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BIOLOGY 1A03 Chapter Notes - Chapter 14: Dna Ligase, Telomerase, Telomere

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14 Sep 2016

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BIOLOGY 1A03 Full Course Notes

BIOLOGY 1A03 Full Course Notes

Document Summary

The bacteriophage that they studied was t2. Viruses attached to the bacteria"s cell wall and injected its dna into the bacteria. Phosphate backbone found in the amino acid cisteine. When blending the bacteria, it does not destroy the bacteria. Using a virus that had a dna genome however some viruses have rna. Semi-conservative mechanism: old strands separate and becomes a template strand, hybrid created with n15 and n14, half low density and half intermediate density. Conservative: entirely new double helix, high density and low density. Dispersive: alternative pieces of dna (parent and daughter, all intermediate density. Removal of n15 before placing it in n14. N15 is found in the parental strand and therefore the parental strand would be at the bottom and the daughter strand will be floating at the top. Origin of replication: a specific sequence of dna that will occur on the plasmid; the location of the beginning of replication.

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Related Questions

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.

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_____________ are physical microscopicevidence of crossing over.

a. Recombinants

e. DNA double strand breaks

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In a 3 point cross, the recombinantphenotypes corresponding to ___________ are the leastfrequent.

a. single crossover events

b. F2 recombinants

c. double crossover events

d. parental recombinants

e. F1 recombinants

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You perform a tetrad analysis experiment inNeurospora to determine the position of a single gene (z) relativeto the centromere. You obtained the following numbers of tetradsordered from the top of the ascus to the bottom. What is the mapdistance between the z gene and the centromere in map units? 379 z+z+ z+ z+ z z z z 421 z z z z z+ z+ z+ z+ 65 z z z+ z+ z z z+ z+ 35z+ z+ z z z+ z+ z z 54 z+ z+ z z z z z+ z+ 46 z z z+ z+ z+ z+ z z(Note: if the genotypes don't align well in the online assignment,see the PDF that was uploaded as well)

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Injection of mice with Streptococcuspneumoniae that grow as smooth colonies in culture results in_________.

a. none of the other answers

b. chronic pneumonia

c. no phenotype

d. embryonic lethality

e. death of the mice

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The transforming principle studied by Avery,MacLeod, and McCarty could be destroyed or removed by_________.

c. ultracentrifugation

e. shear forces.

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The Hershey-Chase Waring blender experimentdetermined that _________________________.

a. bacteriophage replicate by semi-conservativeDNA synthesis.

b. all of the other answers are correct.

c. bacteriophage integrate into the hostgenome, becoming insensitive to external shear forces.

d. radiolabelled bacteriophage DNA enters hostcells during infection but that radiolabelled proteins do not

e. bacteriophage are made up of roughly equalamounts of protein and DNA using radioactive sulphur andphosphorous.

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All DNA polymerases synthesyze DNA in a__________.

a. N-terminal to C-terminal direction.

b. 5' to 3' direction.

d. 3' to 5' direction.

e. C-terminal to N-terminal direction.C-terminal to N-terminal direction.

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Meselson and Stahl used a _______ isotope of____________ to disprove the conservative and dispersive models ofDNA replication.

a. heavy, sulphur

b. radioactive, nitrogen

c. heavy, nitrogen

d. radioactive, sulphur

e. radioactive, phosphate

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DNA replication is initiated at________________ of replication.

a. semi-conservative sites

d. bidirectional sites

e. termination sites

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In bacteria, initiation of DNA replicationrequires ____________________.

a. primase to synthesize a short DNAprimer.

b. nuclear DNA binding proteins.

c. a replication fork.

d. a terminal 3'OH group on DNA.

e. primase to synthesize a short RNAprimer.

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DMC1 protein facilitates strand invasionduring _______.

a. double strand break repair.

b. resolution of holliday junctions.

c. branch migration.

d. meiotic recombination.

e. mitotic recombination.

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Resolution of holliday junctions in the sameplane results in _______.

a. genetic loss.

b. meiotic crossover.

c. failed crossover.

d. homoduplex formation.

e. bidirectional fork migration.