BIO130H1 Lecture Notes - Lecture 5: Archaea, 18S Ribosomal Rna, Transcription Factor Ii H

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Stranded-directed mismatch repair in eukaryotes: this is a post polymerase error repair process. (after dna polymerase and before ligase) Okazaki fragments: mutl (after locating new strand and error) will cut out the entire section of dna and allow dna polymerase to re-synthesize. In e. coli, (prokaryotes), you look at new synthesized strand through the presence of methylated adenine. Only parent strands will have methylated adenine (new strands does not) Damaged dna: even after synthesis, dna can get damaged and need repair, defects in repair mechanisms have been linked to a variety of human diseases, examples, breast cancer, colon cancer, skin cancer. Dna can be damaged by: oxidation, radiation (uv, heat, chemicals, as well as other cell stressors. Thymine dimer is when two neighbouring pyrimidines covalently bond instead of the normal complimentary hydrogen bonds. (commonly caused by uv) Depurination: purine (guanine, adenine) reacts with water (very rare), and release the purine, which will result in a phosphate sugar missing a base.

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.

Question 4

Using the terms provided, complete the statements below. Some terms may apply more than once, while others may not apply at all. (15 points)

Guanine

histone

Double helix

Adenine

conservative

cytosine

Okazaki fragments

Thymine

introns

Leading strand

RNA polymerase

5-prime to 3-prime

Uracil

chromatin

DNA ligase

Semi-conservative

promoter

Lagging strand

DNA polymerase

nucleosomes

Electron transport chain

enhancer

Watson and Crick determined that DNA exists in the form of a _______________, where two antiparallel chains of nucleotides wind around each other. The nitrogenous bases project to the interior where they hydrogen bond in specific pairs, ___________ with __________ and _____________ with ____________.Meselson and Stahl demonstrated that DNA is replicated by the _______________ model in which the parent molecule unwinds and each strand serves as a template for synthesis of a new strand. Synthesis of DNA is carried out by _________________, which builds the new strands in the ___________ direction. While the ______________ grows continuously; the _______________ is built in short sections called _________________, which will eventually be joined together by __________________. Eukaryotic __________________ is composed of DNA and _____________ proteins that bind together forming ________________________, the basic units of DNA packaging.

Question 5

In the table below, predict (yes or no) whether or not the E. coli lac operon will be transcriptionally active in the presence or absence of glucose or lactose as indicated and respond to questions "a" and "b."

Lactose	Glucose	Lac expression?
No	Yes
Yes	Yes
Yes	No

Explain each of your answers in terms of the molecular mechanisms that are known to underlie the regulation of the lac operon. Which mechanism is considered to be negative control and which is considered to be positive control? Explain.

Question 6

Use the genetic code table in your textbook to aid you in answering the questions below. (15 points)

a.) Use the genetic code table to deduce the amino acid sequence of a protein encoded by the mRNA shown here:

5âAUGAUUGGAGGUUUGAUCGGGCAAUAGGGGUUUCAGUAAAUG3â

b.) Explain how the above sequence in "a" illustrates that redundancy of the genetic code.

c.) Explain what would happen if a mutational event caused the underlined "G" to be changed to a "U"? What is the name for this kind of mutation?

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BIO130H1 Lecture Notes - Lecture 5: Archaea, 18S Ribosomal Rna, Transcription Factor Ii H

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