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Department
Biological Sciences
Course
55-213
Professor
Hubberstey
Semester
Winter

Description
Lab #1 Bacterial Tranformation • E.Coli is used for recombinant DNA research • It is a normal inhabitant of the human colon and is a common soil bacterium • Easily gorwn in lab by either o Suspension in a liquid media that provides nutrients such as luria broth (LB) o Suspension on semi-solid media containing nutrients such as LB agar  On this media, bacteria will continue to divide until it forms a colony of bacteria with each member of colony being a clone of the other Characteristics of E.Coli • Contains extra chromsonsmal circular ds DNA known as plasmids (1,000-2,000bp) • Strains used in labs are non-pathogenic , thus pose minimal risk to lab personal • Grow at an optimal temperature of 37 degrees with no special requirement for nitrogen, oxygen or light • Undergo direct cell division with one parental cell producing two identically cells and will continue this process given enough nutrientsn and room for growth Bacterial transFORmation – manipulating the plasmids to change the bacteria such that the plasmids carry foreign DNA that alters the phenotype of the bacteria • Allows biologist to propagate , express and isolate recombinant DNA molecules • Can be used to introduce new genes from other species into a species that normally doesn’t have those genes. o Ex. plant and animals can be made resistant to certain disease by this process. Competency – in order to take up the plasmid DNA, E. coli needs to be manipulated Involves a change in structure and permeability of the cell membrane that allows the plasmid to enter the cell and can be achieved by the addition of chemicals followed by heat and cold treatments or by the use of electric pulses. E. Coli normally cannot grow: 1. In the presence of antibiotic ampicillain 2. Nor can it break down sugar like molecule “x-gal” pBluescript + (pBS+) - is a DNA plasmid that contains 2 genes: 1. Amipcilan resistance 2. Lac Z N-terminal region of the enzyme β galactosidase Ampicilan resistance • After undergoing transformation, bacteria which successfully take up new DNA will be identified by their ability to grow in the prescenve of ampicillian • Form colonies that are clones of the original bacterium that was transformed. 1 Lac Z • Encodes the N-terminal region of the β galactosidase • β galactosidase hydrolyzes lactose into galactose and glucose • Lac Z has been used as a reporter gene since the β galactosidase substrate X-gal produces a dark blue precipitate on enzygametic hydrolysis • Thus, when bacterium has transformed with a plasmid containing the lac Z gene it will produce dark blue colonies when grown on media containing X-gal • The plasmid pBS+ has been made so it is useful for inserting foreign DNA molecules into the plasmid • pBS+ contains multiple cloning sites within the lac Z gene • when foreign piece of DNA are inserted into and the multiple cloning sites, the lac Z gene is disruptive and becomes non-functional. • Thus plasmids containing inserts of foreign DNA in the multiple cloning sites will produce white colonies on media contain X-gal Competent cells prepared by CaCl2 method E.Coli + CaCl2 • Treating it with CaCl2 is thought to produce holes in the bacterial membrane/cell wall that allow them to take up the DNA. • These cells are very delicate and thawing them inappropriately or harsh treatment can kill them • Incubation on ice is needed because: • This is the time you are giving DNA a chance to stick the cells such that the DNA is ready to enter the cell. • Entry of DNA is thought to require a heat shock step • Elevated temperature possible increases the fluidic movement of the cell membrane bringing the holes punched by CaCl2 closer or make them larger • Exact mechanism is unclear. Phenotypic delay – time period in which the cells recuperate from the harsh treatment, thereby allowing the plasmid to replicate and begin synthesis of any genes they carry Questions for Lab #1 1. Observe, draw and describe each of the 5 plates carefully and write down observations for each one. 2. How much bacterial growth do you see on each, relatively speaking? What color are the bacteria? Are all the colonies the same? If not, explain why this may be the case. Are there plates with excess or no growth. If so, why? LB Plate A = had 2 genes (lac Z and amp) • growth on the plate 2 • White lawn because there is no x-gal to breakdown • White lawn - Produces white colonies • No amp – no selection for growth, every cell can grown • No selection for amp resistant gene B • Growth onteh plate • White lawn because; o No x-gal to breakdown o B does not contain functional lac Z gene A(LB + amp + x-gal) • Can break the ___ to produce B colony • Beucase it has amp, if the cells on it have it are e?? • Able to grow sicne they have taken up the gene • Blue b/c able to break down x-gal • Growth on the plate • Know how the colony are growing ** o White vs. blue o ___ vs. lawn • Blue colonies because x-gal is broken down and a blue precipitate is formed • LAC Z GENES B (LB + amp+ x-gal) • Growth on the plate • White colonies because B does not contain a functional lac Z gene • It is interrupted at the multiple cloning sites A+B (LB+ amp+ x-gal) • Growth on the plate • White and blue colonies • A has lac Z gene and can break down X-gal to produe blue precipitate • B has no lac Z gene 3. For the mixed plate, explain your results. Which colour of colonies was greater? Describe three possible reasons you observed the results you did. 3 possibilities A is more likely to take up because it is smaller than B, since B has more multiple cloning sites 1. Size of the insert plasmid is larger so it transform at a lower efficiency 2. Any? Cell taking up both plasmids will show blue over white 3. Plasmid A is more concentrated than plasmid B a. Might have been a higher PBS concentration of the plasmid with lac Z gene 4. Count the antibiotic resistant colonies on each plate. Make a quadrant; count 1 of the quadrant and then multiply Count colonies on 1 quadrant and then multiply by 4 3 5. Determine transformation efficiency (number of colonies per ug of DNA). Your demonstrator will go over an example while waiting for phenotypic delay. Amount of DNA used: 0.01 ug Transformation efficiency = count colonies (4 quadrants) # of DNA used (0.01 ug) 1 ng X 10 ul = 10ng = 0.01 ug ul 6. Why was it necessary to wait for phenotypic delay prior to plating the bacteria. • Phenotypic delay is necessary because bacteria cannot uptake a gene and begin expressing it immediately there needs to be a certain amount of time to allow for uptake gene expression • The cells were put in CaCl2 treatment, which puntered holes o The cell needs to recuperate following the CaCl2 treatment Answers to Lab #1 1. A-plates – the plasmid contains the intact lac gene and they should be blue. If they are white, they have been contaminated. B-plates- should be all white because a disrupted lac gene is present in the plasmid. A+B- plates – both blue and white colonies should be observed but in different proportions. 2. A-plates – bacteria should be blue. B-plates- bacteria should be all white. A+B- plates – both blue and white colonies in different proportions. The colonies are not the same and this may be because: • Depending on the size of the plasmid, bacteria may have difficulty taking up the plasmid (physical factors). • Some plasmids can do different things to the bacteria and some can multiply faster than others • Concentration may be different; there could be less plasmid in A than B and therefore fewer colonies observed. 3. The mixed plate had both white and blue colonies and in different groups, the proportions varied. • Bacteria can have both plasmids; it still has a plasmid with intact lac gene so you can still see blue colonies and white as well. • This is also because the concentrations of the different plasmids are different. • Also depends on the size of the plasmid and the ability of the bacteria to take it up. 4. Count the number of antibiotic resistant colonies – yeah right  5. Example for determination of transformation efficiency (number of colonies per ug of DNA) 6. It was necessary to wait for phenotypic delay prior to plating the bacteria. The bacteria were given a new genotype. We had given the bacteria amp gene to create resistance to ampicilin. We were waiting for the proteins to be synthesized and the gene to be expresses. There is a change in the genotype instantly but there is a delay before the change in the phenotype happens. E. coli notes • Very easy to induce competency • Small force to the cell membrane of the bacteria • Naturally susceptible to many antibiotics 4 • The plasmid contains ampicilin resistant gene and Lac Z • Beta galactosidase cannot reduce the sugar (non functional) • Within the lac Z gene, there is a MCS with many restriction enzymes sites; that is how the insertion is inserted into the 2 plasmid. Uninserted plasmid • Complete Lac Z gene, produces lactosidase products, breaks down X-gal and gives blue color • Blue colonies are formed when no foreign DNA is inserted in the MCS of the lac Z gene • Lac Z needs to be present to get blue color Inserted plasmid • White color As soon as insertion is added, blue color disappears Lab #2 Isolation of E.Coli genomic DNA To isolate nucleic acid 1. Disrupt the bacterial cells a. Gram negative bacteria like E.Coli are susceptible to lysis by detergent alone b. Depending on the cell wall i. Physical methods ii. Combo of detergents and hydrolytic enzymes such as lysozomes To isolate DNA from cell lysate use the Marmur method which employs chloroform to remove protein components of the lysate. • E.Coli is suspended in saline ethylene diamine tetra acetate (EDTA) • Cells were prepared by centrifugation of E.Coli broth culture when it was in logarithmic or early stationary phase of growth. SDS = sodium dodecyl sulfate. • Lysis is evident when there is rapid increase in viscosity • May also be possible to detect change in cell suspension from turbid to opalescent Centrifugation should yield 2 phases: 1. An upper aqueous phase 2. An bottom chloroform phase with a zone of cell debris • Take less of the aqueous phase rather than risk contamination with the chloroform phase • Aqueous layer contains DNA • Might be necessary to take some of the cell debris that is not compacted with the aqueous phase Questions fo
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