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55-213 lab exam review.pdf

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University of Windsor
Biological Sciences

Lab Exam Review NOTE THIS REVIEW DOES NOT CONTAIN ALL THE INFORMATION YOU ARE RESPONSIBLE FOR STUDYING. IT IS MERELY A GUIDELINE TO HELP YOU PREPARE FOR THE EXAM Questions? Email me at [email protected] Lab Exam is all multiple choice questions! Lab 1 Bacterial Transformation Bacterial Transformation: Changing bacteria such that its plasmids carry novel genetic information or foreign DNA that alters the phenotype of the bacteria. This process allows molecular biologists to propagate, express, and isolate recombinant DNA molecules. Competency: Involves changing the structure and the permeability of the cell membrane that allows the plasmid to enter the cell (addition of chemicals followed by heat and cold treatments or electric pulses) Objective: Transform the E.Coli and now analyze your results to see how efficient the transformation was Previously: 3 tubes containing unknown DNA plasmids: A, B, A+B (10ng/10ul) Three dishes: LB agar + X gal+ Amp Two dishes: LB agar only Competent E.coli cells (holes in the membrane so they can take up the plasmid DNA) Added 100ul of E.coli cells to each of the three tubes containing 10ul each, incubated on ice so DNA would coat the E.coli cells, heat shocked so the DNA would enter the cells, put on ice to close the pores in the membrane, added 500ul LB broth to each tube and incubated to allow them to recover Then plated 100ul of the solution onto the plate NOTE: Normally E.coli cells can’t grow in the presence of antibiotic ampicillin and it can’t break down the sugar like substrate X-gal When it is transformed with pBS+, contains gene that encodes for amp resistance and lacZ gene which encodes for N terminal region of B-gal (which breaks down sugars) Reporter gene: the characteristics they confer on organisms expressing them are easily identified and measured, or because they are selectable markers. Reporter genes are often used as an indication of whether a certain gene has been taken up by or expressed in the cell or organism population, Successful transformation means functional B gal which breaks down X gal and will form blue colonies A vector is a DNA molecule used as a vehicle to artificially carry foreign genetic material into another cell, where it can be replicated and/or expressed. A vector containing foreign DNA is termed recombinant DNA A plasmid is a small DNA molecule that is physically separate from, and can replicate independently of, chromosomal DNA within a cell. In this lab its pBluescript + (pBS+) Phenotypic delay: a time period allowing the plasmid to replicate and to begin synthesis of any genes they carry pBS+: also contains MCS (many restriction sites) within lacZ gene, if foreign pieces of DNA are inserted into this site, disrupts lacZ expression and no B gal therefore can’t break down substrate X gal and will produce white colonies Transformation efficiency = quantitative value that describes how effective you were at getting your plasmid DNA into your E.coli bacteria, how many cells were transformed per microgram ug of plasmid DNA used 1) Determine the concentration of DNA in the sample per microlitre 2) How much DNA is present: Concentration X volume = DNA in ng 3) Convert to micrograms: 1 ug = 1000 ng, 1 ng = 0.001 ug 4) Calculate total volume 5) Determine the amount of plasmid DNA you plated: If calculated 0.001 ug in 1000ul then 0.0001 ug in 100ul 6) Transformation efficiency: # of colonies counted/ ug DNA 6 No8mal preparation of competent cells can yield transformation efficiency ranging from 10 to 10 cfu/μg DNA Example: If given a question with the following information: Mixed 10 ul plasmid DNA with a concentration of 10ng/ul with 100ul competent cells, 500ul Luria Broth and then plated 100ul onto an agar plate. 24 hours later counted 200 colonies A) How much DNA did you start with: 10ng/10ul =1ng/1ul 1ng/1ul x 10ul =10ng Convert to micorgrams (1ng=0.001ug): 10ng=0.01 ug B) Total volume (plasmid DNA, competent cells, Luria Broth) = 10+100+500= 610 ul How many microgram DNA in this total volume: 0.01ug in 610 solution= 0.01/610=0.000016ug/ul Then plated 100ul of this solution so 0.000016ug/ul X 100ul= 0.0016ug C) Calculate TE: 200 colonies/0.0016ug DNA = 125 000 colonies/ug DNA Explaining the mixed plate results: Blue= no foreign insert in MCS White= foreign DNA insert in MCS 1) In the sample where you mixed A (let’s say this is the smaller plasmid forms blue) and B (let’s say this is the bigger plasmid with the insert, forms white colonies) you have the following options of the plasmids entering the pores of the competent Ecoli cells: Only A Only B A+B (if competing to enter the pores of the competent Ecoli cells A will be more successful because its smaller) Neither A or B Therefore, more blue colonies will appear 2) Started with the same concentration of stock solutions (ng per microlitre) of plasmid A and B but if one plasmid is larger then there will be less of that plasmid in the same volume of stock solution. For example if you have one container and fill it with golf balls (smaller plasmid) or tennis balls (bigger plasmid) you'll have more golf balls in that set container/volume so you start off with more 3) Plasmid that forms blue colonies is smaller because it doesn’t have an insert, therefore will more easily enter the pores of the competent Ecoli cells. Lab 2: Isolation of E.coli Genomic DNA Objective: To isolate genomic DNA from bacterial cells using the Marmur method Marmur method: Is a method of DNA extraction employing the use of cholorform to remove protein from the lysate Why isolate genomic DNA? : molecular biology studies like looking at DNA structure, DNA- protein interactions, DNA hybridization and genetic studies First step: Resuspended E.coli cells in saline-EDTA (Already done) Function: chelating agent, i.e. its ability to "sequester" metal ions such as Ca and Fe . After being bound by EDTA, metal ions remain in solution but exhibit diminished reactivity, ion depletion is commonly used to deactivate metal-dependent enzymes to suppress damage to DNA In this lab: When we break open the cells, the DNA is exposed to Dnases, in order to protect the DNA from being degraded we need to make these nucleases inactive and this is accomplished through use with EDTA Second Step: Added lysozyme to the E.coli suspension and incubated for 20 min at 37 Function: Lysozyme is abundant in a number of secretions, such as tears, saliva, human milk, and mucus; The enzyme functions by attac
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