Budding Yeast Saccharomyces cerevisiae as a Model ' Pg739753' Omit 747749, mating type switching

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Budding Yeast (Saccharomyces cerevisiae) as a
Model . (Pg739-753. Omit 747-749, mating
type switching) Budding yeast Human cells
25 of human disease genes have ortholog in
yeast Orthologs in yeast XP, RAS- Cancer
genes Orthologs Not in yeast p53, dystrophin
Technical Advantages/Comparison
s Single-celled multicellular 6000
genes 30,000 genes Growth cheap
Expensive Haploid diploid phases! Diploids
only Double cell number in 2 hr Double in 12-24
hr. Molecular Manipulations! (see
later) http//genome-www.standford.edu/saccharomyc
es/
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Chromosomes
Budding yeast Human cells
16 linear chromosomes 23 linear chromosomes
22 autosomes 1 sex chromosome
Same scale
(cant see em)
Easily visible in light microscope
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Fig A.2 Yeast Chromosomes
Agarose gel karyotype
Fig A.3 Yeast Chromosomes- Comparison of genetic
and physical map
2 points Major Point Genetic and physical maps
similar. . .
..Recombination random and proportional
to distance. Minor Exceptions Example Mat-thr4
region more crossovers (recombination) makes
genetic distance larger than physical distance.
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The Yeast Genome- Factoids
-13MB of DNA sequence- completed in 1996 Whats a
gene??? Defining an ORF is not so easy
ATG.. ..TGA
On the evolution of the yeast genome..Fig A.5
Fig A. 4 ORF size
gene families arose by duplication of
chromosomal regions.
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KEY Technical Trick Pioneered in Yeast Gene
Targeting
In vitro
What this looks like in a yeast cell
In vivo- in yeast cells
Well talk tetrads later..
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Bit of yeast biology Fig A8
Mating of Yeast Cells-when an a meets an alpha
a haploid
a haploid
Zygote- diploid formed
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Fig A13 Mating Response
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The Mating pathway- a Signal Transduction Process
Biochemistry
Genetics
MATa cell
MATa cell
Pheromone
Gpa1
STE2
GTP binding proteins
GPA1
P
Ste11
Protein kinase cascade
Ste5
P
STE4
Ste7
Fus3
P
STE5
P
Ste12
STE11
Active transcription factor (like CRP for lac
operon)
STE7
FUS3, KSS1
Cell cycle arrest
STE2- receptor- positive feedback FAR1- cell
cycle arrest FUS1- cell fusion SST2-negative
feedback
STE12
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The MAT locus - a model for cell
identity
HMRa
HMLa
MAT
Transcribed? NO YES
NO
Cell type Meiosis?
MATa haploid a
NO Mata haploid a
NO MATa/MATa diploid a/a
YES MATa/MATa diploid a NO
How Cell Identity is Determined in Yeast
Combinations!
hsg, RME, a1
a1a2 a2 a1 mcm1 mcm1
asg
asg
asg
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A Genetic View of Chromosome Behavior in
Mitosis and Meiosis
a
a
a
a
a
a
Mitosis
a
a
Diploid cell
a
a
a
a
a
a
a
Meiosis
a
a
a
a
a
a
a
Meiosis I
a
a
a
a
a
a
Meiosis II
a
a
a
a
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his3-
HIS3
his3-
HIS3-
a
his3-
Meiosis
HIS3
a
a
a
a
a
a
a
Meiosis I
a
a
a
a
a
a
a
a
a
a
a his3- a HIS3 a his3- a
HIS3
a
Meiosis
a
a
a
a
a
leu2-
LEU2
a
a
Meiosis I
a
a
a
a
a
a
a
a
a
a
a leu2- a leu2- a
LEU2 a LEU2
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Yeast S114
Yeast cells genetic analysis of tetrads or
random spores
MATa MATa
sporulate starve, they do meiosis
Tetrad with single cells (with nuclei)
Dissect
Agar plate w/ 1 cell Not to scale!)
Grow few days
Agar plates w/ 106 cells per colony
a a a a
a a a a
a a a a
Genotypes
Genotype 11 (22)
22!!!!
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A simple assay of
mating.
MATa ura1
Agar plate with streaked yeast cells on rich
media
MATa ura3
MATa ura1
MATa ura3
Rich media No selection
Grow overnight- they mate if they can!
Select Ura by replica plating
Haploid streaks
MATa/MATa ura1/ ura3/
MATa/MATa ura1/ura1
MATa/MATa ura1/ ura3/
Agar plate without uracil Selects for Ura cells
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Mendelian segregation of genetic
markers
Case 1 of 2 loci that are not linked MAT and
HIS3 on different chromosomes, or are far away
on same chromosome.
MAT HIS
Sporulate (meiosis)
MATa his3-
MATa his3
a a a a
- -
Recombinant
X
MATa
MATa HIS3
parental
parental
Recombinant
All 4 equally likely 1111
In tetrads this is more complicated, but overall
outcome the same random segregation
Case 2 of 2 loci that are tightly linked MAT and
LEU2 are on same chromosome and near to each other
MAT LEU2
MATa leu2-
Sporulate (meiosis)
MATa leu2-
a a a a a a
- - -
parental
X
MATa
MATa LEU2
parental
parental
Parentalsgtgtrecombinants Means that very few
crossovers between MATa and leu2.
parental
Recombinant
Recombinant
(board)
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• Lets isolate and characterize mating mutants
• Mutagenize cells with EMS, or X-rays, or UV
• Screen for mating defect
• Determine if we have mutations in several genes
• Determine recessive and dominance
• Perform epistasis/order of function tests
• Form the pathway

MATa Ste ura3-
mutagenize
MATa ura1- tester
Plate and grow colonies
MATa Ste ura3-
Test Mating
Strain 1 Strain 2
Grow one day
Growth Growth No growth
Grow 1 day
Replica to plate without uracil
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We need conditional
mutants. To see if SteA and SteB (two mutants)
have mutations in the same or different gene, we
need to make a diploid cell, and then test for
complementation and for recombination. Yet, the
very nature of SteA and SteB is that they
cannot form diploids! Solution Isolate
conditional mutants Generate alleles that have a
permissive temperature (say 23C) and a
restrictive temperature (say 37C)
MATa ura1 tester
MATa Ste ura3
mutagenize
Plate and grow colonies
Wildtype Conditional mutant Non-conditional mutant
Grow permissive temperature
Grow restrictive temperature
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• Lets analyze some sterile mutants.
• Suppose we isolate 2 mutants that each fail to
  mate.
• Call them steA and steB
• We wish to know
• Are they dominant or recessive?
• Are mutations in the same gene or in different
  genes?
• How about double mutants to determine order of
  function?

If 2 mutations in same gene, if both recessive,
then we make a diploid
Case 1 A and B in same gene
Case 2 A and B in different genes
OR
MATa ste1-1
MATa ste2-1
MATa ste1-2
MATa ste1-1

• Complementation tests
• Recombination tests

• Mates
• 1/4 haploids mate

• Does not mate
• All haploids fail to mate

One more tiny problem to solveMATa STE1
MATa STE1
Fails to mate because cells are MATa and MATa!
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Solution Make an a/a or a/a diploid
MATa
Have
MATa
MATa
Need
MATa
Solution Mitotic recombination to homozygous
MAT locus
How Use Another Genetic Trick of Mitotic
Recombination
CryS is dominant to CryR Drug crytoplurine kills
CryS/CryS and CryS/CryR cells, only
CryR/CryR survive.
CryS MATa
CryR MATa
Select CryR/CryR cells
CryR MATa
CryS MATa
CryR MATa
CryS MATa
How can this happen?
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Mitotic Recombination
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CryR MATa
CryS MATa
DNA replication
1
Crossover rare, spontaneous 1 in 105 cells in
mitotic cells (1 in 10 cells in meiotic cells!)
2
3
4
1 3 cosegregate
2 4 cosegregate
AND
CryR MATa
CryS MATa
CryR MATa
CryS MATa
(Note All DNA distal to crossover becomes same
on two homologs)
Or
1 4 cosegregate
2 3 cosegregate
Why recessive mutations (/-) can become
homozygous by mitotic recombination. Consider
Rb/Rb- or Brca1/Brca1-.(board)
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Back to storyline Recessive or Dominant? MATa
steA and MATa steB. To determine dominance
and recessiveness, do complementation test Get
both a mutant gene and wildtype gene in same cell
and test mating. MATa CryS steA ura1 his3 x
MATa CryR Ste ura3 his3
Select Ura diploids
MATa/MATa CryS/CryR steA/ ura1/ ura3/
Select CryR to homozygous MAT
MATa/MATa CryR/CryR steA/ ura1/ ura3/
his3/his3
Test mating to MATa his7 tester
Form His cells? Yes? Then SteA recessive NO?
then SteA is dominant.
(MATa/MATa/MATa CryR/CryR/Cry ste1//
ura1// ura3// his3/his3/ //his7)
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SteA and SteB One Gene or Two Genes?
What you can do is perform complementation test,
then infer if you have one gene or two genes.
Complementation test or Recombination test
Complementation test Lets say steA is
recessive, and steB is recessive. Lets call
steA ste1-1. Lets call steB ste1-1 or
ste2-1 Mate MATa steA ura1 his3 x MATa steB
ura3 his3
(Both ura1/ ura3/ his3/his3)
Case 1 2 mutations, same gene
Case 2 2 mutations, 2 genes
OR
MATa ste1-1
MATa ste2-1
MATa ste1-2
MATa ste1-1
Infer
Infer

• Mates

• Does not mate

Can we test if mutations are in different genes
if one is dominant? -by Complementation? -by
Recombination?
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What you can do is perform complementation test,
then infer if you have one gene or two genes.
Mate MATa steA ura1his3 x MATa steB ura3 his3
Case 1 2 mutations, same gene
Case 2 2mutations, 2 genes
OR
MATa ste1-1
MATa ste2-1
MATa ste1-2
MATa ste1-1
(Both ura1/ ura3/ his3/his3)
Meiosis. Isolate haploids. Test mating
ability (see earlier figure)
Mating (Ste phenotype) - (ste1-1) -
(ste1-1) - (ste1-2) - (ste1-2) 04
SteSte-
Mating - STE1 ste2-1 - ste1-1 STE2 -
ste1-1 ste2-1 STE1 STE2 13
SteSte-
Recombination test informative for dominant
mutations.
(Assuming STE1 and STE2 are not linked)
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Odd but possible issues easily resolved by
yeast genetic analysis. Does strain steA have
one mutation (as we have assumed in the previous
analyses), or does it have more than one mutation
involved in sterility? 3 possibilities, all
answered by simple segregation ratios.
1. If steA has only 1 mutation, then
Mate MATa ste1-1 to MATa STE1
Mating Nonmating 1 1
meiosis
MATa ste1-1
MATa STE1
2. If steA has 2 mutations, each of which make
cell mating-defective (steA is actually a
ste3-1 ste4-1 double mutant, for example) !
Mating Nonmating 1 3
MATa ste4-1
MATa ste3-1
3. If steA has 2 mutations, each of which is
needed to make cell mating-defective (Ste-) So,
steA actually has two mutations in genes that
either of which is sufficient for mating!
Gene A
Gene B
Genotype Mating AB a- B
Ab- a- b- -
OR
31-
mating
So-called redundant genes
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Whew..Mutants collected, of genes identified
now determine pathway Genotype Mating
Response - pheromone
pheromone Wildtype -
ste2- - - ste4- - - ste5- - - Ste11
- - - Ste7- - - Fus3-, kss1- - - Probl
em Order of function not possible, all same
sign. Back to drawing board!
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New screen- for mutants that activate pathway
(constitutively)
TRICK Fus1 transcriptionally induced in mating
response Make Fus1lacZ fusion (or Fus3HIS3
fusion- reporter gene). Screen for mutants that
always make at least some Fus1
Genotype Fus1LacZ reporter -
pheromone pheromone Wildtype -
ste2- - - ste4- - - STE4-1ON
ste5- - - ste11- - - STE11-7ON
ste7- - - fus3-, kss1- - - ste12-
- - gpa1-
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Now use these for informative
epistasis tests
Genotype Mating Response - pheromone
pheromone Wildtype -
ste2- - - gpa1- ste4- - - ste2-
gpa1- ste4- gpa1- - - STE4-1ON
STE4-1ON ste5- - - STE4-1ON
ste2 ste5- - - ste11- - - STE11-7ON
STE11-7ON ste5- STE11-7ON
ste7 - - fus3-, kss1- - - ste12 - -

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The Mating pathway
Biochemistry
Genetics
MATa cell
MATa cell
Pheromone
Gpa1
STE2
GTP binding proteins
GPA1
P
Ste11
Protein kinase cascade
Ste5
P
STE4
Ste7
Fus3
P
STE5
P
Ste12
STE11
Active transcription factor (like CAP)
STE7
FUS3, KSS1
Cell cycle arrest
STE2- receptor- positive feedback FAR1- cell
cycle arrest FUS3- cell fusion SST2-negative
feedback
STE12
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Fig A15 Mating Pathway (books view)
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Biochemistry and genetics case of
phosphorylation.
Signal
Gpa1
Ste4 Ste18..
P
Ste11
Ste5
P
Ste7
Fus3
P

• Experiment
• Treat cells with pheromone
• Make cell protein extracts
• Separate proteins on acylamide gel
• visualize with antibody- dont worry about
  details- MCB410

Virtual experiment
No af Ste
Plus af Ste
Plus af ste2-
Plus af ste12-
No af ste11-
No af STE11-1ON
Big proteins
Ste7-P Ste7 (no P)
Conclusion Ste7 downstream of Ste11
Small proteins
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The story of Fus3 and Kss1redundant proteins ??
Biochemistry
Genetics
MATa cell
MATa cell
Pheromone
Gpa1
STE2
GTP binding proteins
GPA1
P
Ste11
Protein kinase cascade
Ste5
P
STE4
Ste7
Fus3
P
Kss1
STE5
P
Ste12
STE11
Active transcription factor (like CAP)
STE7
FUS3, KSS1
Cell cycle arrest
STE2- receptor- positive feedback FAR1- cell
cycle arrest FUS3- cell fusion SST2-negative
feedback
STE12