gene transfer in bacteria and bacteriophage using gene transfer between bacteria as a means for...

Gene Transfer in Bacteria and Gene Transfer in Bacteria and BacteriophageBacteriophage

Using Gene Transfer Between Using Gene Transfer Between Bacteria As a Means for Studying Bacteria As a Means for Studying Bacterial GenesBacterial Genes

Types of Types of Traits StudiedTraits Studied

• For bacteriaFor bacteria-need for nutrients-need for nutrients prototropic: can grow on minimal mediumprototropic: can grow on minimal medium

auxotropic: must have specific nutrients added auxotropic: must have specific nutrients added to medium to medium

-morphology of colonies-morphology of colonies-resistance/sensitivity to antibiotics-resistance/sensitivity to antibiotics

• For bacteriophage For bacteriophage -host range -host range (ability to infect specific bacteria) (ability to infect specific bacteria)

-appearance of plaques-appearance of plaques (shows growth)(shows growth)

Testing for Nutritional RequirementsTesting for Nutritional Requirements

Replica plating transfers the pattern of Replica plating transfers the pattern of bacterial colonies to test plates. bacterial colonies to test plates.

DNA of Prokaryotic CellsDNA of Prokaryotic Cells

• Bacterial cells have a single, circular Bacterial cells have a single, circular chromosome and therefore have one chromosome and therefore have one copy of each gene.copy of each gene.

• Partial diploids (merozygotes) can be Partial diploids (merozygotes) can be formed by the introduction of genetic formed by the introduction of genetic material from another cell.material from another cell.

Gene Transfer Processes for Gene Transfer Processes for Bacteria and Their Viruses Bacteria and Their Viruses

1.1. ConjugationConjugation

2.2. TransformationTransformation

3.3. Transduction Transduction

4.4. Infection with bacteriophageInfection with bacteriophage


1. Conjugation1. Conjugation

Transfer of DNA from Transfer of DNA from one bacterial cell one bacterial cell to anotherto another

Donor cell (FDonor cell (F++ or Hfr) transfers DNA to or Hfr) transfers DNA to recipient cell (Frecipient cell (F--))

Conjugation Conjugation

Genetic Analyses Using Conjugation Genetic Analyses Using Conjugation

A.A. Determining linkage from interrupted Determining linkage from interrupted mating experimentsmating experiments

B.B. Determining gene order from gradient Determining gene order from gradient of transferof transfer

C.C. Higher-resolution mapping by Higher-resolution mapping by recombination frequencyrecombination frequency


A.A. Determining linkage from Determining linkage from interrupted mating experimentsinterrupted mating experiments

Combine Hfr strain (StrCombine Hfr strain (Strss) and F) and F-- strain. strain.Remove samples at specific time intervals.Remove samples at specific time intervals.Use blender to disrupt mating.Use blender to disrupt mating.Plate on streptomycin to kill donor cells.Plate on streptomycin to kill donor cells.Test recipient cells for genes from Hfr Test recipient cells for genes from Hfr strain.strain.


A. Determining linkage from interrupted A. Determining linkage from interrupted mating experimentsmating experiments

Problem 1, page 2-4Problem 1, page 2-4

Three Hfr strains for Three Hfr strains for E. coliE. coli are mated are mated individually with an auxotrophic Findividually with an auxotrophic F- - strain using strain using an interrupted mating procedure. Approximate an interrupted mating procedure. Approximate times of entry of each gene are listed below. times of entry of each gene are listed below. Determine the map of the Determine the map of the E. coliE. coli chromosome chromosome and show the orientation of the F plasmid in and show the orientation of the F plasmid in each Hfr strain. each Hfr strain.


A. Determining linkage from interrupted A. Determining linkage from interrupted mating experimentsmating experimentsProblem 1, page 2-4Problem 1, page 2-4

Strain 1Strain 1 Strain 2Strain 2 Strain 3Strain 3

laclac++ 3 min3 min argGargG++ 19 min19 min ilvilv++ 5 min5 min

galgal++ 12 min12 min xylxyl++ 30 min30 min xylxyl++ 9 min9 min

hishis++ 39 min39 min ilvilv++ 34 min34 min argGargG++ 20 min20 min

argGargG++ 63 min63 min thrthr++ 51 min51 min hishis++ 44 min44 min

xylxyl++ 74 min74 min laclac++ 59 min59 min galgal++ 71 min71 min

ilvilv++ 78 min78 min galgal++ 68 min68 min laclac++ 80 min80 min

thrthr++ 95 min95 min hishis++ 95 min95 min thrthr++ 88 min88 min


A. Determining linkage from interrupted A. Determining linkage from interrupted mating experimentsmating experiments

Problem 1, page 2-4Problem 1, page 2-4 galgallaclac

hishis

argarg

xylxyl

ilvilv

thrthr 9

27

2411

4

17

8

2

1

3


B. Determining gene order from gradient B. Determining gene order from gradient of transferof transfer

Combine Hfr and F- strains.Combine Hfr and F- strains.Allow for natural disruption of conjugated Allow for natural disruption of conjugated pairs.pairs.Select for earliest transferred marker.Select for earliest transferred marker.Test for markers transferred later in Test for markers transferred later in conjugation.conjugation.


B. B. Determining gene order from gradient of Determining gene order from gradient of transfertransfer


An Hfr strain donates the genes xylAn Hfr strain donates the genes xyl++ pro pro++ laclac+ + and galand gal+ + to an Fto an F-- strain. Recombinants strain. Recombinants are selected forare selected for galgal++. Tests are done to . Tests are done to determine the presence of the other three determine the presence of the other three genes in the galgenes in the gal++ recombinants. What is recombinants. What is the gene order? the gene order?


2. 2. Determining gene order from gradient of transferDetermining gene order from gradient of transfer


Gene order: Gal---Lac---Pro---XylGene order: Gal---Lac---Pro---Xyl

gal+gal+ 100% of strains100% of strains

lac+lac+ 70% of strains70% of strains

pro+pro+ 30% of strains30% of strains

xyl+xyl+ 10% of strains10% of strains

Select for gal+Select for gal+Test for lac+, pro+, xyl+Test for lac+, pro+, xyl+

Recombination to Integrate Recombination to Integrate Transferred GenesTransferred Genes

aabb cc

aa++ bb++ cc++

aa++ bb++cc++

aa bb cc


C. Higher-resolution mapping by C. Higher-resolution mapping by recombination frequency recombination frequency

Combine Hfr and F- strains.Combine Hfr and F- strains.Allow for natural disruption of conjugated Allow for natural disruption of conjugated pairs.pairs.Select for marker that enters LAST. Select for marker that enters LAST. Test for unselected markers. Test for unselected markers.


C. Higher-resolution mapping by C. Higher-resolution mapping by recombination frequency recombination frequency Problem 3, page 2-4Problem 3, page 2-4

An Hfr strain that is metAn Hfr strain that is met++ arg arg++ leu leu++ str strss is is conjugated with an Fconjugated with an F-- strain that is met strain that is met-- arg arg-- leuleu-- str strrr. Interrupted mating studies show that . Interrupted mating studies show that leuleu++ enters last. Recombinants that are leu enters last. Recombinants that are leu++ strstrrr are selected and then tested for the are selected and then tested for the presence of metpresence of met++ and arg and arg++. The following . The following numbers of bacteria are found for each of the numbers of bacteria are found for each of the genotypes listed below. Determine the gene genotypes listed below. Determine the gene order and the distances between the genes in order and the distances between the genes in map units. map units.


C. Higher-resolution mapping by C. Higher-resolution mapping by recombination frequency recombination frequency Problem 3, page 2-4Problem 3, page 2-4

leuleu++ met met-- arg arg-- 5050

leuleu++ met met++ arg arg-- 8080

leuleu++ met met++ arg arg++ 370370

leuleu++ met met-- arg arg++ 00

Select for leuSelect for leu++

Test for metTest for met++, arg, arg++


C. Higher-resolution mapping by C. Higher-resolution mapping by recombination frequency recombination frequency


leuleu++ met met+ + arg arg++

leuleu-- metmet--argarg --

HfrHfr

FF--



Smallest number of offspring represents 4 Smallest number of offspring represents 4 crossovers, identifies middle gene. crossovers, identifies middle gene.

Genotype will be leuGenotype will be leu++ met met-- arg arg++. .



HfrHfr

FF--



Recombination between leu and met givesRecombination between leu and met gives

leuleu++ met met-- arg arg-- offspring. offspring.



HfrHfr

FF--



Recombination between met and arg gives leuRecombination between met and arg gives leu++ met met++ argarg-- offspring. offspring.


leuleu-- metmet--

argarg --

HfrHfr

FF--



Leu Leu met met 5050 = .1 = 10 map units = .1 = 10 map units

500500

Met Met arg arg 8080 = .16 = 16 map units = .16 = 16 map units

500500

leu met argleu met arg10 map units10 map units 16 map units16 map units

Gene Transfer Processes for Gene Transfer Processes for Bacteria and Their VirusesBacteria and Their Viruses

2. Transformation2. Transformation

DNA taken up from external environmentDNA taken up from external environment

Genetic Analysis Using Genetic Analysis Using Transformation Transformation

Determining genetic distance with Determining genetic distance with transformation mappingtransformation mapping

Transform bacteria with DNA containing two markers Transform bacteria with DNA containing two markers (eg. his(eg. his--, met, met--) in addition to penicillin sensitivity.) in addition to penicillin sensitivity.

Select transformants on minimal medium + penicillin Select transformants on minimal medium + penicillin to kill non-transformants.to kill non-transformants.

Plate survivors on complete medium to test for hisPlate survivors on complete medium to test for his--, , metmet--

..




DNA is isolated from E. coli strain A (hisDNA is isolated from E. coli strain A (his- - metmet-- pen penss) ) and used to transform strain B (hisand used to transform strain B (his+ + metmet++ pen penss). ). Transformants are selected on minimal medium + Transformants are selected on minimal medium + penicillin to kill hispenicillin to kill his+ + metmet+ + cells and survivors are plated cells and survivors are plated on complete medium. The classes and numbers of on complete medium. The classes and numbers of cells obtained are listed below. Determine the cells obtained are listed below. Determine the recombination frequency between the his and met recombination frequency between the his and met genes. genes.




Rf = Rf = number of single transformantsnumber of single transformants

total number of transformantstotal number of transformants



hishis++ met

met ++

hishis-- met met--

hishis-- met met+ + 3535

hishis++ met met-- 2727

hishis-- met met-- 194194


Determining genetic distance with transformation mappingDetermining genetic distance with transformation mapping

Single transformants, hisSingle transformants, his-- met met++ and his and his++ met met--, , represent crossovers between the genes.represent crossovers between the genes.

hishis++ met

met ++

hishis-- met met--

hishis++ met

met ++

hishis-- met met--




Rf = Rf = 35 + 2735 + 27 = = 6262 = .24 = 24 map units = .24 = 24 map units 256 256256 256

Gene Transfer Processes for Gene Transfer Processes for Bacteria and Their VirusesBacteria and Their Viruses

3. Transduction 3. Transduction Transfer of Transfer of bacterial genes bacterial genes with a with a bacteriophage bacteriophage

TransductionTransduction

Genetic Analysis Using Transduction Genetic Analysis Using Transduction

Determining cotransduction frequency Determining cotransduction frequency with three-factor transduction.with three-factor transduction.

Cotransduction frequency = tendency Cotransduction frequency = tendency for genes to be transferred together on for genes to be transferred together on same piece of transducing DNAsame piece of transducing DNA

Genetic Analysis Using Transduction Genetic Analysis Using Transduction

Three-factor transduction:Three-factor transduction: Transducing bacteriophage are used to transfer Transducing bacteriophage are used to transfer

DNA with three markers to bacterial cells.DNA with three markers to bacterial cells. Bacteria are selected for one of the markers Bacteria are selected for one of the markers

and tested for the presence of the other two and tested for the presence of the other two markers.markers.

Gene order and cotransduction frequency can Gene order and cotransduction frequency can be determined.be determined.


Three-factor transductionThree-factor transduction

Problem 6, Page 2-5Problem 6, Page 2-5

Transducing phages that infected an ATransducing phages that infected an A++BB++CC++ cell cell are used to infect an Aare used to infect an A--BB--CC-- cell. Transductants cell. Transductants receiving the Areceiving the A++ marker were tested for the marker were tested for the presence of Bpresence of B+ + andand CC++. The classes and numbers . The classes and numbers of transductants observed is shown below. of transductants observed is shown below. Determine the gene order and the Determine the gene order and the cotransduction frequencies for Acotransduction frequencies for A++ with B with B++ and A and A++ with Cwith C++. .


Three-factor transductionThree-factor transduction


AA++ BB++ C C++ 4545

AA++ BB++ C C-- 8080

AA++ BB-- C C++ 11

AA++ BB-- C C-- 300300

Select for ASelect for A++

Test for BTest for B+ + andand CC++

Genetic Analysis Using TransductionGenetic Analysis Using TransductionProblem 6, page 2-5Problem 6, page 2-5

A+ B+ C+A+ B+ C+

AA--BB--

CC--

Smallest number of offspring represents 4 Smallest number of offspring represents 4 crossovers, identifies middle gene. Genotype crossovers, identifies middle gene. Genotype will be Awill be A++ B B-- C C++. .

Genetic Analysis Using Genetic Analysis Using TransductionTransduction


Cotransduction of A and BCotransduction of A and B

AA++BB++CC++ 45 45

AA++BB++CC-- 8080

125/426 = .29125/426 = .29

Cotransduction of A and CCotransduction of A and C

AA++BB++CC++ 45 45

AA++BB--CC++ 1 1

46/426 = .1146/426 = .11

Genetic Analysis Using Genetic Analysis Using TransductionTransduction


Cotransduction of A and B = .29Cotransduction of A and B = .29

Cotransduction of A and C = .11Cotransduction of A and C = .11

The higher the cotransduction frequency, the The higher the cotransduction frequency, the closer the genes are to each other.closer the genes are to each other.

Therefore A and B are closer than A and C.Therefore A and B are closer than A and C.


4. Infection with 4. Infection with bacteriophagebacteriophage

In a mixed infection,In a mixed infection,recombination can be recombination can be detected between detected between bacteriophage carrying bacteriophage carrying different genes.different genes.


Infection with bacteriophageInfection with bacteriophage

Infect bacteria with bacteriophage of two different Infect bacteria with bacteriophage of two different genotypes. genotypes.

Recombination can occur between bacteriophage genes.Recombination can occur between bacteriophage genes. Determine genotypes of resulting bacteriophage.Determine genotypes of resulting bacteriophage.

Rf = Rf = number of recombinant plaquesnumber of recombinant plaques total number of plaquestotal number of plaques



lawn of bacterial cellslawn of bacterial cells

Plaque for Plaque for one genotypeone genotype

Plaque for Plaque for alternatealternategenotypegenotype

Genetic Analysis for Infection Genetic Analysis for Infection With BacteriophageWith Bacteriophage

rraa-- hh++

hh--rraa++

Parental TypesParental Types

rraa- - hh++

rraa+ + hh--

Recombinant TypesRecombinant Types

rraa- - hh--

rraa+ + hh++

XX



Problem 5, Page 2-5Problem 5, Page 2-5 Three different bacteriophage T2 strains carrying Three different bacteriophage T2 strains carrying

mutations in the r gene (rmutations in the r gene (raa, r, rb b and rand rcc) were each ) were each involved in a cross rinvolved in a cross r--

xxhh++ X r X r++xxhh--, where x=a, b or , where x=a, b or

c. The numbers of bacteriophage of each type are c. The numbers of bacteriophage of each type are listed below. Give any one of four possible linkage listed below. Give any one of four possible linkage maps for these genes. maps for these genes.


4. Infection with bacteriophage4. Infection with bacteriophage


rr--x x hh++ rr++

x x hh-- rr++x x hh++ rr--

x x hh--

rr--a a hh++ x r x r++

a a hh-- 340340 420420 120120 120120

rr--b b hh++ x r x r++

b b hh-- 320320 560560 6060 6060

rr--c c hh++ x r x r++

c c hh-- 390390 590590 88 1212



Rf = Rf = number of recombinant plaquesnumber of recombinant plaques

total number of plaquestotal number of plaques

Rf = Rf = 120 + 120120 + 120 = = 240240 = .24 = 24 map units = .24 = 24 map units

1000 10001000 1000



One possible map:One possible map:

hh rrcc rrbb rraa

22 1010 1212

gene transfer in bacteria and bacteriophage using gene transfer between bacteria as a means for...

Documents

conjugation transfer

auxotrophic f strain

gal test

interrupted mating procedure

f strains

hfr strains

mingenetic analyses

donor cell f