaki fragments

C. Primase, removes RNA primers

D. DNA polymerase III, builds most of the DNA on the leading and lagging strand

QUESTION 5. Which of the following contributes to DNA replication accuracy? To be marked correct, you'll need to select all true statements, as there may be more than one correct answer.

A. The capacity to proofread improves DNA replication accuracy.

B. A slow rate of DNA replication contributes to accuracy.

C. The use of primase improves DNA replication accuracy.

D. DNA replication accuracy is improved thanks to complementary base pairing rules.

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.

BIOL 2000 Lecture Notes - Telomerase, Telomere, Chromosome

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