BIOA01H3 Lecture Notes - Lecture 23: Coiled Coil, Nucleosome, Chromatin

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16 Dec 2016

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Dna stores genetic information encodes for proteins that provide structure and do much of the work of the cell. Genetic information is stored in genes: turning on a gene, or gene expression, ultimately causes an effect on the organism. Gene regulation controls when this process should occur. Transfer of material from dead virulent to live non-virulent strain that made it virulent. Dna is extracted from heat-killed virulent cells, along with trace amounts of rna and protein. Three samples, one treated with protease, one with dnase and one with rnase. Only extracts treated with the enzyme that destroys dna were unable to transform nonvirulent bacteria. A binds with t by 2 hydrogen bonds, c binds with g by 3 hydrogen bonds. The sugar-phosphate backbones wind around the outside of the molecule and the bases point inward. Hydrogen bonding and base stacking contribute to the stability of dna. One is from mom and one is from dad in each pair.

Related Questions

Cystic fibrosis is caused by mutations in the CFTR gene, whichencodes for a chloride channel. You are studying cystic fibrosispatients to determine what mutation they possess in the CFTR gene.The difference between the mutant and wild type CFTR genes can beuncovered by examining the CFTR:

(1 point - partial possible)

DNA

mRNA

protein

You decide to identify the CFTR mutation by analyzing thegenomic DNA of your patients compared to genomic DNA of healthyindividuals. You specifically are looking to see whether a specificgene truncation has occurred in the patients. You will determinethis using hybridization techniques. Which of the following willallow you to accomplish this?

(1 point - partial possible)

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

Using an RNA probe complementary to the region removed by thetruncation.

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

Using an DNA probe complementary to the region removed by thetruncation.

You would like to ensure that this experiment (to determinewhether patients have a specific CFTR gen truncation usinghybridization) is properly controlled. Which of the followingsamples must you include?

(1 point - partial possible)

The genome of a healthy individual who does not have CFTR.

The genome of a CFTR patient known to have the specifictruncation you are trying to identify.

The genome of a CFTR patient with a point mutation in a separateregion of the gene but without the disease truncation.

The genome of a healthy individual married to a CFTR patientwith the specific truncation you are trying to identify.

The genome of a patient with muscular dystrophy, which can bedue to a trucation in the dystrophin gene.

To conduct the hybridization experiment, you are trying todecide between using a DNA or RNA probe. Which would be a betterchoice and why?

(1 point)

An RNA probe because RNA has uracil bases.

An RNA probe because it could also be used in a translationexperiment.

A DNA probe because it is more stable than RNA.

A DNA probe because RNA cannot bind to DNA.

Which of the following will lower the Tm of a given DNAstrand?

(1 point - partial possible)

Increasing the percentage of GC base pairs.

Raising the pH of the solution from extremely acidic toneutral.

Decreasing the salt concentration in the DNA solution from 50mMNaCl to 5mM NaCl.

None of the above.

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

(1 point)

A.No, because only DNA can be labeled with radioactivity.

B.No, because the RNA genome would not enter the bacteria uponinfection.

C. No, because while DNA and RNA nucleotides are similar, theyare not identical.

D. Yes, because DNA and RNA nucleotides are similar.

Griffith and Avery's transformation experiments allowed us toidentify that DNA is our genetic information. Which of thefollowing scenarios would result in bacterial cells that arecapable of killing mice upon injection?

(1 point - partial possible)

A. Heat killed non-virulent bacteria is added to a live virulentbacteria strain.

B. Heat killed virulent bacteria is added to a heat killednon-virulent bacteria strain.

C. A heat killed virulent bacteria that is treated with anuclease, is then added to a non-virulent bacteria strain.

D.Heat killed mouse cells are added to a non-virulentbacteria.

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

(1 point - partial possible)

A. Random DNA mutations generally won't affect RNA and proteinfunction.

B. It is faster to duplicate the genome when these arepresent.

C. The existence of introns can lead to multiple variations ofproteins encoded by a single gene.

D. Abundant non-coding genome is a protective feature againstthe consequences of transposition of mobile elements.

bronzewater-buffalo862

Question 1

1 pts

Cystic fibrosis is caused by nonsense and missense mutations in the CFTR gene, which encodes for a chloride channel. You are studying cystic fibrosis patients to determine what mutation they possess in the CFTR gene. The difference between the mutant and wild type CFTR genes can be uncovered by examining the CFTR:

	DNA
	mRNA
	protein
	tRNA

Question 2

1 pts

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.

Question 3

1 pts

You would like to ensure that this experiment (to determine whether patients have a specific CFTR gene truncation using hybridization) is properly controlled. Which of the following samples must you test?

	The genomic DNA of a healthy individual who does not have cystic fibrosis.
	The genome of a CFTR patient known to have the specific truncation you are trying to identify.
	The genome of a CFTR patient with a missense mutation but full length gene.
	The genome of a healthy individual married to a CFTR patient with the specific truncation you are trying to identify.
	The genome of a patient with muscular dystrophy, which can be due to a trucation in the dystrophin gene.

Question 4

1 pts

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.

Question 5

1 pts

Which of the following will lower the Tm of a given DNA strand?

	Increasing the percentage of GC base pairs.
	Raising the pH of the solution from neutral to basic.
	Decreasing the buffer concentration from 50mM NaCl to 5mM NaCl.
	None of the above.

Question 6

1 pts

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.

Question 7

1 pts

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.

Question 8

1 pts

Griffith and Avery's transformation experiments allowed us to identify that DNA is our genetic information. Which of the following scenarios would result in bacterial cells that are capable of killing mice upon injection?

	Heat killed non-virulent bacteria is added to a live virulent bacteria strain.
	Heat killed virulent bacteria is added to a heat killed non-virulent bacteria strain.
	A heat killed virulent bacteria that is treated with a nuclease, is then added to a non-virulent bacteria strain.
	Heat killed mouse cells are added to a non-virulent bacteria.

Question 9

1 pts

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.

Question 10

1 pts

Andrew Murray's sister, Andrea, is adding to her brother's work on chromosomes. She is using cells that are unable to synthesize adenine (âade) and histidine (âhis). The plasmid she is currently working with consists of an origin of replication and the Ade gene.

Following her transformation of the plasmid into her yeast, what media will the cells be plated on to select for cells that have picked up the plasmid?

	Media containing histidine
	Media containing adenine
	Media lacking adenine
	Media lacking histidine

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2.Conclusion:

What have you learned from your study?

Interpret your data/ results:

Do your results support of disprove your hypothesis?

Compare and contrast your result with those of previous research.

Significance of the results: âBig Pictureâ

What is the global impact of your results?

What questions remain unanswered?

What kinds of experiments can come out of your study?

i have lab for biology and i need help with the Lab Report

the lab name is Bacterial Transformation Lab

1. Introduction:

Introduce the background and the nature of the problem your experiment is investigating.

Summarize the current knowledge of your subject.

Use examples of other research that have examined the same or a similar topic as your experiments.

Ending your introduction:

State your hypothesis

State your predictions

Give a brief statement about how you tested your hypothesis.

please help me with it is important for my lab class tomorrow

that is the lab i put the info please help me for the intro and for number 2 please be clear please

i would be happy if you help me it is due tommorrow i realy need your help on the intro and number 2 please be clear and show clear work hope you do it because it is due tomorrow

Bacterial Transformation Lab

Genetic engineering is one of the newest applications of basic knowledge in biology. It involves the isolation of specific genes from one organism and the insertion of those genes into a second organism of the same, or even different, species. Remember that a gene is a piece of DNA which codes for a product, usually a protein, which gives an organism a particular trait. The development of these techniques has generated considerable excitement and answered many questions, but it has also raised questions. Many scientists and others see genetic engineering as offering opportunities to make life better, for example, by exploring how genes are regulated, by curing diseases of genetic origin, by adding desirable traits to agricultural crops such as the ability to fix nitrogen in corn and pest resistance in cotton, and by genetically altering bacteria with genes enabling them to digest oil spills.

For almost 100 years, it has been possible to isolate DNA from cells, but only in the last 40 years or so has its function been understood. Your textbook discusses the work of such investigators as Fred Griffith and Oswald Avery, who demonstrated that if DNA was isolated from a pneumonia-causing strain of bacteria and added to cultures of nonvirulent bacteria, the recipient cells were transformed into virulent types that caused pneumonia. These transformation experiments were the first to show that genes could be artificially transferred in the laboratory. Since the time of those experiments (Avery's work was done in the 1940s), our knowledge of gene structure, function and control has increased astronomically. The application of this knowledge has created the field of genetic engineering.

Modern genetic engineering techniques depend on three critical factors:

A suitable host organism in which to insert the foreign gene;

A vector to carry foreign genes into the host; and

A means of isolating the host cells that have taken up the foreign gene.

The host organism

The most commonly used host organism is the gut bacterium Escherichia coli (E. coli). Its chromosome is a single large circular DNA molecule, called the genomic DNA, containing about 4 x 10⁶ base pairs. Not every cell will take up foreign DNA when exposed to it, but E. coli grows rapidly, dividing in less than 20 minutes, so that a single cell can give rise to a billion genetically identical descendants in a matter of 10 hours. Thus, a few transformed cells can provide a large number of offspring within a day. The strain we are using is K-12.

The vector

A vector is a DNA molecule used as a vehicle to carry genes from one organism to another. This DNA molecule can be modified by gene-splicing techniques using restriction enzymes and ligases so that it carries the genes of interest (essentially a cut-and-paste operation). Sometimes viruses are used as vectors, but for E. coli, plasmids are often used. A plasmid is a small circular DNA molecule containing 1000 to 200,000 base pairs. Plasmids exist naturally in many strains of E. coli and other bacteria. Plasmids replicate as the cell grows, independent of replication of the genomic DNA. These plasmids are transmitted to each of the product cells during cell division. Plasmids carrying genetic information that is beneficial to the host cell are maintained in a given population by natural selection. Plasmids frequently carry genes that confer antibiotic resistance, an obvious benefit.

When E. coli cells are exposed to plasmids or any other foreign DNA, some of the cells, called competent cells, will take up the DNA. We can induce cells to become competent by treating them with divalent cations, such as Ca⁺⁺, followed by a brief exposure to high temperature, called a heat shock treatment. Exactly how this treatment causes cells to become competent is not understood, but it is described as making the membranes "leaky" so that large DNA molecules (like plasmids), which would not normally pass through the cell membrane, can enter the cell. This treatment does not make all cells competent, but it greatly increases the number of competent cells. Once in a cell, the plasmids replicate and are passed to the product cells during cell division.

The selective medium

The last requirement for genetic engineering is a means of isolating host cells that have taken up the gene of interest. The techniques used depend on the growth requirements of the host organism. For example, each species of bacteria has particular nutrient requirements. The composition of the growth medium (agar or broth) can be modified to allow growth of only certain bacteria -- for example, by adding or removing nutrients or other chemicals from the medium mixture. This modified medium is called a selective medium, because it selects for the growth of only some species. By introducing an antibiotic such as ampicillin into the medium in which the bacteria are grown, you will select for growth of only those bacteria carrying the antibiotic-resistance gene on newly acquired plasmids. In a matter of days, billions of antibiotic-resistant cells, all with the plasmid, can be grown.

Genetic engineers use a modification of this technique to insert a foreign gene into E. coli. In simplified terms, the plasmid (already carrying an ampicillin resistance gene) is modified by inserting the gene of interest. The gene of interest has been isolated from another source; for example, the gene for human insulin has been isolated from the human genome. The modified plasmid, carrying the inserted gene, is then mixed with competent E. coli and the bacteria are placed in a medium containing ampicillin. Only those cells that have the plasmid will grow in the medium, and they are also the cells carrying the insulin gene (in their plasmids). These selected cells can now be used to manufacture insulin, which can be used for insulin therapy by diabetics.

The genes

In the following experiment, you will insert a plasmid into E. coli. The plasmid you will use carries the gene bla, which codes for a protein (beta-lactamase) that is secreted by the bacteria, inactivating the ampicillin in the agar and allowing for bacterial growth. The plasmid also carries an indicator gene. The indicator gene we are using was isolated from a bioluminescent jellyfish, Aequorea victoria. It codes for Green Fluorescent Protein (GFP).

The GFP gene has been modified to include a regulator system that can be used to control the expression of the gene. Gene expression in all organisms is carefully regulated to allow for adaptation to differing conditions and to prevent wasteful overproduction of unneeded proteins. The genes involved in the breakdown of different food sources are good examples of highly regulated genes. For example, the sugar arabinose is both an energy source and a carbon source for bacteria. The bacterial genes that make digestive enzymes to break down arabinose for food are not expressed when arabinose is not in the environment. But, when arabinose is present, these genes are turned on. When the arabinose runs out, the genes are turned off.

Arabinose initiates transcription of these genes by promoting the binding of RNA polymerase. In the genetically engineered pGLO DNA, some of the genes involved in the breakdown of arabinose have been replaced by the jellyfish gene GFP. When bacteria with this plasmid are grown in the presence of arabinose, the GFP gene is turned on and the bacteria glow green in UV light. When arabinose is absent form the growth medium, the GFP gene remains turned off and the colonies appear white. This is an example of the central molecular framework of biology in action: DNA ? RNA ? protein ? trait.

In summary, you will render E. coli cells competent to take up plasmid DNA. The plasmid carries two identifiable phenotypic markers:

The beta-lactamase or bla gene codes for resistance to the antibiotic ampicillin. We will use this to select for bacteria that have taken up the plasmid.

The pGLO gene codes for Green Fluorescent Protein (GFP), which is switched âONâ if arabinose is present.

Safety Procedure

For your own safety and the safety of others, it is essential that you follow all instructions about handling bacteria and disposal of materials. When you finish work for the day, swab your table with disinfectant. WASH YOUR HANDS thoroughly with soap before leaving the lab for any reason, and at the end of the day after you have cleaned your table. Report any spills to your instructor.

List of Materials

Prepared agar plates (2 LB, 2 LB/amp, 1 LB/amp/ara)

Transformation solution (50 mM CaCl₂, pH 7.4)

LB broth

1 pack of inoculation loops

Sterile transfer pipets

1 foam microtube holder w/ tubes

E. coli starter plate

1 red biohazard bag for loop disposal (common use â Loops/pipets only - NO PAPER!!!)

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BIOA01H3 Lecture Notes - Lecture 23: Coiled Coil, Nucleosome, Chromatin

BIOA01H3 Full Course Notes

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