Post on 11-Jan-2016
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
Gene Transfer Processes for Gene Transfer Processes for Bacteria and Their Viruses Bacteria and Their Viruses
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
Genetic Analyses Using Conjugation Genetic Analyses Using Conjugation
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.
Genetic Analyses Using Conjugation Genetic Analyses Using Conjugation
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.
Genetic Analyses Using Conjugation Genetic Analyses Using Conjugation
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
Genetic Analyses Using Conjugation Genetic Analyses Using Conjugation
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
Genetic Analyses Using Conjugation Genetic Analyses Using Conjugation
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.
Genetic Analyses Using Conjugation Genetic Analyses Using Conjugation
B. B. Determining gene order from gradient of Determining gene order from gradient of transfertransfer
Problem 2, page 2-4Problem 2, page 2-4
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?
Genetic Analyses Using Conjugation Genetic Analyses Using Conjugation
2. 2. Determining gene order from gradient of transferDetermining gene order from gradient of transfer
Problem 2, page 2-4Problem 2, page 2-4
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
Genetic Analyses Using Conjugation Genetic Analyses Using Conjugation
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.
Genetic Analyses Using Conjugation Genetic Analyses Using Conjugation
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.
Genetic Analyses Using Conjugation Genetic Analyses Using Conjugation
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++
Genetic Analyses Using Conjugation Genetic Analyses Using Conjugation
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++
leuleu-- metmet--argarg --
HfrHfr
FF--
Genetic Analyses Using Conjugation Genetic Analyses Using Conjugation
Problem 3, page 2-4Problem 3, page 2-4
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++. .
leuleu++ met met+ + arg arg++
leuleu-- metmet--argarg --
HfrHfr
FF--
Genetic Analyses Using Conjugation Genetic Analyses Using Conjugation
Problem 3, page 2-4Problem 3, page 2-4
Recombination between leu and met givesRecombination between leu and met gives
leuleu++ met met-- arg arg-- offspring. offspring.
leuleu++ met met+ + arg arg++
leuleu-- metmet--argarg --
HfrHfr
FF--
Genetic Analyses Using Conjugation Genetic Analyses Using Conjugation
Problem 3, page 2-4Problem 3, page 2-4
Recombination between met and arg gives leuRecombination between met and arg gives leu++ met met++ argarg-- offspring. offspring.
leuleu++ met met+ + arg arg++
leuleu-- metmet--
argarg --
HfrHfr
FF--
Genetic Analyses Using Conjugation Genetic Analyses Using Conjugation
Problem 3, page 2-4Problem 3, page 2-4
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--
..
Genetic Analysis Using Genetic Analysis Using Transformation Transformation
Determining genetic distance with Determining genetic distance with transformation mappingtransformation mapping
Problem 4, page 2-5Problem 4, page 2-5
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.
Genetic Analysis Using Genetic Analysis Using Transformation Transformation
Determining genetic distance with Determining genetic distance with transformation mappingtransformation mapping
Problem 4, page 2-5Problem 4, page 2-5
Rf = Rf = number of single transformantsnumber of single transformants
total number of transformantstotal number of transformants
Genetic Analysis Using Genetic Analysis Using Transformation Transformation
Determining genetic distance with Determining genetic distance with transformation mappingtransformation mapping
hishis++ met
met ++
hishis-- met met--
hishis-- met met+ + 3535
hishis++ met met-- 2727
hishis-- met met-- 194194
Genetic Analysis Using Genetic Analysis Using Transformation Transformation
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--
Genetic Analysis Using Genetic Analysis Using Transformation Transformation
Determining genetic distance with Determining genetic distance with transformation mappingtransformation mapping
Problem 4, page 2-5Problem 4, page 2-5
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.
Gene Transfer Processes for Gene Transfer Processes for Bacteria and Their Viruses Bacteria and Their Viruses
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++. .
Gene Transfer Processes for Gene Transfer Processes for Bacteria and Their Viruses Bacteria and Their Viruses
Three-factor transductionThree-factor transduction
Problem 6, Page 2-5Problem 6, Page 2-5
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
Problem 6, page 2-5Problem 6, page 2-5
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
Problem 6, page 2-5Problem 6, page 2-5
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.
Gene Transfer Processes for Gene Transfer Processes for Bacteria and Their Viruses Bacteria and Their Viruses
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.
Gene Transfer Processes for Gene Transfer Processes for Bacteria and Their Viruses Bacteria and Their Viruses
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
Gene Transfer Processes for Gene Transfer Processes for Bacteria and Their Viruses Bacteria and Their Viruses
Infection with bacteriophageInfection with bacteriophage
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
Gene Transfer Processes for Gene Transfer Processes for Bacteria and Their Viruses Bacteria and Their Viruses
Infection with bacteriophageInfection with bacteriophage
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.
Gene Transfer Processes for Gene Transfer Processes for Bacteria and Their Viruses Bacteria and Their Viruses
4. Infection with bacteriophage4. Infection with bacteriophage
Problem 5, Page 2-5Problem 5, Page 2-5
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
Gene Transfer Processes for Gene Transfer Processes for Bacteria and Their Viruses Bacteria and Their Viruses
Infection with bacteriophageInfection with bacteriophage
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
Gene Transfer Processes for Gene Transfer Processes for Bacteria and Their Viruses Bacteria and Their Viruses
Infection with bacteriophageInfection with bacteriophage
One possible map:One possible map:
hh rrcc rrbb rraa
22 1010 1212