Exam 2 - University Of Illinois 2/Exam 2 2002.pdf- Exam 2 - Write your name at the top of this cover...

Mcbio 316: Exam 2A&C ANSWERS Name ______ANSWERS___________________Fall 2002

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Mcbio 316- Exam 2 -

Write your name at the top of this cover sheet and oneach page.

Write legibly -- if we cannot read your writing we cannot give you creditfor your answer. If you change your response, make sure to clearlycross out the "incorrect" answer.

Your answers should be clear and concise. You should be able toanswer each of the questions in the space provided below the question.If you need more space (for example, because you crossed out yourinitial response), you may use the back of the page.

These experiments should require only mutants, phage, media, andgenetic approaches -- No physical, biochemical, or recombinant DNAtechniques will be accepted.

This exam is worth 100 points. The number in parenthesis indicates thenumber of points for each question.

Please double-check to make sure you have answered all questionsbefore turning in the exam.


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(5) 1). A new proB mutation was mapped against a set of proB deletionmutations. The region removed by each deletion mutation is indicated by anopen box below the proB map. The results indicating whether or notrecombinants were obtained are shown to the right of each deletion.

Based on these results, where does the new proB mutation map with respectto the deletion intervals?

ANSWER: The mutation maps in deletion interval #9 where #10 does notoverlap because it cannot repair any deletion that removes this interval, butit can repair all the other deletions.


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(5)2). Given a ser+ met+ Hfr strain where F is integrated in the chromosome asshown below.

a). How would you isolate an F' met+ from this Hfr? (The Hfr strain andrecipient strain can carry any genetic markers you choose, but be sure toindicate what they are.)

ANSWER: Mate the Hfr with an F- met- recA StrR recipient, selecting Met+StrR. The recA mutation in the recipient will prevent inheritance of Met+directly via the Hfr, and the StrR provides a counter-selection against thedonor.

b). Indicate the composition of the medium you would need to use(including auxotrophic supplements).

ANSWER: To select for Met+, the donor + recipient should be plated onminimal glucose medium without methionine. The medium should alsocontain streptomycin to prevent growth of the donor cells.

ser+ met+F


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(15)3). A new mutant was isolated that is StrR and unable to use acetate as acarbon source (ace). To determine where the mutation maps, it was matedwith the four different StrS ace+ Hfr donor strains shown below.[Arrowheads indicate the location and direction of transfer from each differentHfr.]

a).What is the selection for exconjugants in this experiment?

ANSWER: Growth on acetate as a sole C-source.b.)What is the counterselection against the donor cells in this experiment?

ANSWER: Streptomycin resistance.c.)Given the results in the following table, where does the ace mutationmap? [Indicate the map position in minutes.]

Donor Strain Ace+ ColoniesHfr1 1000Hfr2 5Hfr3 1000Hfr4 80

ANSWER: About 95 min, in the region transferred early between Hfr-1 and Hfr-3.

thrA (0 min.)

Hfr 3

Hfr 2Hfr 1

Hfr 4

pyrC (23 min.)

cysA (50 min.)

cysG (73 min.)

metA (90 min.)


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(10)4.) Complementation analysis was done on six mutants that lack theoninesynthetase activity. The order of the mutational sites is not known. Theresults are shown below:

thr-1 thr-2 thr-3 thr-4 thr-5 thr-6thr-1 - + + - - -thr-2 - - + + -thr-3 - + + -thr-4 - - -thr-5 - -thr-6 -

a.)How many complementation groups are represented?ANSWER: At least two complementation groups: Group 1 = thr-1, thr-4, thr-5; Group 2 = thr-2, thr-3; Ungrouped = thr-6b.)Suggest an explanation for the results for thr-6.

ANSWER: thr-6 fails to complement all of the other mutants. Thiscould be either due to a trans-dominant negative phenotype of this mutant(e.g. due to a missense mutation that poisons threonine synthetase) or acis-dominant negative phenotype caused by the mutation (e.g. due to anamber mutation that prevents expression of downstream genes). In eithercase, it is impossible to determine whether the thr-6 mutation is withinone of the two complementation groups described by the other mutationsorwhether it is in a different complementation group.

c.)What is a simple experiment that would provide more information onthe mechnanism of action of thr-6?

ANSWER: Do a complementation analysis with thr6 and the wild-type thrgenes. If thr6 is dominant to WT by poisioning the enzyme complex thecomplementation will be negative. If the wild-type is dominant it suggeststhat thr6 works in cis (by decreasing promoter activity or by nonsensemutation that is polar on the other thr alleles).


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(5)5). Three Arg- mutants were isolated and crossfeeding tests were done asshown below.

a). Based upon these crossfeeding results, indicate the order of the A,B, and C genes in the arginine biosynthesis pathway shown below.

b). Briefly explain your rationale for the proposed order.

ANSWER: The A mutant accumulates the diffusible intermediateornithine which is excreted and can be used by the C mutant. The B mutantaccumulates the diffusible intermediate citrulline which excreted and canbe used by the A and C mutants.

B- A-

C-

Minimal Media No Arg

Glutamate Ornithine Citrulline ArginineC A B


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(15)6). Yanofsky isolated two trpA mutants (trpA8 and trpA17). To determine theorder of the two trpA mutants relative to the trpE gene, he did the P1 transductionexperiment described below.

Donor: trpA17 trpE+

Recipient: trpA8 trpE -

Selected marker: trpA+ 100 coloniesScored markers: trpA+ trpE+ 10 colonies; trpA+ trpE - 90 colonies

Based upon these results, what is the most likely gene order? [i.e., Determine ifthe order is trpE trpA8 trpA17 or trpE trpA17 trpA8).] Make sure you draw clearlythe genetic events that support your answer (ie no guessing).

ANSWER: The order is trpE trpA17 trpA8. In three factor crosses therarest class indicates that four crossovers were required for inheritance.So, to obtain trpA+ trpE+ four crossovers are required.

E+ A17- A8+

E- A17+ A8-

4 Crossovers RequiredFor TrpE+ TrpA+

E+ A17-A8+

E- A17+A8-

Only 2 Crossovers Required For TrpE+ TrpA+


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(10)7.) The drawing below shows the region of the chromosome containing thearg gene after DNA replication. The arrows indicate homologous sequences.

a).Draw the genetic event that shows how duplication of the regioncontaing the arg + gene could occur.

b). Spontaneous duplication of the arg + region occurs at a rate of 10-3 inSalmonella typhimurium. Describe a genetic experiment that would allowyou to isolate a strain that contains the duplication of the arg +region of thechromosome. This can be done by using a S. typhimurium strain thatcontains a KanR cassette in the arg + gene and bacteriophage P22HT. Besure to clearly draw out the genetic events necessary and indicate thecomposition of the medium. (use next page if necessary)

arg+

arg+

arg+

arg+

arg+ arg+


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ANSWER: Select on Minimal Media + Kanamycin (no Arg).

arg + arg+

kan R

arg ’ ‘arg

arg+kan R

arg ’ ‘arg


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(20)8).The trpA gene is required for the biosynthesis of the amino acidtryptophan. trpA mutants require tryptophan for growth. You have made aderivative of a pir-dependent vector that has two fragments that code for thebeginning and end of the trpA gene fused to each other as shown in the figurebelow. The vector also carries a gene that makes cells resistant to tetracycline(tetA) and a gene (sacB) that makes cells sensitive to 5 % sucrose. Expressionof both tetA and sacB is driven by promoters on the plasmid and are expressedin cells.

a). Describe how you could select for integration of this plasmid into thechromosome of a trpA+ cell. Be sure to draw clearly the genetic event(s)that are required.ANSWER: Plate on Minimal Media + Tetracycline.

0 200 1500 1600

0 1600

tetA+ sacB+

ori R6K

tetA+ sacB+

ori R6K

trpA+

TrpA+ TetR SacS


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b). Would the resulting strain be Trp+ or Trp- ? Be sure your drawingsupports your answer.ANSWER: Trp+

c). How could you select for cells that have lost the plasmid?MM + Trp –Tc + 5% Sucrose.

d). Draw out the genetic events that could occur during loss of theplasmid. Indicate the structure of the trpA gene left behind. (Hint: Thereare two different ways the plasmid can excise). What would be thetryptophan phenotype of each type?

tetA+sacB+

ori R6K

trpA+

TrpA+ TetS SacR

trpA+

tetA+sacB+

ori R6K

trpA+

TrpA- TetS SacR

trpA-


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(15)9). The DnaA protein of E. coli, encoded by the dnaA gene, is required forinitiation of E. coli chromosomal DNA synthesis. Mutants [dnaA (TS)] have beenisolated that can synthesize DNA at 30o C but not at 42o C. However, if a cultureof cells containing the dnaA (TS) mutation and a copy of the F plasmid in thecell, it is possible to isolate rare strains that grow normally at 42o C.

a). What is the likely reason why the strain with the dnaA (TS) cannotgrow at 42o C?

ANSWER: Most likely the DnaA protein is required for initiation ofelongation of DNA synthesis. It is inactive at 42oC so the cells die.

b). What is the likely reason (at the molecular level) that the rare strainscontaining the dnaA (TS) mutation and the F plasmid can grow at 42o C?

ANSWER: F can integrate into the chromosome to form Hfrs. Hfrformation is rare and so is the ability of the F+ dnaA TS strain to growat 42oC. Thus it is likely that when F integrates into the chromosome,the TS defect is alleviated. The F plasmid has its own origin ofreplication which it uses to replicate as a plasmid. When Fintegrates to form an Hfr, its origin of replication can be used toinitiate DNA synthesis. DNA synthesis initiated from the F origin ofreplication allows the replication complexes to replicate thechromosomal DNA in Hfr strains. [This is called integrativesuppression and has been used to isolate Hfrs]

c). Briefly describe a genetic test to explain your answer to b. You canuse any strains of E. coli you like.ANSWER: If Hfrs are being made, one could purify a colony that grewat 42oC and test it for the ability to act as an Hfr donor in matingexperiments. One could use an F- recipient with multiple selectablemarkers located around the chromosome so that transfer of anyregion of the chromosome could be detected. Only using onemarker is not a good idea because it may be a late marker and not bedetected. One would also want to use counterselection markers atdifferent regions to be sure that early transfer of the sensitive markerdoes not affect the results.

Exam 2 - University Of Illinois 2/Exam 2 2002.pdf- Exam 2 - Write your name at the top of this cover...

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