Gene Transfer in Bacteria and Bacteriophage Using Gene Transfer Between Bacteria As a Means for Studying Bacterial Genes

Types of Traits Studied

• For bacteria -need for nutrients prototropic: can grow on minimal medium auxotropic: must have specific nutrients added to medium -morphology of colonies -resistance/sensitivity to antibiotics • For bacteriophage -host range (ability to infect specific bacteria) -appearance of plaques (shows growth)

Testing for Nutritional Requirements Replica plating transfers the pattern of bacterial colonies to test plates.

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

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

Gene Transfer Processes for Bacteria and Their Viruses 1. Conjugation 2. Transformation 3. Transduction 4. Infection with bacteriophage

Gene Transfer Processes for Bacteria and Their Viruses 1. Conjugation Transfer of DNA from one bacterial cell to another Donor cell (F + or Hfr) transfers DNA to recipient cell (F )

Conjugation

Genetic Analyses Using Conjugation A. Determining linkage from interrupted mating experiments B. Determining gene order from gradient of transfer C. Higher-resolution mapping by recombination frequency

Genetic Analyses Using Conjugation A. Determining linkage from interrupted mating experiments Combine Hfr strain (Str s ) and F strain.

Remove samples at specific time intervals.

Use blender to disrupt mating.

Plate on streptomycin to kill donor cells.

Test recipient cells for genes from Hfr strain.

Genetic Analyses Using Conjugation A. Determining linkage from interrupted mating experiments Problem 1, page 2-4 Three Hfr strains for

E. coli

are mated individually with an auxotrophic F strain using an interrupted mating procedure. Approximate times of entry of each gene are listed below. Determine the map of the

E. coli

chromosome and show the orientation of the F plasmid in each Hfr strain.

Genetic Analyses Using Conjugation A. Determining linkage from interrupted mating experiments Problem 1, page 2-4 Strain 1 lac + gal + his + argG + xyl + ilv + thr + 3 min Strain 2 argG 12 min xyl + + 39 min ilv + 63 min thr + 74 min lac + 78 min gal + 95 min his + Strain 3 19 min ilv + 30 min xyl + 34 min argG + 51 min his + 59 min gal + 68 min lac + 95 min thr + 5 min 9 min 20 min 44 min 71 min 80 min 88 min

Genetic Analyses Using Conjugation A. Determining linkage from interrupted mating experiments Problem 1, page 2-4 thr 8 1 lac 9 ilv 17 3 4 xyl 11 arg 2 24 gal 27 his

Genetic Analyses Using Conjugation B. Determining gene order from gradient of transfer Combine Hfr and F- strains.

Allow for natural disruption of conjugated pairs.

Select for earliest transferred marker.

Test for markers transferred later in conjugation.

Genetic Analyses Using Conjugation B. Determining gene order from gradient of transfer Problem 2, page 2-4 An Hfr strain donates the genes xyl + lac + and gal + to an F pro + strain. Recombinants are selected for gal + . Tests are done to determine the presence of the other three genes in the gal + recombinants. What is the gene order?

Genetic Analyses Using Conjugation 2. Determining gene order from gradient of transfer Problem 2, page 2-4 gal+ lac+ pro+ xyl+ 100% of strains 70% of strains 30% of strains 10% of strains Select for gal+ Test for lac+, pro+, xyl+ Gene order: Gal---Lac---Pro---Xyl

Recombination to Integrate Transferred Genes a + b + c + a b c a a + b c b + c +

Genetic Analyses Using Conjugation C. Higher-resolution mapping by recombination frequency Combine Hfr and F- strains.

Allow for natural disruption of conjugated pairs.

Select for marker that enters LAST. Test for unselected markers.

Genetic Analyses Using Conjugation C. Higher-resolution mapping by recombination frequency Problem 3, page 2-4 An Hfr strain that is met + conjugated with an F arg + leu + str s strain that is met is arg leu str r . Interrupted mating studies show that leu + str r enters last. Recombinants that are leu are selected and then tested for the + presence of met + and arg + . The following numbers of bacteria are found for each of the genotypes listed below. Determine the gene order and the distances between the genes in map units.

Genetic Analyses Using Conjugation C. Higher-resolution mapping by recombination frequency Problem 3, page 2-4 leu + met arg leu + met + arg leu + met + arg + leu + met arg + 50 80 370 0 Select for leu + Test for met + , arg +

Genetic Analyses Using Conjugation C. Higher-resolution mapping by recombination frequency Problem 3, page 2-4 leu + met met + arg + Hfr F -

Genetic Analyses Using Conjugation Problem 3, page 2-4 Smallest number of offspring represents 4 crossovers, identifies middle gene. Genotype will be leu + met arg + . leu + met + met arg + Hfr F -

Genetic Analyses Using Conjugation Problem 3, page 2-4 Recombination between leu and met gives leu + met arg offspring.

leu + met + met arg + Hfr F -

Genetic Analyses Using Conjugation Problem 3, page 2-4 Recombination between met and arg gives leu + arg offspring.

met + leu + met + met arg + Hfr F -

Genetic Analyses Using Conjugation Problem 3, page 2-4 Leu  met 50 = .1 = 10 map units 500 Met  arg 80 = .16 = 16 map units 500 leu met arg 10 map units 16 map units

Gene Transfer Processes for Bacteria and Their Viruses 2. Transformation DNA taken up from external environment

Genetic Analysis Using Transformation Determining genetic distance with transformation mapping Transform bacteria with DNA containing two markers (eg. his , met ) in addition to penicillin sensitivity.

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

Plate survivors on complete medium to test for his , met .

Genetic Analysis Using Transformation Determining genetic distance with transformation mapping Problem 4, page 2-5 DNA is isolated from E. coli strain A (his and used to transform strain B (his + met + met pen s ) pen s ). Transformants are selected on minimal medium + penicillin to kill his + met + cells and survivors are plated on complete medium. The classes and numbers of cells obtained are listed below. Determine the recombination frequency between the his and met genes.

Genetic Analysis Using Transformation Determining genetic distance with transformation mapping Problem 4, page 2-5 Rf = number of single transformants total number of transformants

Genetic Analysis Using Transformation Determining genetic distance with transformation mapping his met his met + his + met his met 35 27 194

Genetic Analysis Using Transformation Determining genetic distance with transformation mapping Single transformants, his met + and his + met , represent crossovers between the genes.

his met his met -

Genetic Analysis Using Transformation Determining genetic distance with transformation mapping Problem 4, page 2-5 Rf = 35 + 27 = 62 = .24 = 24 map units 256 256

Gene Transfer Processes for Bacteria and Their Viruses 3. Transduction Transfer of bacterial genes with a bacteriophage

Transduction

Genetic Analysis Using Transduction Determining cotransduction frequency with three-factor transduction.

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

Genetic Analysis Using Transduction Three-factor transduction: Transducing bacteriophage are used to transfer DNA with three markers to bacterial cells.

Bacteria are selected for one of the markers and tested for the presence of the other two markers.

Gene order and cotransduction frequency can be determined.

Gene Transfer Processes for Bacteria and Their Viruses Three-factor transduction Problem 6, Page 2-5 Transducing phages that infected an A + B + C + cell are used to infect an A B C cell. Transductants receiving the A + marker were tested for the presence of B + and C + . The classes and numbers of transductants observed is shown below. Determine the gene order and the cotransduction frequencies for A + with B + and A + with C + .

Gene Transfer Processes for Bacteria and Their Viruses Three-factor transduction Problem 6, Page 2-5 A + B + C + A + B + C A + B C + A + B C 45 80 1 300 Select for A + Test for B + and C +

Genetic Analysis Using Transduction Problem 6, page 2-5 Smallest number of offspring represents 4 crossovers, identifies middle gene. Genotype will be A + B C + . A+ B+ C+ A B C -

Genetic Analysis Using Transduction

Problem 6, page 2-5 Cotransduction of A and B A + B + C + A + B + C 45 80 125/426 = .29

Cotransduction of A and C A + B + C + 45 A + B C + 1 46/426 = .11

Genetic Analysis Using Transduction

Problem 6, page 2-5 Cotransduction of A and B = .29

Cotransduction of A and C = .11

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

Therefore A and B are closer than A and C.

Gene Transfer Processes for Bacteria and Their Viruses 4. Infection with bacteriophage In a mixed infection, recombination can be detected between bacteriophage carrying different genes.

Gene Transfer Processes for Bacteria and Their Viruses Infection with bacteriophage Infect bacteria with bacteriophage of two different genotypes. Recombination can occur between bacteriophage genes.

Determine genotypes of resulting bacteriophage.

Rf = number of recombinant plaques total number of plaques

Gene Transfer Processes for Bacteria and Their Viruses Infection with bacteriophage lawn of bacterial cells Plaque for one genotype Plaque for alternate genotype

Genetic Analysis for Infection With Bacteriophage Parental Types r a h + r a + h r a r a + X h + h Recombinant Types r a h r a + h +

Gene Transfer Processes for Bacteria and Their Viruses Infection with bacteriophage Problem 5, Page 2-5 Three different bacteriophage T2 strains x carrying mutations in the r gene (r a , r b and r c ) were each involved in a cross r h + X r + x h , where x=a, b or c. The numbers of bacteriophage of each type are listed below. Give any one of four possible linkage maps for these genes.

Gene Transfer Processes for Bacteria and Their Viruses 4. Infection with bacteriophage Problem 5, Page 2-5 r a h + x r + a h r b h + x r + b h r c h + x r + c h r x h + 340 320 390 r + x h 420 560 590 r + x h + 120 60 8 r x h 120 60 12

Gene Transfer Processes for Bacteria and Their Viruses Infection with bacteriophage Rf = number of recombinant plaques total number of plaques Rf = 120 + 120 = 240 = .24 = 24 map units 1000 1000

Gene Transfer Processes for Bacteria and Their Viruses Infection with bacteriophage One possible map: h 2 r c 10 r b 12 r a