HOUR EXAM I BIOLOGY 422 FALL, In the spirit of the honor code, I pledge that I have neither given nor received help on this exam.

Transcription

1 Name First Last (Please Print) PID Number - HOUR EXAM I BIOLOGY 422 FALL, 2011 In the spirit of the honor code, I pledge that I have neither given nor received help on this exam. 1 Signature

2 1. (14 points) Below is a diagram of a bacterium. Is this a gram-negative or a gram-positive organism? (circle one) Label the indicated parts on the drawing. B A C F D What is the composition in terms of major macromolecules of each of the parts you labeled on the drawing above (be sure to indicate more than one type of macromolecule if more than one type plays a major role)? The letters below should correspond to the letters above. A. B. C. D. E. If this bacterium were of the other type with respect to the Gram stain, what would be 3 major differences in the cell structure? (Be careful not to use the same difference more than once)

3 2. (5 points) Where in the cell would you expect to find each of the following? (Note: not present is a possible answer). Cytochromes Nucleus Beta-galactosidase Lactose permease CheW 3. Bacterial Growth (15 points) Use the following growth curves to answer parts A, B, C, D. Graphs may be utilized more than once. A. Bacteria X is microaerophile. At time 0, you grow Bacteria X in a shaking incubator in rich media. At time A, you transfer the test tube containing the culture to static conditions in an aerobic incubator. Which growth curve best describes the growth kinetics of Bacteria X in this scenario? 3

4 B. Bacteria X contains only one gene that synthesizes an enzyme that directly inactivates reactive oxygen species. At time 0, you grow Bacteria X in static conditions in an aerobic incubator. At time A, you introduce an inhibitor that blocks activity of this enzyme. Which growth curve best describes the growth kinetics of Bacteria X in this scenario? C. Bacteria Y requires iron as a growth factor. At time 0, you grow Bacteria Y in rich media. At time A, you add an iron chelator that immediately binds up all available iron in the culture. Bacteria Y does not produce siderophores that can outcompete this chelator. What growth curve best describes the growth kinetics of Bacteria Y this scenario? D. Bacteria Z contains the lac operon and is therefore able to utilize lactose as an energy source. At time 0, you grow bacteria Z in minimal media containing 2% glucose. At time A, you add 0.2% lactose into the culture. What growth curve best describes the growth kinetics of Bacteria Z this scenario? E. Chemostat culture systems mimic the growth conditions that bacteria in the intestinal tract are exposed to. Keeping in mind how the flow rate of the media affects cell mass in a chemostat, explain why diarrhea might be a protective mechanism for the host against pathogenic enteric bacteria? 4. Gene Regulation (11 points) Pseudobacter spificus is a novel commensal organism you have discovered in 90% of the world s chickens. It has Green and Blue enzymes that are harmless to the host but the bacteria s Red enzyme generates a deadly toxin. P. spificus has three operons that control which enzymes are produced. Red operon Regulatory sequence for Red Red-enzyme Repressor of Blue RoB Green operon Regulatory sequence for Green RoR Blue operon Regulatory sequence for Blue AoG Green-enzyme Repressor of Red Blue-enzyme Activator of Green A major chicken producer in the US finds a mutant strain of P. spificus that always produces the Red- enzyme. This red only phenotype results in massive animal deaths. 4

5 A. (2 pt) Which mutation could cause a red only phenotype? Circle all possible correct answers. A) RoB - B) red-enzyme - C) AoG - D) None of the above B. (4 points) How could you test whether your answer to part A was in fact correct? (Assume you cannot sequence these regions; you have available competent always red bacteria and clones of any gene or region you wish). C. (5pts) You isolate a mutant which is never green. Name two possible mutations which could produce this phenotype. Design a transformation based experiment that could determine where the mutation is. (Assume you cannot sequence these regions; you have available competent greenminus bacteria and clones of any gene or region you wish). 5

6 5. Chemotaxis A. (8pts) Shown below are the components of the chemotaxis system in E. coli. Fill in the table below with the phenotype of the following mutants. Mutation Phenotype (type of motion, two word max) CheA that cannot be dephosphorylated (cannot donate phosphate) CheZ -- CheA that cannot be phosphorylated MCP that cannot bind CheY 6. Conjugation A. (1pt) Genetic maps of E.coli are in units of minutes. Why? B. (1pt) When a donor cell carrying a single copy of a plasmid transfers plasmid DNA to a recipient cell, why doesn t the donor cell loose the plasmid? 6

7 Time After mixing strains C. (4pt) You have just done a conjugation experiment. The F+ donor strain has the following phenotype: amp R, trp --, mal +, thi +, rif S The F- recipient species had the following phenotype: amp S, trp+, mal --, thi --, rif R While allowing the two strains to conjugate, you took samples every 15min and get the following results: Number of Colonies with Indicated Phenotype Amp R Trp + Mal + Thi + Rif R 0 min min min min min Based on these results, what are the positions of the genes on the donor plasmid? (indicate the approximate positions on the line) _orit 7. Transformation A. (1pt) It takes a large amount of energy to undergo natural transformation. Name one reason a bacterium would benefit from natural transformation. B. (4pt) You just did a transformation with the following: Plasmid containing genes encoding: his + arg + mal + amp R Competent E. coli containing phenotype: his -- trp -- leu + arg -- thr + lac + mal -- ara + neo R amp S Now you need to use an agar plate to select for transformants. Starting with an agar plate containing only minimal media, circle what else you MUST add to the plate. Put an X through anything you should NOT add to the plate. Amino acids vitamins sugars antibiotics threonine (thr) biotin (bio) glucose (glu) streptomycin (sm) leucine (leu) thiamine (thi) lactose (lac) rifampicin (rif) histidine (his) maltose (mal) ampicillin (amp) tryptophan (trp) nucleic acid bases arabinose (ara) tetracycline (tet) arginine (arg) adenine (ade) neomycin (neo) cytosine (cyt) kanamycin (kan) 7

8 C. (6 points) You perform a transformation using as your source of donor DNA an E. coli strain which is leu -, mal +, ara +. The recipient is leu +, mal -, ara -. You select for cells which are mal + and for cells which are trp +. You want to determine the gene order and relative distance. You score the transformants for all 3 traits and obtain the following results: % of selected cells which are Selection leu + Mal + ara + Mal ara Draw a map showing the order and relative position of the genes: Why didn t you select for leu + transformants? 8. (10 points) Regulation of the enzymes for the synthesis of amino acids Tryptophan is an amino acid. The genes for tryptophan biosynthesis trpedcba are found in an operon and are regulated by a repressor and by attenuation. For each E. coli mutant shown in the table below, put a checkmark indicating the amount of gene mrna produced if the bacteria were grown in the stated medium. o: operator o c : operator constitutive P; tryptophan promoter EDCBA: genes for tryptophan biosynthetic enzymes 1 - mrna leader sequence region 1 R: tryptophan repressor gene 2 - mrna leader sequence region mrna leader sequence region mrna leader sequence region 4 + means the region is present, - means the region is absent. Where two genotypes are shown in a box the upper one is on the chromosome and the lower one is on a plasmid. E. coli Genotype R + P + o EDCBA R + P + o c EDCBA R - P + o EDCBA R + P + o c EDCBA R + P + o c E -- DCBA R - P + o EDCBA Level of trpe mrna produced in minimal medium with glucose and excess tryptophan None Low High Level of trpe mrna produced in minimal medium with glucose only None Low High 8

9 9. Transposon mutagenesis: Use the reagent shelf below to answer the following questions (= inverted repeat): TnA: > Amp r Transposase < TnB: > Kan r Transposase < TnC: > Amp r < Transposase TnD: > Kan r < Transposase Your lab studies the pathogenesis of uropathogenic E. coli (UPEC). You are interested in studying genes required for persistence and growth in urine. As you conduct growth studies in urine, you notice that one of your strains produces cellular aggregates when grown in urine but not in other media. This UPEC strain is also resistant to ampicillin. You decide to perform a transposon mutagenesis to determine what genes are involved in producing these cellular aggregates in urine. A. (4 points) Which transposon from the reagent shelf would you use for the transposon mutagenesis? Why? B. (6 points) Name two ways in which you can introduce your transposon into your UPEC strain of interest. Explain one in detail

10 C. (2 points) How would you select for mutants with the transposon you used in part A? D. (4 points) From this screen, you obtain 50 mutants of 10,000 that you screened that are unable to produce cellular aggregates in urine. Name two types of genes that you may identify that could be required for cellular aggregation E. (4 points) What type of experiment would you design to test one of your hypotheses from part D? State which hypothesis you are testing. 10