Question be used in blue/white screening without the use

Question 1: What isyour hypothesis for the cause of reduced cell growth and low enzyme production? The production of glucose by beta-glucosidase is a possible explanationfor why the enzyme yield is low.   The target gene is under control of a T7 promoter, and theT7 polymerase is under control of the lac operator.  Maximal transcription of the lac operonoccurs when glucose is absent and lactose is present in the medium.

In theseconditions, cyclic AMP levels increase. cAMP binds to the catabolite activator protein (CAP), which binds to thelac operon and enhances transcription. However, when glucose is present and used as the main carbon source, cAMPconcentration decreases, also reducing transcription from the lac operon.   Beta-glucosidase catalyzes hydrolysis of glycosidic bonds inbeta-D-glucosides and oligosaccharides, resulting in the release of an alcoholand glucose molecule.  Therefore, themore beta-glucosidase is expressed, the more glucose will be present in thecell. This acts as a negative feedback loop: The presence of glucose inhibitsthe transcription of the T7 polymerase from the lac operon, which in turninhibits the transcription of the target gene under the T7 operon.

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 A second possibility is that high concentrations ofbeta-glucosidase is toxic to E. coli,which will hinder growth and limit plasmid copy number in the cell.   Question 2: Whichplasmids would you build to test your hypothesis and engineer a potentialsolution? Describe the composition and function of these plasmids. Low yield may be caused by that fact that beta-glucosidaseis under the control of a lac-TR7 system, and the glucose produced by ourtarget gene will negatively regulate expression under the lac operon.  I would test this hypothesis using a plasmidthat does not place beta-glucosidase under the control of the T7 promoter.   There are only two strains in my freezer that are compatiblewith my available plasmids: BL21(DE3), which is used for T7 expression systems,or NEB10b, which can be used in blue/white screening without the use of IPTG orlactose.  Since I want to avoid the T7system, I selected to use NED10b for my experiments. NEB10b uses alpha-complementation of the b-galactosidasegene.

  Alpha-complementation is where thebacterial strain only has the lacZ-omega fragment of the beta-galactosidasegene, and is therefore nonfunctional. Function can be restored to beta-galactosidase by transforming the cellwith a plasmid containing the lacZ-alpha fragment.  The target gene is inserted within thelacZ-alpha region, thereby disrupting the lacZ-alpha gene and acting as aselection marker.  If a plain plasmid istransformed into your cells, your cells will produce functionalbeta-galactosidase, and adding X-gal to your medium will turn the colonies blue(beta-galactosidase cleaves the X-gal and causes a color change).  If the engineered plasmid is transformed intothe host, beta-galactosidase will be nonfunctional (since the lacZ-alpha geneis disrupted), and in the presence of X-gal the colonies will remain white,thereby allowing the selection of successfully engineered strains.

    Out of the plasmidsavailable for NEB10b, I chose to go with two options in order to perform myexperiments: Plasmids 6 and 7(pJ23101-B0034-GFP, one containing the ColE1origin of replication, and the other containing the RSF1030).  The plasmid structure consists an origin ofreplication (ColE1 or RSF1030), a CmR resistance gene, a B0034 BioBrick(ribosome binding site), a lacZ-alpha gene, and a gene to express greenfluorescence protein (GFP) for detection methods.   I chose to use plasmids 6 and 7 because theycontain the same antibiotic resistance gene, allowing for a more directcomparison when performing experiments. Using different resistance genes would create an additional variable inmy experiments and could skew results.

   I decided not to select B0064-containing plasmids becauseits ribosomal binding site efficiency is only 35% as strong as B0034’s.  Naturally, it made sense to use a moreefficient RBS in order to increase translation rates.  I chose not to select plasmids with PSC101 orR6ky origins of replication because they have stringent control, meaning thatplasmid replication is dependent on host replication and plasmid copy numbersare thereby limited.   Plasmid 6 contains the ColE1 ori, which produces ~15-20plasmid copies per cell.  Plasmid 7,containing RSF1030, produces >100 plasmid copies per cell.  By testing these plasmids against each other,I can test whether beta-glucosidase is toxic to E.

coli at higher levels: If growth is stunted using the plasmid 6(copy number >100), then it would appear that beta-glucosidase is toxic to E. coli at high levels.  If that is the case, I would proceed withusing Plasmid 7.  Yield may not be ashigh since my copy number is lower, but I would avoid killing my hosts withoverexpression of beta-glucosidase.    Question 3: Describehow you would physically construct these plasmids using a cloning technique ofyour choosing.

   First I would amplify my target gene using PCR.  I would design primers that contained thesame cut sites as the plasmid’s cut sites where I want to insert my gene (theinsertion will be in the lacZ-alpha gene segment).  I will use different endonuclease cut siteson the 5′ and 3′ ends in order to prevent re-circularization of the vectorafter it has been cut.  The annealingtemperature of my PCR reaction would be determined using the meltingtemperature of the segment of the primer that hybridizes to the gene’s ORF.  After PCR I would run my PCR product througha gel in order to confirm that my gene is the correct size and that theamplification worked.   After amplifying my insert, I will digest both my targetgene and plasmid using the selected restriction endonucleases.  The digestion will produce complementaryoverhangs of DNA that will permit the ligation of the insert into the plasmid.  After my digestion I will run the insert andplasmid DNA through a gel in order to isolate my insert and plasma DNA, as wellas to ensure that the digestion worked and my gene and plasmid fragments arethe correct size.

  I would then purify mybands away from the gel and quantify the concentration of DNA. I would then perform a ligation reaction using ligase and a1:3 plasmid:insert ratio.  This reactionwill ligate the beta-glucosidase gene into my plasmid, disrupting the plasmid’slacZ-alpha gene.    Upon transformation into E.

coli and growth on chloramphenicol-containing medium, I can select which colonies were successfully transformed byselecting white colonies (blue colonies will contain non-engineered plasmids),and quantify gene expression by measuring the intensity of fluorescence fromthe GFP gene in the plasmid.