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Western University
Biology 2290F/G

Sample Exam Questions Zabulionis Unit FALL/12 It is in your best interest to try the questions first before looking at the answers; this gives you a chance to see how your thinking works for these types of questions. 1. Ampicillin plates are prepared to a final "working concentration" of 100 µg/mL. If the stock solution of amp is 10 mg/mL, how much antibiotic stock solution would be mixed with how much medium to achieve the working concentration in 1 litre final volume? (Show your work.) (3 marks) -7 2a. If an overnight culture was diluted to 10 and then 278 colonies were counted after plating 100µL, what is the estimated number of cells per mL in this overnight culture? (Show your work for part marks.) (4 marks) 2b. What assumption is made in determining the number of cells per mL in this overnight culture? (1 mark) 3. Some mutants of E. coli have twice as many pores (adhesion zones) on the outer surface of the cell. Would you expect such strains to have more, fewer, or the same number of transformants as normal strains using the calcium chloride/ice/heat method used in class? Why? (4 marks) 4. Given the data below, propose a biological model (it can be in 2 parts) to explain these results: Growth (number of transformants) on LB + ampicillin plates after transformation of E. coli JM101 cells with pGREEN plasmid that codes for ampicillin resistance and increasing concentration of Ca2l CaCl Concentration: Growth (number of transformants) 2 0 mM 0 25 mM 24 50 mM 40 100 mM 150 200 mM 30 (6 marks) Sample Exam Questions Zabulionis Unit FALL/12 ANSWERS: 1. use C1V 1 C V2 2 C1= 10 mg/mL = 10,000 µL/mL V 1 ? C = 100 µg/mL 2 V 2 1 L = 1,000 mL V 1 100 µg/mL X 1,000 mL = 10 mL. 10,000 µL/mL Therefore, 10 mL of ampicillin stock solution will be added to 990 mL of medium. 3 2a. 278 cells/100µL = 2780 cells/mL = 2.78 X 10 cells/mL. cells were diluted 10 ; therefore multiply by 10 3 7 10 2.78 X 10 cells/ mL X 10 = 2.78 X 10 cells/ mL 2b. Each colony represents one living cell in the overnight culture. 3. Mutants with twice as many pores present a larger number of entry points for plasmid molecules than normal cells since it is assumed that plasmids enter the cells through the adhesi
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