BIO130H1 Lecture Notes - Lecture 5: Telomerase, Proofreading, Telomere

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Leading strand synthesized continuously from single rna primer. Okazaki fragments = rna primer + dna. Dna synthesis goes in 5" to 3" direction. Telomeres also protects dna from degradation and keeps chromosomes from sticking together. Telomerase adds new repeat units of dna to. Lengthens 3" end of overhanging strand, allows dna polymerase to use newly synthesized segment as template to return. 5" end of complementary strand to previous length. Rna template in telomerase determines repetitive sequence that added to 3" end of parental strand. Rna template from telomerase resembles reverse transcriptase + generates g rich ends. Adds nucleotides to 3" ends of parental strands template. Lagging strand cannot finish off template because doesn"t have a surface for a primer to provide oh group. Problems at right hand side of linear chromosome. Leading strand can duplicate sequence all the way to end. But lagging strand can"t because no more template to put primer on.

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.

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.

Please answer all

Why does a template-dependent DNA polymerase require a primer to initiate DNA synthesis?

	DNA polymerase require 5' phosphate to add a new nucleotide
	DNA polymerase require 3' hydroxyl group to add a new nucleotide

DNA polymerase require energy by hydrolizing the primers

Which of the following enzymes is template independent?

	Reverse transcriptase
	Klenow fragment

	DNA polymerase
	Terminal deoxynucleotidyl transferase

RNA polymerase

The natural function of a DNA ligase enzyme is:

	Joining together DNA sequences with compatible cohesive ends from two different organisms to yield a recombinant DNA molecule
	Sealing single stranded nicks in double stranded DNA molecules

	Joining together two blunt ended fragments of DNA within a cell
	Joining together two single stranded DNA molecules

The natural function of 3' -> 5' exonuclease activity of a DNA polymerase is to:

	Remove the 5' end of the DNA strand that is being copied
	Remove damaged nucleotide from the template strand

	Remove nucleotides from the end of DNA to generate blunt ends
	Remove incorrect nucleotides from the newly synthesized DNA

Which technique is used to resolve the different sizes of DNA fragments?

	Gene cloning
	PCR

	DNA sequencing
	Gel electrophoresis

Which of the following vectors does not contain bacterial origin of replication (oriC)?

	plasmid
	cosmid

	bacterial artificial chromosome
	lambda vector

Which process of bacteriophage is not used for gene cloning?

	Concatemer formation
	Lytic cycle

	Lysogenic cycle
	viral packaging

What vector would be best suited for creating a contig of bovine (cattle) chromosome 10?

	lambda vector
	Cosmid vector

	P1 vector
	Plasmid

E. coli takes up plasmid DNA by which of the following methods?

	Transduction
	Transformation

	Conjugation
	Transfection

The following times and temperatures are an example of the steps for PCR. You can use the Figure to help you answer the following questions.

Why is the first step is carried out at 94Â°C?

	To degrade the template
	To denature the template

	To activate the polymerase
	To facilitate the primer binding

What happens in the reaction when the temperature shifts to 55Â°C during cycling?

	the DNA polymerase will carry out DNA synthesis by extending the annealed primers
	the DNA polymerase will finish DNA synthesis

	the primers anneal to the single-stranded regions of the DNA
	the primers anneal to the double-stranded regions of the DNA

During cycling, what occurs when the temperature is at 72Â°C?

	the primers anneal to the double-stranded regions of the DNA
	the DNA polymerase will carry out DNA synthesis by extending the annealed primers

	the primers anneal to the single-stranded regions of the DNA
	the DNA polymerase will finish DNA synthesis

You designed a set of primers to amplify 1 kb DNA with polymerase chain reaction. By which cycle of polymerase chain reaction, we would expect to see the first double stranded DNA with expected size?

	The end of cycle 1
	The end of cycle 2

	The end of cycle3
	The end of cycle 4

1. Which of the following is the correct sequence of events (from start to finish) in cellular fatty acid oxidation?

A. Fatty acid activationÃ carnitine shuttleÃ beta-oxidation

B. Beta-oxidationÃ carnitine shuttleÃ Fatty acid activation

C. Fatty acid activationÃ carnitine shuttleÃ beta-oxidation

D. Carnitine shuttleÃ Fatty acid activationÃ beta-oxidation

2. What reaction is catalyzed by lipoprotein lipase, or lipase?

A. Hydrolysis of TAGs into fatty acids and glycerol

B. Hydrolysis of TAGs into amino acids and glycerol

C. Hydrolysis of cholesterol in to fatty acid and water

D. Hydrolysis of cholesterol into acetyl-CoA

3. How do fatty acids of different carbon lengths cross into the mitochondrial matrix for oxidation?

A. Small FAs of 10 carbons or fewer can freely diffuse through the membrane

B. Large FAs are cleaved into two smaller units that are actively transported across the membrane

C. A membrane-bound transferase catalyzes the conversion of long-chain FAs and FA-carnitine esters, which are transported across the membrane via a specific carrier

D. The breakdown of large FAs starts outside the matrix, allowing for the smaller particles to pass easily through the membrane

4. How can any RNA nucleotide be easily distinguished from a DNA nucleotide?

A. RNA nucleotides have a unique phosphodiester bond linkage

B. RNA nucleotides have uracil nitrogenous bases

C. RNA nucleotides have an -OH on the 2â carbon of the ribose

D. RNA nucleotides cannot be easily distinguished from DNA nucleotides

5. Which of the following statement is true concerning the sequence pTACGTAACGTA?

A. This is an RNA sequence

B. In order to have a more rigid structure in a double-stranded molecule, the sequence given would need more adenosine

C. If the complementary strand were present, the number of purines would be equal to the number of pyrimidines

D. The amount of purines is equal to the amount to pyrimidines

6. Complementary base pairing is considered which level of structure?

A. Secondary

B. Tertiary

C. Quaternary

D. Primary

7. Which of the following is/are required for the polymerization reaction catalyzed by DNA polymerase III?

I. RNA primer

II. DNA primer

III. RNA template

IV. DNA template

V. NTPs

VI. dNTPs

VII. Magnesium

VIII. Calcium

A. II, IV, V, and VII

B. II, IV, VI, and VII

C. I, IV, V, and VIII

D. I, IV, VI, and VII

8. DNA polymerase I, from E. coli, has three different catalytic sites. Which of the following is NOT one of them?

A. RNA-dependent RNA polymerase

B. RNA-dependent DNA polymerase

C. DNA-dependent RNA polymerase

D. DNA-dependent DNA polymerase

9. Eukaryotes solve the problem of replicating the ends of their linear chromosomes by:

A. Using a protein as primer

B. Enzyme activity that uses an RNA template and reverse transcription in a series of repeat cycles

C. All of the listed solution

D. Terminal redundancy whereby each end of the chromosome is duplicated allowing recombination

BIO130H1 Lecture Notes - Lecture 5: Telomerase, Proofreading, Telomere

BIO130H1 Full Course Notes

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