BIOL 401 Lecture 14: EMSA

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greenfly749

23 Mar 2017

School

Binghamton University

Department

Biology

Course

BIOL 401

Professor

Steven P Tammariello

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Document Summary

Electrophoretic mobility shift assay: technique that verifies protein-dna binding. Looks at transcription factors, particularly at dimers binding to dna. Protein binding to labeled dna creates a larger complex and shifts band up in gel. This all has to do with the molecular mass of biomolecules running through a gel-it"s shifting up higher. Remember that just because a cell is producing a protein, it doesn"t mean the aren"t always active! protein is being used in the cell. If no binding occurs the probe will be at the bottom of the gel. This overall process is very important because it tells us what proteins bind to what sequences of dna. This shift shows something is binding to our dna. Theoretically, if three bands form, there could be more transcription factors binding. He could give a question to explain what"s happening. Hsp: proteins found in organisms placed at a high temperature.

Related Questions

You further investigate the regulatory proteins that bind to Regions 1, 2 and 3 using gel shift assays. A gel shift assay uses labeled DNA as a target for regulatory protein binding.

A labeled DNA fragment that fails to bind a protein will migrate quickly on an electrophoretic gel.

A labeled DNA fragment that does bind a protein will migrate more slowly because the DNA + protein complex is larger than the DNA alone. Thus, the label associated with the DNA will âshiftâ to a higher location on the gel, as shown in this diagram.

To investigate regulatory protein binding to Regions 1, 2, and 3, you will use labeled DNA fragments with some of the same point mutations as in Part B, incubate them with proteins extracted from heart or lung tissue, and conduct a gel shift assay. How would gel shift assay results be affected by these mutations?

Select the two correct answers.

1. Region 2 binds protein from lung tissue. If Region 2 is mutated, then the DNA will not shift on the gel when incubated with protein from lung tissue.

2. Regions 1 and 3 bind protein from heart tissue. If both Regions 1 and 3 are mutated, then the DNA will not shift on the gel when incubated with protein from heart tissue.

3. Region 2 binds protein from heart tissue. If Region 2 is mutated, then the DNA will not shift on the gel when incubated with protein from heart tissue.

4. Regions 1 and 3 bind protein from lung tissue. If both Regions 1 and 3 are mutated, then the DNA will not shift on the gel when incubated with protein from lung tissue.

viridiancod415

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

tealcockroach755

HELP ASAP;

Please help me understand my lab results, so be detailed to clearly understand . INclude answers to these questions what role does ampicilin (amp) and arabinose (ara) play in the experiment. Because GFP was isolated from jelly fisht, what modification to the GFP DNA had to be made to result in expression in E. coli?

BACKGROUND INFORMATION

Transformation of GFP and the pGLO Plasmid

GFP is a commonly used reporter protein in research labs, as the fluorescence creates a marker protein that can be used in many types of cell biology and biochemical studies. In basic research, GFP is often fused to a specific target protein of interest, creating a chimeric reporter protein. GFP has been used as a reporter protein to study blood vessel and tumor progression in mice, brain activity in mice, and malaria eradication in mosquitoes for instance (see website mentioned in notes).

In the first week you will use a procedure to genetically transform bacteria with the gene that codes for Green Fluorescent Protein. Following the transformation procedure, the bacteria express their newly acquired jellyfish gene and produce the fluorescent protein, which causes them to glow a brilliant green color under ultraviolet light. During the second week you will purify the protein away from all of the other proteins produced by the bacteria using chromatography, and during the third week you will evaluate the success of the purification procedure and estimate the size (molecular weight) of GFP by using sodium dodecylsulfate polyacrylamide gel electrophoresis (SDS-PAGE).

Fundamental to the process of genetic transformation of bacteria are plasmids. In addition to one large chromosome, bacteria naturally contain one or more small circular pieces of DNA called plasmids. Plasmid DNA usually contains genes for one or more traits that may be beneficial to bacterial survival. In nature, bacteria can transfer plasmids back and forth allowing them to share these beneficial genes. This natural mechanism allows bacteria to adapt to new environments. The occurrence of bacterial resistance to antibiotics is due in large part to the transmission of plasmids which contain antibiotic resistance genes.

The pGLO plasmid (Fig. 5, not a naturally occurring plasmid) contains the gene for GFP, and also contains the gene for the enzyme Î²-lactamase (bla), which provides resistance to the antibiotic ampicillin. The Î²-lactamase protein is produced and secreted by bacteria that contain the plasmid. Î²-Lactamase inactivates the ampicillin present in the LB nutrient agar, allowing bacterial growth. Only transformed bacteria that contain the plasmid and express Î²-lactamase can survive on plates that contain ampicillin. Only a very small percentage of the cells take up the plasmid DNA and are transformed. Untransformed cells cannot grow on the ampicillin selection plates. pGLO also incorporates a special gene regulation system, which can be used to control expression of the fluorescent protein in transformed cells. The gene for GFP can be switched on in transformed cells by adding the sugar arabinose to the cellsâ nutrient medium. Selection for cells that have been transformed with pGLO DNA is accomplished by growth on antibiotic plates. Transformed cells will appear white (wild-type phenotype) on plates not containing arabinose, and fluorescent green when arabinose is included in the nutrient agar medium.

for the first week section of lab

To each of two microfuge tubes, add 250ul of transformation buffer. Keep these tubes on ice.

Using a sterile loop, pick one colony of bacteria from the LB plate.

Transfer this colony to one of the microfuge tubes. Gently vortex the tube until the colony is dispersed. Repeat for the second tube.

Add 0.08ug of DNA to one of these tubes (+DNA tube).

Incubate both tubes on ice for 10 minutes.

Incubate both tubes at 42Â°C for 50 seconds.

Incubate both tubes on ice for 2 minutes.

Add 250ul of LB broth to each tube and incubate at room temperature for 10 minutes.

Flick each tube gently. Add 100ul of this transformation mix to the appropriate plates.

Results:

+DNA: LB/amp--> 1 colony, Glow: uncertain

+DNA: LB/amp/ara--> 9 colonies, Glow: yes

-DNA: LB/amp--> 7 colonies, GLow: no

-DNA: LB--> many colonies, Glow: no

BIOL 401 Lecture 14: EMSA

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