ArabidopsisArabidopsis Experiments: Experiments:

I.I. Forward Genetic Screen (Ethylene Insensitive Mutants): Forward Genetic Screen (Ethylene Insensitive Mutants): • requires thinking.requires thinking.

II. Reverse Genetic Screen / PCR Genotyping (HII. Reverse Genetic Screen / PCR Genotyping (H++- ATPase Mutants):- ATPase Mutants):• requires scoring F2 and thinking.requires scoring F2 and thinking.

What Next?What Next?…experiment I…experiment I

Backcross to wild-type,Backcross to wild-type,

what might the F1 and F2 tell us?what might the F1 and F2 tell us?

Complementation tests?Complementation tests?

Given Given etr1, etr2, ers1, ers2, ein4, etr1, etr2, ers1, ers2, ein4, ctr1, ein2, ein3, eil1ctr1, ein2, ein3, eil1 and and erf1erf1 homozygous plants, and wt homozygous plants, and wt plants; devise a plan to describe plants; devise a plan to describe the genetic nature of the 12 long the genetic nature of the 12 long hypocotyl mutants you found.hypocotyl mutants you found.

dominantdominant

recessiverecessive

Thought Experiments…Thought Experiments…

What Next?What Next?…experiment II…experiment II

L t T5’ 3’

5’ 3’heterozygote

L t T

5’ 3’5’ 3’homozygote

wt

L t T5’ 3’

5’ 3’

homozygotemutant

Genetic AnalysisGenetic AnalysisF2 Segregation (Friday)F2 Segregation (Friday)

…what next? …what next?

1 : 2 : 1

TT Tt

Tt tt

T t

T

t

Not Lethal

1 wt : 2 het

TT Tt

Tt tt

T t

T

t

Lethal

1 wt : 1 het

TT Tt

Tt tt

T t

T

t

GametophyteLethal

How would you confirm / extend F2 results?

Genetic Selection

...the process that establishes conditions in which only the desired genotype will grow.

Selective Media: what might this be?

Genetic Screen

• A system that allows the identification of rare mutations in large scale searches,

– unlike a selection, undesired genotypes are present, the screen provides a way of “screening” them out.

The (Awesome) Power of Bacterial Genetics

... is the potential for studying rare events.

Liquid Cultures,

• 109cells/microliter,

Colonies on Agar,• 107+ cells/colony

Counting Bacteria

(Serial) Dilution is the Solution

10-3 10-510-4

Extra Credit: On another piece of paper, answer the dilution problems on the last page of your handout (2 pts), due Thursday, 13th.

Bacteria Phenotypes

• colony “morphology”,

– large, small, shiny, dull, round or irregular,

– resistance to bactericidal agents,

– vital dyes,

• auxitrophs,

– unable to synthesize or use raw materials from the growth

media.

Prototroph…a cell that is capable of growing on a defined, minimal

media,

– can synthesize all essential organic compounds,

– usually considered the ‘wild-type’ strain.

Auxotrophs…a cell that requires a substance for growth that can be

synthesized by a wild-type cell,

his- ...can’t synthesize histidine (his+ = wt) leu- ...can’t synthesize leucine (leu+ = wt)

arg- ...can’t synthesize arginine (his+ = wt)bio- ...can’t synthesize biotin (bio+ = wt)

Bacterial Nomenclature

• genes not specifically referred to are considered wild-type,

– Strain A: met bio (require methionine and biotin)

– Strain B: thr leu thi

• bacteriacide resistance is a gain of function,

– Strain C: strA (can grow in the presence of strptomycin).

Conjugation

...temporary fusion of two single-celled organisms for the transfer of genetic material,

…the transfer of genetic material is unidirectional.

F+ Cells(F for Fertility)

… F+ cells donate genetic material.

… F- cells receive genetic material,

…there is no reciprocal transfer.

F- Cells(F for Fertility)

F Pilus

…a filamentlike projection from the surface of a bacterium.

F+

F-

F Factor…a plasmid whose presence confers F+, or

donor ability.

F Pilus Attaches to F- Cell

F Factor Replicates During Binary Fission

Properties of the F Factor

• Can replicate its own DNA,

• Carries genes required for the synthesis of pili,

• F+ and F- cells can conjugate,– the F factor is copied to the F- cell, resulting in two F+ cells,

• F+ cells do not conjugate with F+ cells,

• F Factor sometimes integrates into the bacterial chromosome creating Hfr cells.

Hfr Cells

F factor

Bacterial Chromosome

Inserted F plasmid

...F factor integration site,

...host (bacteria chromosome) integration site.

F’Cells

• an F factor from an Hfr cell excises out of the bacterial genome and returns to plasmid form,

• often carries one or more bacterial genes along,

• F’cells behave like an F+ cells,

– merizygote: partially diploid for genes copied on the F’plasmid,

• F’plasmids can be easily constructed using molecular biology techniques (i.e.vectors).

Transfer of lac+pro+ from a F' to an F- strain.

• Strain Sex Genotype

• CSH23 F’ lac+ proA+ proB+ (lacpro) supE spc thi

• CSH50 F- ara (lacpro) strA thi

strA: confers resistance to streptomycin

spc: confers resistance to spectinomycin indicates a deletion of the genes in parentheses

lac: cannot utilize lactose as a carbon source

pro: indicates a requirement for proline

thi; indicates a requirement for thiamine

supE: suppresses nonsense mutations

ara: cannot utilize arabinose as a carbon source.

•

Strain F’ genotype Chromosome Genotype

CSH23 F’lac+ proA+ proB+ (lacpro)supE spc thi

x

CSH 50: ara (lacpro)strA thi

Conjugation

Recombinant Strain: F’lac+ proA+ proB+ ara (lacpro)strA thi

Procedure I:• Day 0: Overnight cultures of the CSH23 and CSH50 will be set up in L broth (a rich medium).

• Day 1: These cultures will be diluted and grown at 37o until the donor culture is 2-3 X 108 cell/ml. What is the quickest way to quickly determine #cells per ml? (This will be done for you.)

Prepare a mating mixture by mixing 1.0 ml of each culture together in a small flask. Rotate at 30 rpms in a 37o shaking incubator for 60 minutes.

At the end of the incubation…

Do serial dilutions:• Fill 6 tubes with 4.5 ml of sterile saline. Transfer 0.5 ml of the undiluted mating culeture to one

of the tubes. This is a 10-1 dilution. • Next make serial dilutions of 10-2, 10-3, 10-4, 10-5 & 10-6. Always change pipets and mix well

between dilutions.

Procedure II:• Plate: 0.1 ml of a 10-2, 10-3 and 10-4 dilution onto minimal + glucose + streptomycin +

thiamine. • Plate: 0.1 ml of a 10-5 and 10-6 dilution onto a MacConkey + streptomycin plates. [A

MacConkey plate is considered a rich media. It has lactose as well as other carbon sources. The phenol red dye is present to differentiate lac+ colonies (red) from lac- colonies (white).]

• Controls: • Plate: 0.1 ml of a 10-1 dilution of donor (CSH23) cells on minimal + glucose + strep +

thiamine plates. Repeat for the recipient (CSH50) cells. • Plate: 0.1 ml of a 10-5 dilution of the recipient on a MacConkey + strep plate.

• Plate: 0.1 ml of a 10-1 dilution of donor on a MacConkey + strep plate.

• Place all plates at 37o overnight.

• Day 2: Remove the plates from the incubator the next day and count the number of white-clear colonies on the MacConkey plates (optional but easier). Store plates at 4oC. NOTE: MacConkey color reactions fade after several days or rapidly in the cold, so plates need to be scored soon after incubation.

What’s Growing?…mate in rich, transfer to…

• Plate: minimal + glucose + streptomycin + thiamine:

– CSH23 yes / no– CSH50 yes / no– exconjugate yes / no

• Plate: MacConkey (rich) + streptomycin plates:

– CSH23 yes / no– CSH50 yes / no– exconjugate yes / no

CSH23 F’lac+ proA+ proB+ (lacpro)supE spc thi

x

CSH 50: ara (lacpro)strA thi

F’lac+ proA+ proB+ ara (lacpro)strA thi

Extra Credit

• On another piece of paper, answer the dilution problems on the last page of your handout (2 pts), due with your abstract on Weds.