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ic interaction and interpretation of genetic intera synthetic pathway/ genes acting in different steps. r genes in a genetic pathway - studies on yeast mat -pheromone response stasis analysis using null mutations- The GAP story stasis analysis using gf mutations - The Ras suppre ncer and synergistic effect between two alleles - -The Ras pathway. rstanding at molecular level/biochemical level. -Limitation of genetics

Genetic interaction and interpretation of genetic interactions - Biosynthetic pathway/ genes acting in different steps. -Order genes in a genetic pathway

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Genetic interaction and interpretation of genetic interactions

- Biosynthetic pathway/ genes acting in different steps.

-Order genes in a genetic pathway - studies on yeast mating -pheromone response

- Epistasis analysis using null mutations- The GAP story

- Epistasis analysis using gf mutations - The Ras suppressors

-Enhancer and synergistic effect between two alleles - -The Ras pathway.

-Understanding at molecular level/biochemical level. -Limitation of genetics

Pheromone response in yeast

QuickTime™ and aPhoto - JPEG decompressor

are needed to see this picture.

Adopted from Hartwell et al. Genetics

Genetics study of mating response

1. How do you want to start it?

2. What is the assay (phenotype) for the screen?

Makay and Manning 1978, Hartwell 1980

Isolated mutations in 12 genes that cause the sterile phenotype. They named them ste1-12. All with the non-response phenotype.

Let us go back to 1970s to re-live the genetics of yeast mating response

Normal plates with yeast lawn + a factor, no growth

spread

+ EMSSte1Ste2Ste3Ste4ste5

Makay and Manning 1978

Hartwell, 1980, get to ste 12 by isolating Ts alleles.

Now what would you do?

If we assume all these genes act in the same pathway.

Assuming ste1-12 mutations are all loss-of-function mutations, can we use them to determine the order of gene actions? A: yes. B: No.

- If you made a double mutant containing ste1 and ste2, what would you see? What would you learn?

A test

Signal Gene A signal response

Gene A has a loss-of-function mutation

If Gene A is a positive factor, the mutation should enhance the signal response

A: Yes.B: No

Signal Gene A signal response

Gene A has a gain-of-function (hyperactive) mutation

If Gene A is a negative factor, the mutation should enhance the signal response

A: Yes.B: No

when two genes act in the same pathway, mutants with opposite phenotypes mean one gene act as a positive regulator, the other is the negative regulator.

Genetic epistasis

Epistatic: one effect masks the other

Epistasis is used to learn about the order of gene action

- indirect- need to learn biochemistry to understand the molecular action- important to verify and biochemical assumption

Epistasis can only be done with two different mutant phenotypes

Genes in the same pathway: Mutations with opposite mutant phenotypes

Blinder et al. 1989 Cell:

Mating constitutive = haploid lethal (please go read the paper)

Use cleaver sector synthetic lethal screen isolated many haploid lethal mutations that are unable to mate.

Several mutations define the Ga gene.

always want to mateUnable to mate ?

About trimeric G protein

receptor

RGS

GTPGDP

Pi

GTPGDP

or , who interacts with the downstream target?

?

?

Who activates the target, or

mating response

G (-): hyper response, always want to mate

G

G (-): No mating response

mating responseG

G(-) & G(-): No mating response

mating responseGA: G

mating responseGGB:C: not sure

Getting tricky

G (-): constitutive mating

Gnon-mating (ste mutants)

G (-) + G (-): no mating.

TargetGA:

Target

G

B:

C: Could be either

Epistasis with lf mutations

TargetGReceptor

lflf

lf lf

Target

GReceptor

lf

lflflf

is required for the target activation

About trimeric G protein

receptor

RGS

GTPGDP

Pi

GTPGDP

TargetGA:

epistasis analysis using loss-of-function mutations: genetic control of programmed cell death

Gene A Gene B

ced-9(+)

ced-9(-)

ced-9(+)

ced-9(-)

ced-4(+)

ced-4(+)

ced-4(-)

ced-4(-)

Phenotype

Normal programmed cell death

Extra cell deaths

Cells that normally die survive

Cells that normally die survive

Conclusion: ced-4(-) phenotype is epistatic to that of ced-9(-)

Results

ced-9 ced-4 cell death

OnOffOff

OffOnOff

No deathExtra deathNo death

Death signalLinear model

ced-9Factor X

Deathsignal

OnOffOff

OffOnOff

OffOnOff

ced-4cell death

No deathExtra deathNo death

Parallel model

Molecular Actions

CED-4No death signal

CED-4 is inactive

Death signal

CED-9

Activates CED-3 caspase for the killing

CED-9

A: linearB: parallel

A second example of epistasis analysis using loss-of-function mutations: dauer larva formation

Gene A Gene B

age-1(+)

age-1(+)

age-1(-)

age-1(-)

daf-18(+)

daf-18(-)

daf-18(+)

daf-18(-)

Phenotype

When starved, worms become dauer larvae

Defective in dauer formation

Constitutive dauer formation

Defective in dauer formation

Conclusion: daf-18(-) phenotype is epistatic to that of age-1

Results

age-1 daf-18 Dauer formation

OnOffOff

OffOnOff

No dauerConstitutive dauerNo dauer

Linear Model signal

Factor Xsignal

OnOffOff

OffOnOff

OnOffOn

Parallel Model

age-1

daf-18Dauer formation

AGE-1PI3Kinase

DAF-18PTEN

PIP3

Binds and activates AKT kinases to prevent dauer formation

signal

+ P

- P

PIP2

Molecular actions A: linearB: parallel

What if you only have positive factors?

- in case of mating response, you only have ste genes, whose lf mutants are non-maters.

You can use gain-of-function mutations (hypermorph).

- gf mutations often generate opposite mutant phenotypes as that of lf mutations in the same gene.

When a hyperactive (gf) mutant phenotype is the winner

Conclusion: the let-60(gf) phenotype is epistatic to that of sem-5(lf)

Gene A Gene B

sem-5(+)

sem-5(+)

sem-5(lf)

sem-5(lf)

let-60 (+)

let-60 (gf)

let-60 (+)

let-60 (gf)

Phenotype in vulval induction

3 of the 6 precursor cells are induced

Extra cells induced (Multivulva)

Less than 3 cells induced (Vulvaless)

Extra cells induced (Multivulva)

Results

Sem-5(GRB2)

let-60(Ras)

Vulval induction

OnOffOff (lf)Off (lf)

OnOffOffOn (gf)

InductionNo induction No induction induction

Model andexplanation

signal

+ -++/-

EGFR SOS

In a given precursor cell

signal pathway function

No parallel model

When a hyperactive (gf) mutant phenotype is the loser

Conclusion: the lin-45 (lf) phenotype is epistatic to that of let-60 (gf)

Gene A Gene B

lin-45 (+)

lin-45(+)

lin-45 (lf)

lin-45 (lf)

let-60 (+)

let-60 (gf)

let-60 (+)

let-60 (gf)

Phenotype in vulval induction

3 of the 6 precursor cells are induced

Extra cells induced (Multivulva)

Less than 3 cells induced (Vulvaless)

Less than 3 cells induced (Vulvaless)

Results

lin-45(Raf)

let-60(Ras)

Vulval induction

OnOffOff (lf)OnOff (lf)

OnOffOnOn (gf) On (gf)

InductionNo inductionNo induction InductionNo induction

Model Asignal

+ -+-+

In a given precursor cell

lin-45/RafGene X

Vulval

induction

OnOnOff (lf)OnOff (lf)

OnOffOnOn (gf) On (gf)

InductionNo inductionNo induction InductionNo induction

signal

+-+-+

let-60/Ras

OnOffOffOnOff

Model B

Biochemistry:Ras directly binds to raf for its activation

Why do we do epistasis analysis?

- provide a critical guide for biochemical analysis

-Add significance to relationship based on biochemical functions