7.02 Microbial Genetics in Lab Quiz. Fall, September 27, 2001 ANSWER KEY

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1 7.02 Microbial Genetics in Lab Quiz Fall, 2001 September 27, 2001 ANSWER KEY This quiz contains 4 questions worth a total of 48 points. Be sure to write your name, Bench letter and Undergraduate TA s (UTA) name on EACH page (2 points) GOOD LUCK! 1

2 1. (12 points) What was the purpose of the following reagents in the Genetics module? A complete answer will BRIEFLY describe when it was used in the module (what experiment) and what it is/how it works/what it does. MacConkey, Arabinose, Kanamycin (Mac Ara Kan) Plates: Mac Ara Kan plates were used to select/screen for putative Ara- mutants after transposon mutagenesis. Kanamycin selects for colonies that have received a transposon insertion anywhere in the bacterial genome. MacConkey and Arabinose were used to screen the KanR colonies for their Ara+/Ara- phenotype. MacConkey is an indicator dye that is sensitive to ph changes. When cells can metabolize arabinose (Ara+), the ph drops, and the MacConkey dye makes the colonies red. When cells cannot metabolize arabinose (Ara-), they use amino acids as a carbon source instead. This raises the ph, and the colonies appear white/translucent. Plasmid pnk: Plasmid pnk was used during the transposon mutagenesis experiment. pnk is a plasmid that contains an IPTG-inducible transposase gene; this allows us to control the expression of transposase, and thus the hopping of the minitn10 transposon into the genome. We needed to provide transposase on a plasmid because mini- Tn10 does not encode this enzyme. We could also stabilize our ara::lacz fusions by moving them into a strain that did not contain pnk. (Stabilization would be impossible if minitn10 contained its own transposase.) Bacterial strain LE392: LE392 was used to determine the titer of λ1205. λ1205 contains an amber mutation in a gene required for DNA replication; this amber mutation usually prevents λ1205 from lysing bacterial cells. The LE392 strain contains an amber suppressor that compensates for the amber mutation in the phage, and therefore DNA replication and bacterial cell lysis are restored. Note: LE392 does NOT restore lysogeny to λ1205!! 2

3 2. (12 points) A wild-type E. coli strain was grown in minimal medium containing glucose and lactose as carbon sources. Over time you followed the growth of the culture and obtained the growth curve shown below. At t1, t2 and t3 an aliquot of cells was obtained and assayed for β-galactosidase activity. At what time is β-galactosidase activity lowest? At what time is ß-galactosidase activity highest? Justify your answer. (Hint: Use diagrams of the lac promoter/operator region, including the appropriate repressors and activators.) Answer: β-gal activity was lowest at t1, and highest at t3. Justification: 1. When glucose and lactose are present (t1) cells will utilize glucose and the lac operon will not be induced. Thus, no β-galactosidase protein is made. Lactose will be utilized by the cell (i.e. the lac operon will be transcribed) only after glucose has been used up (t3). This phenomenon is called catabolite repression. 2. In the presence of glucose and lactose the lac operon is not transcribed for two reasons: a. Lactose cannot enter the cell. Thus, the lac repressor LacI will bind to the promoter region of the lac operon and prevent binding of the E. coli transcription machinery (RNAP). b. CAP cannot bind to the promoter, as cyclic AMP levels are low due to high levels of glucose present in the medium. 3. At t3, all the glucose in the medium will have been used up, the lac operon will be transcribed and β-galactosidase will be produced. In the presence of lactose, the lac operon is transcribed for two reasons: a. The lac repressor LacI cannot bind to the promoter, as allolactose binds to LacI and prevents it from binding to the operator site. b. Due to the lack of glucose, cyclic AMP levels are high, allowing CAP to bind to the promoter. This is necessary for recruiting the E. coli transcriptional machinery to the promoter. These two events allow the transcription of the lacz gene, which encodes the enzyme ß-galactosidase. PLEASE NOTE A DRAWING OF THE PROMOTER UNDER THE DIFFERENT CONDITIONS WAS ALSO AN ACCEPTABLE ANSWER. 3

4 3. (12 points) a) You have reason to believe that E. coli can metabolize sweetose, a sugar you recently discovered. Based on what you learned in 7.02 describe a strategy (in flow-chart form) to identify mutants defective in metabolizing this sugar. In order to identify mutants involved in the metabolism of sweetose, I would begin by mutagenizing the cells with transposon mutagenesis. (Other strategies of mutagenesis would work, but each has its own challenges.) E. coli BW140(Lac-)/plasmid pnk cells (Gene encoding transposase is found on plasmid pnk and is inducible by IPTG) Infect with λ1205 carrying mini-tn10 transposon (λ1205 is a defective phage that serves as a delivery vehicle for mini-tn10; mini-tn10 contains Kan R gene and lacz) Select for mutants that have received a transposon on plates containing kanamycin Screen for mutants that can no longer utlize sweetose as a result of transposon insertion. The screen and selection can be done on MacConkey Sweetose Kan plates. I will look for mutants that are no longer able to ferment sweetose. These colonies will be white because they are metabolizing amino acids in the media, raising ph, which turns the ph sensitive dye white. (You could also patch colonies on M9 Glucose and M9 sweetose plates and look for growth. Cells that grow on M9 Glucose and not on M9 sweetose are Sweetose-. However, it is very important that you don t plate your initial transposon mutagenesis mixture directly on M9 sweetose - You will select against the mutant you are interested in identifying.) (Identifying the regulation of genes in response to the presence of sweetose through lacz activity is not necessary at this point, as it doesn t directly indicate disruption of a gene involved in sweetose metabolism.) b) How would you changes this scheme if you knew that sweetose metabolism was essential in the life cycle of E. coli? If metabolism of sweetose is essential in the cell, a null mutation (which is what a transposon usually generates) would probably be lethal. In order to study mutations in genes that are involved in an essential process, it is necessary to study hypomorphs or conditional mutants. These types of mutations are generated by milder mutagens, such as chemical or UV mutagens. It is then important to screen these mutants under 4

5 permissive and non-permissive conditions (for conditional mutants) and look for a sweetose- mutant, or to look for reduced function (possibly reflected by slower growth rates) in sweetose metabolism when compared to wild type cells (for hypomorphs.) 4. (12 points) You want to determine the cotransduction frequency of markers a, b, and c using P1 cotransduction mapping. To do this, Debbie has given you a suspension of E. coli that contains 10 9 colony forming units/ml. Debbie has also made a P1 lysate with a titer of 10 9 pfu/ml. The 7.02 staff has determined that you ll get the most cotransductants if you do the following: 1. Use an MOI of Plate 0.1 ml of the P1/cell mix per plate 3. Plate 10 5 cells per plate. a) By what factor will you need to dilute the cells? (Show all calculations!) You want to plate 10 5 cells per plate, and 0.1 ml of cells per plate. You have a stock of cells with a titer of 10 9 cfu/ml. Therefore: 1. want: 10 5 cells/0.1 ml 10 6 cells/ml (cfu) 2. have: 109 cells/ml /ml /ml 4. Therefore, you must dilute the cells by a factor of 10 3 (make a 1/1000 dilution). b) By what factor will you need to dilute the phage stock? (Show all calculations!) 1. Want: MOI = have pfu = MOI = pfu cfu = 10 6 (from part a)` pfu = 5x10 3 cfu 2x10 5 pfu 3. pfu= 0.2 x 10 6 = 2x Therefore, you need to dilute the phage stock 1/5000 (2x10-4 ) c) By using an MOI of 0.2, a high percentage of cells will receive no phage. Why did Debbie nevertheless insist that you use a low MOI rather than a high one, i.e. MOI=10? If the MOI is too high, a high percentage of E. coli cells will receive more than one P1 phage. However, only a small percentage of phage in the P1 lysate will be transducing phage. Thus, if 5

6 two phage infect a bacterium, it is highly likely that one of the phage is wild-type. The bacterium will be lysed, and the transductant cannot be recovered. In the highly unlikely event that two transducing phage enter the bacterium simultaneously, mixed recombination between the two phage and the host genome could skew the outcome of the recombination mapping. 6