Bio5LAManual12f

# 2 be able to list the two steps in genetic

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Unformatted text preview: /10 ml x 1/100 = 1/1000 0.1 ml/10 ml = 1/100 1/100,000 = 10-5 Choose plate B, C or D and count the colonies. You want to choose the plate that has 25 - 250 colonies per plate. Some plates may not have enough colonies to count and others may have too many. Remember, 1 colony = 1 cell. 4. Calculate the # of cells available for transformation in each sample (3 drops of competent E. coli cells). Let B = # of cells on plate B; then # Cells per 3 drops = B x 103 Let C = # of cells on plate C; then # Cells per 3 drops = C x 104 Let D = # of cells on plate D; then # Cells per 3 drops = D x 105 5. Calculate the transformation efficiency. Biology 05LA – Fall Quarter 2012 Lab 7 – Page 8 Ask: What fraction of the available cells were transformed? You know: # of transformed cells per 3 drops of competent cells (Step 2) # of viable E. coli cells per 3 drops of competent cells (Step 4) Therefore the transformation efficiency (TE) is: TE = TE = # of transformed cells # of viable competent E. coli cells # of transformed cells per 3 drops # of cells per 3 drops Learning Goals/Intended Outcomes: 1) Be able to define the term “genetic transformation”. 2) Be able to list the two steps in genetic transformation and to identify which step is associated with a change in genotype or a change in phenotype. 3) Be able to explain what plasmids are and how they are involved in genetic transformation. 4) Be able to explain why the E. coli cells used in our experiments are said to be “competent”. 5) Relative to the primary exercise on genetic transformation, you should be able to name the gene that we attempted to insert into the E. coli cells, name the protein that it codes for, and give the function of that protein. 6) Be able to explain how and why ampicillin was used in these experiments. 7) For each of the steps of the transformation protocol listed below, you should be able to explain how that step facilitated the transformation process. a) cold incubation after the addition of the plasmid b) the heat shock c) the outgrowth step (two considerations here) 8) At the end of the transformation exercise, we ended up with the 4 plates listed below. You should be able to explain what each of these plates contributed to our interpretation of the experiment. a) +DNA / AMP b) +DNA / LB c) -DNA / AMP d) -DNA / LB 9) Be able to define the term “transformation efficiency.” 10) Be able to explain how we knew whether or not the plasmid used in the Special Exercise contained added DNA. Bio 05LA – Fall Quarter 2012 Lab 8 Lab 8: Genetic Testing Laboratory – Detection of Alu Sequences by PCR * It is estimated that there are 30,000–50,000 individual genes in the human genome. The true power of PCR is the ability to target and make millions of copies of (or amplify) a specific piece of DNA (or gene) out of a complete genome. In this activity, you will amplify a region within your chromosome 16. Amplifying the Target Sequence The recipe for a PCR amplification of DNA contains a simple mixture of ingredients. To replicate a piece of DNA, the reaction mixture requires the following components: 1. DNA template — containing the intact sequence of DNA to be amplified which in this case is genomic DNA that will be extracted from your cheek cells. 2. Individual deoxynucleotides (A, T, G, and C) — raw material of DNA 3. DNA polymerase — an enzyme that assembles the nucleotides into a new DNA chain 4. Magnesium ions — a cofactor (catalyst) required by DNA polymerase to create the DNA chain 5. Oligonucleotide primers — pieces of DNA complementary to the template that tell DNA polymerase exactly where to start making copies 6. Salt buffer — provides the optimum ionic environment and pH for the PCR reaction The two DNA primers provided in this kit are designed to flank a DNA sequence within your genome and thus provide the exact start signal for the DNA polymerase to “zero in on” and begin synthesizing (replicating) copies of that target DNA. Taq DNA polymerase extends the annealed primers by “reading” the template strand and synthesizing the complementary sequence. In this way, Taq polymerase replicates the two template DNA strands. PCR amplification includes three main steps, a denaturation step, an annealing step, and an extension step (summarized in Figure 1). In denaturation, the reaction mixture is heated to 94°C for 1 minute, which results in the melting or separation of the double-stranded DNA template into two single stranded molecules. The DNA templates must be separated before the polymerase can generate a new copy. The high temperature required to melt the DNA strands normally would destroy the activity of most enzymes, but Taq polymerase is stable and active at high temperature. During the annealing step, the oligonucleotide primers “anneal to” or find their complementary sequences on the two single-stranded template strands of DNA. In these annealed positions, they can act as primers for Taq DNA polymerase. Binding of the primers to their template sequences is also...
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