Microbiology and Immunology 2500A/B Lecture Notes - Lecture 20: Rna-Dependent Rna Polymerase, Baltimore Classification, Escherichia Coli

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lemonwater-buffalo91

29 Sep 2017

School

Western University

Department

Microbiology and Immunology

Course

Microbiology and Immunology 2500A/B

Professor

Dimitrios J Dikeakos, John Mc Cormick, Kelly Lynn Summers

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Document Summary

Hershey-chase experiment or proof the genome is key! Nucleic acid or protein (shell): allow bacteriophage to infect with radioactive precursors to dna or to protein, blend, separate, radioactive phosphorus could be detected in next generation of bacteriophage. This experiment determined that the genome is key: experiment done in parallel. Step 1: virus infect the bacteria: genome gets inside the bacteria cell, genome is exposed. Step 2: use a blender to actually separate the virus from the surface of the e. coli: this normally never happens, never detaches after its genome is inserted. Modern day hershey-chase experiment or proof the genome is key! Nucleic acid or protein (shell): bacteriophage mixed with cyanide dye (binds viral genomes, upon infection into bacterial host; dye leaves bacteriophage and enters host. Discovery and isolation of reverse transcription: 7 (seven groups) Key concept: viral genome must make mrna; this mrna must be read by host ribosomes. mrna is read in the 5" to 3" direction.

Related Questions

1. In order to demonstrate that DNA (and not protein) was thegenetic material, bacteriophage T4 were labeled with either 32P or35S and allowed to infect E. coli.

a. What does 32 P label?

b. What does 35S label?

c. The bacteriophage that were produced fromthis infection retained which label 32P or 35S?

d. How did this experiment show that DNA was the geneticmaterial?

2. Briefly, discuss the pros and cons of having a nucleoid (asbacteria do) versus a double nuclear membranesurrounding the DNA (as in eukaryotes).

3. Histones and other DNA binding proteins usually have a netpositive charge. Explain

scarletdog803

2.)

a.) Are viruses that infect Archaea different from other bacterial viruses?

Yes, they do not have protein coats because of the extreme conditions where they live.

No, they consist of nucleic acid and protein just like other bacterial viruses.

Yes, the structure of archaeal viruses is always completely different from other bacterial viruses.

Yes, they have only DNA genomes.

b.) Of the following examples, which is not a common part of many viral life cycles?

The pox virus genome is replicated by DNA polymerase.

Adenoviruses attach to the cell and the nucleocapsid is taken up by the cell.

Pox viruses replicate in the cytoplasm of the cell.

Once an adenovirus virion is inside a cell, host RNA polymerases transcribe early genes

c.) Some viruses are associated with cancer. There are a variety of ways in which viruses may affect a cell in a way that may lead to cancer. Based on your knowledge of the life cycles of viruses and of molecular genetics, what is one way that this might happen?

Viral genetic material inserts into a host genome close to the termination sequence, far downstream of the promoter, and causes a silent mutation.

Viral genetic material inserts into a host genome in a repetitive noncoding region.

Viral genetic material inserts into a host genome, placing a promoter in a new location and turning on a gene that had previously been turned off.

Viral genetic material inserts into a host genome and causes a mutation in a promoter, turning off a gene that had previously been turned on.

d.) Which statement describes the difference between the normal neuron protein PrP^C and the pathogenic version PrP^Sc?

PrP^C is just a protein and the PrP^Sc is a protein with nucleic acid inside.

The PrP^C protein contains more beta sheets while the PrP^Sc protein contains more alpha helices.

The three-dimensional conformation is identical, but the amino acid sequence is different.

The amino acid sequence is identical, but the three-dimensional conformation is different.

Multiple answers to each question might be possible!

You decide to identify the CFTR mutation by analyzing the genomic DNA of your patients compared to healthy individuals. You specifically are looking to see whether a specific 3' gene truncation has occurred in the patients. You will determine this using hybridization techniques with samples from healthy and CF patients. Which of the following will allow you to accomplish this?

	Using an RNA probe complementary to the region not removed by the truncation.
	Using an RNA probe complementary to the region removed by the truncation.
	Using an DNA probe complementary to the region not removed by the truncation.
	Using an DNA probe complementary to the region removed by the truncation.

To conduct the hybridization experiment, you are trying to decide between using a DNA or RNA probe. Which would be ideal to use and why?

	As both are composed of nucleic acids, using either would result in identical results.
	An RNA probe because RNA has uracil bases.
	An RNA probe because it could also be used in a translation experiment.
	A DNA probe because it is more stable than RNA.
	A DNA probe because RNA cannot bind to DNA.

One step of the Hershey/Chase experiment involved blending the virus/cell mixture before centrifugation and probing the pellet for radioactivity. Why was the blending step necessary?

	To collect the bacteria at the bottom of the tube.
	To break open the bacteria to release the genome.
	To separate the bacteria from the bacteriophages.
	To be able to detect the radioactivity.

Imagine Hershey/Chase had used an RNA virus (genome composed of RNA) instead of a DNA virus in their experiment. Would radioactivity still have been found in the pellet?

	No, because only DNA can be labeled with radioactivity.
	No, because the RNA genome would not enter the bacteria upon infection.
	No, because while DNA and RNA nucleotides are similar, they are not identical.
	Yes, because DNA and RNA nucleotides are similar.
	Yes, because genome in any form (DNA, RNA, protein) would be labeled similarly.

The human genome consists mostly of non-coding DNA. Which of the following are benefits of this?

	Random DNA mutations generally won't affect RNA and protein function.
	It is faster to duplicate the genome when these are present.
	The existence of introns can lead to multiple variations of proteins encoded by a single gene.
	It is unlikely transposons would exist in the genome if there was too much protein coding DNA.

Explain the 5â to 3â directionality of a DNA stand.

	It is due to the fact that the free 5â carbon is on one end and the free 3â carbon is on the other
	It is due to the fact that new nucleotide are added to the 5â carbon of the previous nucleotide
	It is due to the fact that there are 3 phosphate groups attached to the 5â carbon
	It is due to the fact the complementary strand is 3â to 5â
	More than one of the above explain the 5â to 3â directionality

You accidentally add a mutant dNTP (which has an H instead of an OH connected to the 3â carbon) to a reaction where DNA is being replicated. Which of the following is true of this mutant dNTP?

	It can be incorporated into DNA strand but cannot form a phosphodiester bond with an incoming wild type dNTP
	It can be incorporated into a DNA strand but cannot base pair with a complementary nucleotide
	It can be incorporated into a DNA strand and can form a phosphodiester bond with an incoming dNTP, but only if it is another mutant dNTP
	It cannot be incorporated into a DNA strand.

Why does DNA polymerase utilize an RNA primer?

	DNA polymerase is unable to initiate strand synthesis but RNA polymerase can
	DNA polymerase can only add a dNTP to an rNTP
	DNA synthesis proceeds in the 3â to 5â when initiating strand synthesis
	Chromosomal DNA contains interspersed RNA fragments
	The RNA primer increases stability of the newly synthesized strand

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Microbiology and Immunology 2500A/B Lecture Notes - Lecture 20: Rna-Dependent Rna Polymerase, Baltimore Classification, Escherichia Coli

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