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1. We can construct our gene library by cloning the PstI fragments into either plasmid pBR322 or phage lambda. Which one should we choose and why?

2. Suppose we want to insert the PstI fragments into the Bam HI site within the gene for tetracycline resistance (tetr) in plasmid pBR322. If both of these restriction enzymes produce cohesive ends, how can this be done?

3. After transforming E.coli with the plasmids, how can we identify the cells that contain a chimeric plasmid (containing DNA insert)?

4. If a million tetracycline-sensitive, ampicillin-resistance clones are grown on nutrients agar plates, how are we going to detect the rare clone or clones that carry the gene for protein P?

5. After selecting a clone carrying the gene for protein P, its ells are propagated to high density in nutrient broth. The chimeric plasmid (with an insert of the gene for protein P) is then extracted and purified from the rest of the cellular DNA. How can we now isolate the gene from the plasmid?

6. How can we demonstrate that the gene we have isolate is indeed the one for protein P?

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Collen VonLv2
28 Sep 2019
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Related questions

ANSWER THE FOLLOWING QUESTIONS(COULD BE MORE THAN ONEANSWER)

19) Describe the roles of M13 genes and their proteins:

a) Genes 1, 4 and 11 produces proteins required for assemblingthe the phage.

b)Genes 3.6,7,8 and 9 produces capsid proteins.

c)Gene 2 nicks the circular phage ssDNA to make it doublestranded by DNA polymerase

d)Gene 5 produces single strand binding proteins and gene 10protein makes M13 SSDNA circular.

e)Protein five bound single stranded circular M13 moved to theE. coli membranne and is released out coated will proteins 3, 6, 7,8 and 9 .

2) What is a phagemid and how does it work?

a)it is a recombinant DNA with a filamentous phage genome and aplasmid genome.

b)it carries two origins of replication: eg., oir f1 and oriColE1

c)May carry a gene for selection: e.g. Ampr gene.

d)May carry a marker gene; e.g, lacZ gene with promoter andregulator

e)May contain MCS, as is pUC18.

3)For storing functional gene for future applications you willneed the following:

a)A vector DNA

b) A cloning site in the vector matching with the ends of a geneto be cloned

c)A host organism capable to hold the recombined vector.

d)A marker gene present in the recombinant vector

e)A functional gene of interest.

24)identify which of the following is true for thebacteriophage lambda?

a)It has two major body parts, the head and the tail.

b)The genome is approximately 49 thousand bases long.

c) It has a linear dsDNA with single stranded 12 bases long costerminals at both

d) The single stranded lambda DNA cos terminals appear to be asfollows: (a) GGGCGGCGACCT and (b)CCCGCCGCTGGA

e)it can accept 7-10 kb long foreign DNA replacing itsnon-essential b2 region.

29) For developing a recombinant lambda DNA with a gene ofinterest and expressing the gene in a host you have to take thefollowing steps:

a)Remove the b2 region from lambda DNA using a restrictionenzyme

b)digest the gene of interest, outside its stop and startcodons, with the same restriction enzyme.

c)Recombine the lambda DNA digest with the digested gene ofinterest carrying complementary overhangs with ligase.

d)Form lambda particles by adding head proteins, tail proteinsand packaging enzynes to the recombined lambda DNA.

e)Add the packaged lambda particles to Agrobacetrium tumefaciensand incubate them in culture.

f)Add the packaged lambda particles to laboratory strains ofEscherichia coli and incubate them in culture.

32)The amino acids coded by our genes differ by thefollowing:

a)Polarity

b) Acidity

c)Basicity

d)Aroma

e)Hydrophobicity

38)What is an M13 phage and what are its specialty?

a)They are rod shaped Escherichia coli bacteriophage.

b)ts genome is made of a single stranded circular DNA.

c)Its DNA is 6400 nucloetides long coding for elevenproteins.

d)After infecting a bacterial cell it starts replicating in 10minutes producing 1,000 phages in an hour

e) soon its kills the host Escherichia coli and gets out to theenvironment.

40)You have a gene with BamH cut sites out side its start andthe stop codons. You can use the following procedures to prepare aclone of the gene in pBR322 and select the transformed E.Coli cellscarrying pBR322, except:

a) Digest both the gene and the plasmid pR322 with BamH1

b) Ligate the fragments using T4 DNA ligase.

c)Transform E coli with the reconstructed plasmid

d)Select the cells transformed cells grow in Lb-agar-ampicilinplates

e)Select the cells transformed cells grow inLB-agar-tetracycline plates

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 β-lactamase (bla), which provides resistance to the antibiotic ampicillin. The β-lactamase protein is produced and secreted by bacteria that contain the plasmid. β-Lactamase inactivates the ampicillin present in the LB nutrient agar, allowing bacterial growth. Only transformed bacteria that contain the plasmid and express β-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 cells’ 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 ‘competent’ (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 ‘recovery’ phase in the absence of ampicillin. During this time the cells begin to produce the β-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 42°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.

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