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midtermkey2 - MIC170 Midterm May 5, 2009 Part I. Choose the...

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MIC170 Midterm May 5, 2009 Part I. Choose the ONE most appropriate answer for each of questions 1~5 . 1. (2 points) The lipid bilayer of the plasma membrane is NOT permeable to glucose 2. (2 points) You are planning to construct a genomic library of the budding yeast S. cerevisiae . If the average size of the genomic DNA fragment in each plasmid is about 10 kb, you would want to make a library containing at least ( ) different plasmids to make sure the complete coverage of the entire genome. 6,000 3. (2 points) You are about to start chemical mutagenesis of the budding yeast S. cerevisiae . As you learned in the class, it would be ideal that a single mutation causes an interesting phenotype. Theoretically , to induce one mutation per cell in S. cerevisiae , the mutation frequency by mutagenesis needs to be about; 6x10 -3 /gene 4 . (3 points) It’s almost 9PM and you want to go home. You have an exponentially growing yeast cell culture, whose current concentration is 1.6 x 10 6 cells/ml. It has been determined that yeast cells in this culture divide every 3 hours. Before leaving the lab tonight, you want to dilute this culture and put it back to the incubator so that it’s concentration will reach again 1.6 x 10 6 cells/ml tomorrow morning at 9AM for your exciting experiment. You need to dilute the culture at 9PM tonight to the concentration of; 1.0 x 10 5 cells/ml 5. (2 points) Most of laboratory strains of the budding yeast S. cerevisiae cannot switch the mating type (heterothallic), because of a mutation in HO Part II. (3 points each) Choose ONE apparently INCORRECT statement for each of questions 6~11 . 6.
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midtermkey2 - MIC170 Midterm May 5, 2009 Part I. Choose the...

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