BIMM 110 Lecture Notes - Lecture 4: Knockout Mouse, Morris Water Navigation Task, Immunofluorescence

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magentawoodchuck427

25 Feb 2017

School

UCSD

Department

Biology/Molec Biol, Microbiol

Course

BIMM 110

Professor

Ella Tour

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Document Summary

Used to detect the presence of a protein of interest in the sample (cells, tissues, organs). First the sample proteins are separated by size using sds-page. Then the separated proteins are transferred on to a membrane. One can then use a probe specific to the protein of interest (such as an antibody specific to your target) to detect the presence/absence of the protein. Example: the question i am asking: does treating with drug x lead to a decrease in amyloid beta in the cortex of mice. I want to measure the amount of protein. So i can do a western blot on the homogenate of the brain cortex of mice treated or not treated with drug x and use antibodies specific to amyloid beta to detect the amount of amyloid beta present. Used to quantitate the amount of protein present in the sample (cells, tissues, organs).

Related Questions

HELP ASAP; Please help me understand my lab results, so be detailed to clearly understand

Then , Please include answers to the question below to include in your response

1. what role does ampicilin (amp) and arabinose (ara) play in the experiment.

2) 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 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.

The transformation procedure involves three main steps. These steps are intended to introduce the plasmid DNA into the E. coli cells and provide an environment for the cells to express their newly acquired genes. 1) Cells are first treated with a solution of CaCl2, which is thought to neutralize the repulsive negative charges of the phosphate backbone of DNA and the phospholipids of the cell membrane, allowing the DNA to adhere to the cells. This is referred to making the cells âcompetentâ (i.e. the cells are capable, or competent, to take-in exogenous DNA). 2) Cells are then subjected to a heat shock, which is thought to increase the permeability of the cell membrane, allowing the cells to take-in the plasmid DNA. 3) Cells are allowed to grow in a ârecoveryâ phase in the absence of ampicillin. During this time the cells begin to produce the Î²-lactamase protein, which will allow them to survive when they are subsequently placed on agar plates containing ampicillin.

For the first week section of lab

To eac

h 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 t

his transformation mix to the appropriate plates.

Results For plates:

+DNA: Has the E. Coli pGlo plasmid and - DNA plates do no have EColi pGLo plasmid .

Amp means ampiclin and Ara means arabinose on agar plates

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

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

--DNA LB/AMP- 1 colony so no colony growth and no glovw

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

Please explain significance of result or thand the reason for the results in scientific terms based on the background information provided and your own knowledge to be detailed. Would these results be expected as wll.

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

1) What is the most efficient and cost-effective way to control infectious diseases? A) variolation C) passive immunotherapy using immunoglobulins B) active immunization by vaccination D) immune testing E) autoimmunization 2) Variolation was first used _______ A) to immunize the Chinese against smallpox. B) to protect individuals against the plague during the Middle Ages. C) to spread smallpox throughout the Native American populations. D) to treat individuals exposed to hepatitis. E) for research purposes in the 20th century. 3) Almost a century after Edward Jenner introduced successful vaccination, Louis Pasteur developed vaccine(s) against A) anthrax. B) human cholera. C) rabies. D) influenza. E) both anthrax and rabies. 4) Pathogens may be attenuated for use in vaccines by ______ A) raising the pathogen for several generations in tissue culture cells. B) genetic manipulation. C) treatment with formaldehyde. D) genetic manipulation coupled with treatment with formaldehyde. E) genetic manipulation and/or raising the pathogen for several generations in tissue culture cells. 5) Which of the following statements regarding an inactivated vaccine is FALSE? A) It can be produced with deactivated whole microorganisms. B) It can be produced from antigenic fragments of a pathogen. C) It is safer than an attenuated vaccine. D) It is made from mutated forms of the pathogen. E) It is made from pathogens that cannot replicate. 6) Which of the following substances is commonly used to inactivate microbes before using them for vaccination? A) formaldehyde B) aluminum C) mineral oil D) saponin E) aluminum phosphate 7) What type of vaccine is the hepatitis B vaccine? A) inactivated whole pathogen B) attenuated vaccine C) toxoid vaccine D) a recombinant vaccine composed of a single antigen of the hepatitis virus E) a vaccine produced by treating the virus with formaldehyde 8) An infectious disease researcher isolates the pathogen responsible for an emerging disease. The microbe is grown in the lab for many generations. A preparation of the laboratory-grown microbe is treated with ionizing radiation and then tested for its potential as a vaccine. What type of vaccine is this? A) attenuated B) subunit C) combination D) toxoid E) inactivated whole 9) OPV, the attenuated oral poliovirus vaccine, is no longer administered in the U.S.A. because it A) is very toxic. B) does not provide good immunity. C) can revert to wild-type virulence. D) can be spread to contacts. E) can be spread to contacts and can revert to wild-type virulence. 10) Hybridomas are factories for antibodies and are produced by _____ A) combining two virus-infected cells. B) fusing plasma cells with myeloma cells. C) repeated culture of a pathogen until it loses its virulence. D) combining a viral infected cell with a bacterial infected cell. E) combining two bacterial infected cells. 11) The study and diagnosis of antigen-antibody interactions in the blood is known as _____ A) immunology. B) hematology. C) serology. D) cytology. E) histology. 12) Which type of antibody assay is represented in this figure? A) an indirect ELISA C) an indirect immunofluorescence assay B) a direct ELISA D) a western blot E) an immunodiffusion assay 13) Titration is a serological procedure that A) identifies the causative microbe of an infectious disease. B) determines the amount of an antibody in the blood. C) must be done before the western blot test to diagnose HIV. D) is used for blood grouping. E) has been replaced by genetic engineering in isolating the antigen of a pathogen. 14) A woman uses a home pregnancy test kit that tests for hCG hormone in urine. She knows this is a type of antibody assay from the kit brochure. Antibodies reacting with the hormone produce two lines on the test strip. What specific type of antibody assay does this represent? A) a direct immunofluoresence test C) an immunochromatographic assay B) a complement fixation test D) an ELISA E) a neutralization assay 15) The complement fixation test uses red blood cells as the target for complement activation. Test serum containing antibodies is combined with a known amount of antigen in a tube, and then the RBCs and antibodies against the RBCs are added. A positive result for the complement fixation test would be A) a line of precipitate near the bottom of the tube. C) loss of color in the tube. B) a cloudy solution in the tube. D) a fluorescent precipitate. E) a solution that is clear due to precipitation of RBCs.

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BIMM 110 Lecture Notes - Lecture 4: Knockout Mouse, Morris Water Navigation Task, Immunofluorescence

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