The Yeast Experiment is the easiest experiment to perform.Why was it left for the end of the term? Why not start off with the easiest experiment?

Although it’s the easiest experiment to do, it’s the most difficult to understandand interpret!

You may be confused, but no worries, you have plenty of time to study this.

You will have to read Article 5 to better understand this experiment.

I will help you with Assignment 4 Today!!

Yeast Switching Model

Summary of Experiment 5

1 2

43

5 6

87

swi5 Tester 4570 Parental Mutants Mutants

1 2 3

4 5 6

7 8 9

Mating on YEPD plates1-8: mutant X Tester9th: tester X parental (LAB 9)

1 2 3

4 5 6

7 8 9

Restreak mated colonies on DS plates to select for diploids and complementation

(LAB 10)

Filter paper

Cut

1 2 3

4 5 6

7 8 9

Β-galactosidase assay

Streak mutants on galactoseplates. 9th square is control

(LAB 11)

Results SectionInitial Phenotype of Mutant

Complementation Test

β-galactosidase Test

Probable Mutation(s)

1 Red or White Red or White Blue or White swi5, swi, she or unknown

2

3

4

5

6

7

8

P

Characterization of probable mutations: will have to give explanations at the genetic level.

You may print this table and use it for your results section if you wish.

Results SectionInitial Phenotype of Mutant

Complementation Test

β-galactosidase Test

Probable Mutation(s)

Red or White Red or White Blue or White swi5, swi, she or unknown

1 Red

2 Red

3 Red

4 Red

5 Red

6 Red

7 Red

8 Red

P White

1) Do you think you have found a she mutant? Why?

2) Do you think you have found a swi5 mutant? Why?

3) Do you think you have found a swi mutant that is not swi5? Why?

4) Have you identified mutants that do not correspond to these classes of mutants? In other words, have you identified mutants that are difficult to classify? Why are they difficult to classify? Any ideas on what they could be?

Only 4 Questions for Assignment 5

Note: The questions are not as easy as they look. Start working on it as soon as you get your results.

Yeast As a Model SystemAdvantages:

Rapid growth, the ease of replica plating, mutant isolation, non-pathogenic, well-defined genetic system, highly versatile DNA transformation system, cost-friendly.

Transformation is carried out directly using short single-stranded synthetic oligos

Strains of Saccharomyces cerevisiae have both stable haploid and diploid state.

Recessive mutations are conveniently manifested in haploid strains

The above studies allow for studying: Analysis of gene regulationStructure-function relationships of proteins

About 30% of disease genes in humans have orthologs in yeast cells.

The Life Cycle of Yeast

Heterothallic Yeast Life Cycle

Homothallic life cycle

Homothallism:

Unstable mating type

Can switch mating type by interconversionof the MAT locus

A single spore can complete the entiresexual reproduction cycle.

Mother specific switching

What does this lineage tell us?

- Only mother cells switch mating type.

In which phase of the cell cycle is the switch

occurring?

M

G2

S

G1

R-POINT

Why switching in only mother cells and not daughter cells?

A single spore can give rise to cells that have differentdevelopmental fates

Differences in daughters may originate from:

Differences in their environment

and/or

Unequal segregation of cell fate determinants during cell division

We know the following: The HO endonuclease is involved

in mating type interconversion.

Therefore, understanding the mechanism of HO expression and

regulation may provide insight into the determinant(s) of

asymmetric segregation.

What is the Asymmetric determinant?

To answer this question, we first need to know what protein(s) are involved in theswitching.

Characteristics of the HO Endonuclease

Gene for Homothallism

An endonuclease that cuts DNA

Expressed only in mother cells

Expression is transient (only during late G1)

HO is a haploid-specific gene (α or a ) and not present in diploids (α/a)

Understanding why HO is confined to mother cells will help understand whyonly mother cells switch mating type.

Mating type interconversion

Mating type switching occurs at too high a frequency to be a mutational event,therefore, it has to be a precise genetic regulation as shown below:

The model proposes that the HMLα and HMRa loci contain "silent copies" of α and a mating type genes. Replicas of either can be copied into MAT, the active locus, where they are expressed.

Mating type locus (MAT)

αsg (α-specific genes)

asg (a-specific genes)

hsg (haploid-specific genes)

RME (Trascriptional repressor of meiosis-specific and sporulation-specificgenes)

Cis and Trans-Acting Factors involved in HO Expression and Switching

2 Regions in the HO promoter are responsible to restricting HO expression to late G1 and to mother cells: URS1 and URS2

Note: Regulation of HO gene expression is MUCH more complicated than depicted above

Switching

• SWI5 required for mother daughter control but is not the asymmetric determinant.

• SWI4 and 6 are required for cell cycle regulation.

HO cis-regulation

• URS1required for Mother daughter control

• URS2 required for cell cycle control

What is the determinant of asymmetric HO expression?

SHE Proteins?ASH1 Proteins?

SHE (Swi5-dependent HO Expression)

ASH1 (Asymmetric Synthesis of HO)

SHE proteins (SHE1-SHE5) are cytoplasmic and SHE1 is similar to the Myo4 proteins.ASH1 is a transcriptional repressor

SHE proteins are involved in transporting ASH1 to daughter cells

Therefore, ASH1 accumulates predominantly in daughter cells in a SHE-dependent manner.

ASH1 represses transcription of HO in daughter cells

She mutants do not express HO because ASH1 is equally present in mother and daughter cells

Mutations in the SHE gene significantly reduces mother cell switching (6%).

Asymmetric accumulation of ASH1 leads to mother cell-specific switchingwhich is SHE-dependent

Symbol Definition

ARG2 A locus or dominant allele

arg2 A locus or recessive allele confering an arginine requirement

ARG2+ The wild-type allele

arg2-9 A specific allele or mutation

Arg+ A strain not requiring arginine

Arg- A strain requiring arginine

Arg2p The protein encoded by ARG2

arg2-1 A specific complete or partial deletion of ARG2

ARG2::LEU2Insertion of the functional LEU2 gene at the ARG2 locus, and ARG2 remains functional anddominant

arg2::LEU2Insertion of the functional LEU2 gene at the ARG2 locus, and arg2 is or becamenonfunctional

can1 Canavanine resistance

CAN1 Canavanine sensitivity

MATα Wild-type allele of the α mating type locus

MATaHO

Wild-type allele of the a mating type locusHomothallic gene encoding the endonuclease involved in mating type switching

Genetic nomenclature

Auxotrophs and Autotrophs

An auxotroph contains mutations that alter the nutritional requirements of an organism. A wild-type organism, containing no mutations, is called a prototroph.

If the strain contains a mutation in a gene that is required for the biosynthesis of a specific amino acid or nucleotide, then the strain must acquire that amino acid or nucleotide from the medium (amino acids are the building blocks of proteins, and without protein, a cell cannot survive).

If the medium lacks that amino acid or nucleotide, the mutant strain will not grow.

Examples of enzymes required in the biosynthetic pathways of amino acids

ADE2:phosphoribosylamino-imidazole-carboxylase (Catalyses step in adenine synthesis) HIS3:imidazoleglycerol-phosphate dehydratase (Catalyses 6th step in histidine synthesis) TRP1:phosphoribosylanthranilate isomerase (Catalyses 3rd step in tryptophan synthesis) ARG2:glutamate N-acetyltransferase (Catalyses 1st step in arginine synthesis)

ADE2

Null mutant is viable and requires adenine. ade2 mutants are blocked at a stage in the adenine biosynthetic pathway that causes red colour intermediate to accumulate giving the cell a red color.

Complementation Analysis

Occasionally, multiple mutations of a single wild type phenotype are observed.

Question: 1. Do any of the mutations occur in a single gene? OR 2. Does each mutation represent one of the several genes necessary for a phenotype to be expressed?

The simplest test to distinguish between the two possibilities is the complementation test.

Two mutants are crossed, and the F1 is analyzed. If the F1 expresses the wild type phenotype then each mutation is in one of two possible genes necessary for the wild type phenotype.

Alternatively, if the F1 does not express the wild type phenotype, but rather a mutant phenotype then both mutations occur in the same gene.

β-galactosidase is an enzyme encoded by the bacterial gene lacZ.

β-galactosidase cleaves the colorless substrate X-gal (5-bromo-4-chloro-3-indolyl-β-galactopyranoside) into galactose and a blue insoluble product of the cleavage.

Therefore, lacZ can be used as a reporter gene (blue/white selection)

X-GAL

The LacZ Reporter Gene

Genotype of haploids

Mutants (UV mutagenesis of strain 4570)

mat del::LEU2, leu2, trp1, ade2-1, his3-11, ura3::C2791::URA3, HO-ADE2, HO-CAN1. C2791=YIplac211+HO/GAL-lacZ plus a mutation in a gene required for HO regulation.

swi5 mutant tester strain:

MATa swi5::LEU2, leu2, trp1, ura3, his3, ade2-1, can1, pCEN TRP1.

Unmutagenized parental 4570 strain:

Mat del::LEU2, leu2, trp1, ade2-1, his3-11, ura3::C2791::URA3, HO-ADE2, HO-CAN1. C2791=YIplac211+HO/GAL-lacZ

Complementation analysis: considerations

What will you expect to see if there is complementation and non-complementation? What are you complementing?

Red colonies means non-complementation.

White colonies means complementation.

You are testing whether your mutation is allelic to swi5.

THE END