HLTB22H3 Lecture Notes - Lecture 7: Growth Medium, Ribosomal Rna, Pentose

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8 Jan 2018

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Genes are also determinants of health, as like many other things. Central dogma of molecular genetics: dna transcription to rrna, mrna, trna, to ribosomes, translated to protein. These are what creates the genes in a human body. Which will stay in cell and in uence cell, be secreted by cell for other reasons. Virus gets into cell into start reproducing: in one beaker, had radioactive phosphorus, other one had s35, p32 marks genetic material. Radioactive material taken into cells: new virus being made which are packed full of radioactive phosphorus and radioactive sulfur; are still within e. coli. Tuesday, november 1, 2016: then took cultures and separated out 2 elements (viruses or e. coli) Checked what was inside each compartment: saw empty phages and in e. coli was the phosphorus labels, radioactive sulphur culture also had sulphur labels, experiment suggested that viruses taken up by cells/proteins taken into cells. Recall: phosphodiester bond: all nucleotides fused together by phosphodiester bonds.

Related Questions

What step in DNA replication precedes the pairing of complementary bases?

	A.	polymerization of DNA.
	B.	formation of the histone core.
	C.	mitotic division.
	D.	separation of the two strands.
	E.	joining of the two strands.

In eukaryotes, transcription occurs in the

	A.	ribosome.
	B.	mitochondria.
	C.	cell membrane.
	D.	cytoplasm.
	E.	nucleus.

Normal human body cells contain how many chromosomes?

	A.	46
	B.	22
	C.	23
	D.	44
	E.	42

Hershey and Chase performed an experiment in which they used radioactive isotopes to phosphorus and sulfur to label different components of bacterial viruses. Their results clearly showed that

	A.	the viruses did not contain protein.
	B.	genetic material normally contains radioactive isotopes.
	C.	the viruses did not contain DNA.
	D.	protein was entering into the infected bacteria.
	E.	DNA was entering into the infected bacteria.

A package of histones with DNA wrapped around them forms spherical structures called

	A.	nucleosomes.
	B.	lysosomes.
	C.	ribosomes.
	D.	nucleoli.
	E.	chromatin.

DNA replication

	A.	does not require proteins.
	B.	is constantly happening in a cell.
	C.	occurs in G1 of interphase.
	D.	takes place in the nucleus of the cell.
	E.	occurs in the cytoplasm of the cell.

The two subunits of the ribosome join during

	A.	promotion of transcription.
	B.	initiation of translation.
	C.	termination of transcription.
	D.	termination of translation.
	E.	elongation.

DNA replication of a single DNA molecule is referred to as semiconservative because

	A.	each of the two DNA molecules will consist of one parental strand and one newly synthesized strand.
	B.	all the DNA strands in the two DNA molecules will have both parental and newly synthesized DNA.
	C.	it results in two nonidentical DNA molecules.
	D.	of the two DNA molecules, one is made up totally of parental DNA, while the other is entirely newly synthesized DNA.
	E.	one DNA strand is replicated continuously, while the other must be replicated discontinuously.

The bases are bonded to what part of the backbone of the DNA molecule?

	A.	the 4â carbon atom of the sugar molecule.
	B.	the 3â carbon atom of the sugar molecule.
	C.	the 5â carbon atom of the sugar molecule.
	D.	the 1â carbon atom of the sugar molecule.
	E.	the 2â carbon atom of the sugar molecule.

1. Select all of the true statements regarding phageconversion.

-The incorporation of viral genesinto the host cell's DNA is an example of phage conversion.

-The expression of the incorporatedgenes and the resulting alteration of the bacterial phenotype is anexample of phage conversion.

-Cholera toxin genes have a viralorigin.

-The cholera toxin is produced bythe virus.

-The cholera toxin itself is transferred to Vibriocholerae by a virus.

2. Viruses grown in media with radioactive sulfur haveradioactive capsids. Why?

Nucleotides contain sulfur

Cysteine amino acids containsulfur

Capsid peptidoglycans containsulfur

Phospholipids contain sulfur

3. Integrase inhibitors work by:

Agglutinating the reversetranscriptase inhibitors

Preventing the incorporation of theviral DNA into the genome

Blocking the final viralassembly

Degrading the viral envelope

4. Once a prophage is formed, the viral genes are reproducedevery time the bacterium multiplies. The genes that encode thelytic cycle are held in check by a repressor protein. Stressinduces the formation of proteases that degrade the repressor. Howwould this affect the prophage?

Apoptosis would be induced.

The virus enters the lyticcycle.

The bacterial genome isdenatured.

The bacterium rejects the viral genes.

5. Can a virus be killed?

No, viruses are non-living andcannot die.

No, viruses are capable ofself-assembly and cannot be permanently killed.

Yes, separating the genes from thecapsid kills it.

Yes, destroying its genome kills it.

6. Which class of drugs would be effective against HIV but notagainst most other pathogenic viruses?

Translation inhibitors that stopassembly of viral proteins.

Nucleotide analogs that haltreplication.

Non-nucleoside reversetranscriptase inhibitors that block reverse transcription.

Receptor antagonists that block viral entry.

7. An unscrupulous malcontent writes a computer program andembeds it in an e-mail that promises lucrative investmentopportunities. The e-mail is sent to thousands of anonymous users,when they open it the program copies itself into the computer'shard drive. The hard drive begins to fill up with copies of theprogram, once it is completely full the computer crashes. Match thecomponents of this story with the components of a viral lifecycle.

1. e-mail containing virus	Virus
2. Innocent user	Binding site onhostcell
3. User's hard drive andCPU	Host cell'sribosomes
4. Program	Uninfected host cell

8. Vaccines work by priming the immune system and inducing theformation of antibodies and memory cells. When the actual virusinfects the host the host's immune system recognizes it andresponds to it. Which feature of viruses helps them thwart thissystem?

	The envelops prevents immune systemrecognition.
	Autocatalytic proteins replicatewithout genetic material.
	The high mutation rate of the viralgenome.
	Integration of viral genescamouflages the virus.
	Shedding the capsid inactivates theantibodies.

greensea-lion551

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.

HLTB22H3 Lecture Notes - Lecture 7: Growth Medium, Ribosomal Rna, Pentose

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