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Review for Practical (Solutions) 1. Name and describe a test that could be used to differentiate between each pair of bacteria listed below. Include descriptions of the observations expected, indicating what positive and negative results would look like. You may use any methods used in class. Test Positive Organism Negative Organism Observation Staphylococcus aureus and Enterococcus faecalis Catalase Blood agar S.aureus S. aureus E. faecalis E. faecalis Bubbles form in hydrogen peroxide. Hydrolysis (clearing) of blood Enterococcus and Enterobacter gram stain Enterococcus Enterobacter Positive stains purple, negative stains pink+ oxidase Micrococcus Leuconostoc oxidase turns the reagent from brown to blue Example: Micrococcus and Leuconostoc 2. Twenty grams of ground beef were added to 50 ml sterile 0.85% saline and thoroughly mixed. The following dilution series was performed, starting with the meat suspension: 1/5, which was then diluted 1/10, and that dilution was diluted 1/20. A spread plate inoculated from the 1/20 dilution produced 133 colonies. a. What was the total dilution? The dilution factor? Dtotal = (20/70)(1/5)(1/10)(1/20) = 20/70,000 = 1/3500; DF = 3500 b. What was the concentration of viable bacteria in the ground beef? C = (133 cfu/0.1 ml)(3500 ml/g) = 4.7 x 106 cfu/g 3. In an Eadie-Hofstee plot the y-intercept is 50, the x-intercept is 0.6, and the slope is -42. What is Vmax and Km from this data? Vmax = 50; Km = 42 4. Explain how you would make a 1 M solution of the triprotic acid phosphoric acid. What would the percentage of phosphoric acid (v/v and w/v) be in this solution? What would the normality be? (H3PO4, MW = 98, d = 1.87) Add 1 mol phosphoric acid to a total volume of 1 L. One mole of phosphoric acid has a mass of 98 g. That’s equivalent to (98 g)/(1.87 g/ml) = 52.4 ml. So, I’d add 52.4 ml to water and bring the volume up to 1 L. Percentages: 98g/1000 ml = 9.8g/100 ml = 9.8%(w/v); 52.4 ml/1000 ml = 5.24 ml/100 ml = 5.24% (v/v) Normality: 1 M solution = 3N 5. Draw a Michaelis-Menton plot that shows two enzymes, one with both a lower Km and a lower Vmax than the other. Under what conditions (should be apparent from your plot) will each one out-compete the other? Enzyme A will outcompete Enzyme B at low substrate concentrations (< 40). Enzyme B outcompetes Enzyme A at higher concentrations. Competition success is based on relative rates (v) of the reactions. 6. What is the difference between a scientific hypothesis and an educated guess? An educated guess is an explanation or prediction that may be based on evidence, opinion, or authority. A scientific hypothesis is an explanation of an observable phenomenon that includes a prediction that can be tested to examine the validity of the hypothesis. Another way to look at it is that a scientific hypothesis must stand up to the rigors of testing while an educated guess just needs to sound reasonable.