Lecture 8

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Lecture 8
• Assignment 2
• Experiment 5
• Yeast genetic screen
• Discussion of Article 5
Assignment 2
• Read the format of the assignment
• Page 1 will be the gel with PCR primers that
you were assigned and performed the PCRs
with. I have checked the data on the web.
Don’t think you can pull a fast one.
• Page 2 The composite will be made from the
following data set.
Assignment 2
Assignment 2 due Nov 10, 2008
Assignment 2
ASSIGNMENT SECTION (Pages 3-4).
This section will have 3 parts and the format will be the following:
Maximum length of 2 pages; 12 point font; 1.5 line spacing;1 inch margins.
Part 1: (2 marks)
Analyzed independently, which of the marker(s) provide important evidence and which do not?
Part 2: (4 marks)
As a detective and a forensic scientist, write a report summarizing the incident, how the analyses were
performed, the results of the analyses and the conclusions of the case. This section should be about
3/4-1 page long. Less than half a page will receive a zero. You must summarize all the information concisely.
Part 3: (4 marks)
In a half a page speculate based on the evidence what happened, and who you think the murderer is.
Experiment 5: Goals
• Isolation and characterization of
mutants.
• Introduction to sterile technique.
Flow chart for yeast experiment
• Week 1: Restreak mutants. These 8 strains
will be your material for the rest of term.
Label well
• Week 2: Mating for complementation
analysis with swi5.
• Week 3: Selection of diploids and growing
strains for beta-galactosidase assay.
• Week 4: Analysis of results
EXPERIMENT 5 (Labs 8, 9, 10,11)
Lab 8: Re-streaking 8 yeast switching mutants:
Pick 8 RED
colonies
EXPERIMENT 5 (Labs 8, 9, 10,11)
Assignment is based on YOUR results. If you make mistakes during lab
it will affect your assignment marks.
1
2
5
4
6
3
8
7
Label the bottom of the plates: don’t write too much on plates,
record extra
information in your lab book.
YOU MUST KEEP TRACK OF ALL 8 MUTANTS FOR 4 WEEKS
Cut
Top
and/or
Bottom
1 2 3 4
5 6 7 8
Filter
paper
POINTERS FOR EXPERIMENT 5 (Labs 8, 9, 10,11)
Sterilize inoculation loop on the flame after every streak.
Please cool the inoculation loop after sterilizing on the flame otherwise you will
literally “fry” your yeast cells.
Keep plates as sterile as possible.
Yeast experiment
• This is perhaps the most difficult
experiment to understand.
• Start off with some basic yeast genetics
• Discuss the yeast life cycle
• Discuss mutagenesis
• Genotype of 4570: mat del::LEU2, leu2,
trp-1, his3-11, ura3::C2791::URA3, HOADE2, HO-CAN1.
C2791=YIplac211+HO/GAL-lacZ
Genetic nomenclature
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-D1
A specific complete or partial deletion of ARG2
ARG2::LEU2
Insertion of the functional LEU2 gene at the ARG2 locus, and ARG2 remains functional and
dominant
arg2::LEU2
Insertion of the functional LEU2 gene at the ARG2 locus, and arg2 is or became
Nonfunctional
can1
Canavanine resistance
CAN1
Canavanine sensitivity
MATa
Wild-type allele of the a mating type locus
MATa
HO
Wild-type allele of the a mating type locus
Homothallic gene encoding the endonuclease involved in mating type switching
Auxotrophs and Autotrophs
An auxotroph contains mutations that alter the nutritional requirements of an
organism. A wild-type organism that contains 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
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.
• Genotype of 4570: mat del::LEU2, leu2,
trp-1, his3-11, ura3::C2791::URA3, HOADE2, HO-CAN1.
C2791=YIplac211+HO/GAL-lacZ
• What do we need to supply in the media
for the strain to grow?
Life cycle genetics
• This was the great advance in
eukaryotic genetics in the 1970’s.
• Why do you think life cycle genetics is
hard to do?
Heterothallic Yeast Life Cycle
Heterothallic life cycle
• Stable haploid mating types a and
alpha.
• Haploids can mate but not sporulate.
• Diploids can sporulate but not mate.
• Sporulation is meiosis and the
encapsulation of the 4 haploid products.
Mating type locus
Homothallic life cycle
Homothallism
• Unstable mating types
• The gene that is required for
homothallism is the wild-type allele of
the HO gene; when yeast are ho (lossof-function allele) they are heterothallic.
Mating type interconversion
Mutagenesis
• Kim mutagenized the yeast strain 4570
with different amount of UV light.
• Genotype of 4570: mat del::LEU2, leu2,
trp-1, his3-11, ura3::C2791::URA3, HOADE2, HO-CAN1.
C2791=YIplac211+HO/GAL-lacZ
Canavanine resistance
• Is canavanine resistance dominant or
recessive to canavanine sensitivity?
Adenosine red phenotype
• In some mutants that can not
synthesize adenosine a red colour
forms. This is due to the accumulation
of a red adenosine pathway
intermediate. The cell grows because
there is adenosine in the media, but is
red because it tries to synthesize its
own adenosine but because it is still
blocked at the same step, the red colour
intermediate accumulates.
• Genotype of 4570: mat del::LEU2, leu2,
trp-1, ade2-1, his3-11,
ura3::C2791::URA3, HO-ADE2, HOCAN1. C2791=YIplac211+HO/GALlacZ
HO regulation
• Haploid specific gene
• You want the diploid to have a stable
MATa/MATa genotype. You do not
want aa and aa diploids that would
mate to create tetraploids.
Mother specific switching
HO switching as a stem cell
lineage
Note that at every division, the daughter cell
can not switch mating type. That is at every
cell division a cell of the same potential is
generated. This is similar to a stem cell
lineage where at each division a stem cell is
generated and a second cell that will enter a
particular developmental pathway.
What does this lineage tell us?
• Only mother cells switch mating type.
• The switch must occur before DNA
replication in the G1 phase of the cell
cycle. Why can I say this?
• It is the expression of HO that regulates
mating type interconversion.
HO expression summary
• HO is a haploid specific gene; it is not
expressed in a/a diploids.
• HO is expressed in mother cells and not
daughter cells.
• HO is expressed at G1 of the cell cycle.
HO cis-regulation
• URS1 required for Mother daughter
control
• URS2 required for cell cycle control
HO trans-acting factors
• Swi5p required for mother daughter
control but is not the asymmetric
determinant.
• Swi4p and 6p are required for cell cycle
regulation.
Cis and Trans-Acting Factors involved in HO Expression and Switching
The screen
• Positive negative: selection/screen
• Rule out mutations in CAN1 and ADE2
• Identifying trans-acting factors for HO
regulation specifically
Positive negative selection
Article 5 discussion
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