Download 1 Lecture 2 Toczyski Learning points 1) Understand what happens

Lecture 2
Toczyski
Learning points
1) Understand what happens when you put a piece of DNA into yeast: That its ends
find homology and recombine.
a) Understand how this could be used to introduce a tag.
b) Understand how this could be used to introduce a promoter.
c) Understand how one can introduce a tag or a single mutation without making any
other changes to the gene (not leaving behind a marker)
2) With this lecture you should start to understand how you would set up a screen
and analyze the progeny.
3) Why might an allele be dominant, and how would one work with it.
4) What are temperature sensitive alleles and why are they employed.
References:
The first of several papers outlining the original CDC screen.
Hartwell, LH, Culotti, J, and Reid, B. Genetic Control of the Cell-division cycle in Yeast,
I. Detection of mutants PNAS 1970 66(2):352-359.
Ordering cdc genes
Hartwell LH. Sequential function of gene products relative to DNA synthesis in the yeast
cell cycle. J Mol Biol. 1976 Jul 15;104(4):803-17
Design of the degron.
Dohmen RJ, Wu P, Varshavsky A. Heat-inducible degron: a method for constructing
temperature-sensitive mutants. Science. 1994 Mar 4;263(5151):1273-6.
A nice example of a genomic screen that couldn’t have been done nonsystematically
Jorgensen P, Nishikawa JL, Breitkreutz BJ, Tyers M. Systematic identification of pathways
that couple cell growth and division in yeast. Science. 2002 Jul 19;297(5580):395-400.
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V) Plasmids.
A) need an origin of replication (called an ars in yeast, for autonomously
replicating sequence) and a marker.
-If the plasmid has the URA3 gene on it you can transform it into a ura3
strain and plate the cells on a plate lacking uracil.
B) If you want a plasmid to segregate well, you must put a centromere on it.
Centromeres and origins (ars) are very simple sequences in yeast (a couple hundred base
pairs). This is very different from higher organisms, which use large complex sequences
that lack strict consensus sequences and are maintained epigenetically. Centromere= the
sequence that is complexed with proteins to form the kinetochore, which attaches to
microtubules. Without a CEN, the plasmid will segregate randomly, and you will often get
cells with many copies of the plasmid
C) If you want to overexpress a gene you can also use a 2µ plasmid. This is a
plasmid which is naturally occurring in many yeast strains and has a complicated method for
maintaining high copy numbers. One can also use strong inducible promoters. The
galactose inducible promoter is one common example.
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VI) Integrations.
Yeast LOVE to do homologous recombination. If you transform yeast with a linear
piece of DNA and the ends of that piece match the sequence of a yeast gene they will
efficiently recombine.
A) Knock outs. If you want to knock out a gene you make a construct (or
generate a PCR product) as follows:
The same rational can be used to make gene fusions.
Common fusions:
-TAP, the tandem affinity tag. This can be used to purify complexes very easily. This has
started to replace previous tags with the same role (e.g. GST, MBP)
-GFP, or other fluorescent proteins. These allow easily visualization in vivo
-the myc or HA epitope. These epitope tags are small peptides (10-15 aa) that have
commercially available antibodies. These can be fused to the ends of proteins. Advantage over
TAP is that they are small. Also, many copies (10) of the epitope can be used for non-abundant
proteins.
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2) Imagine that you would like to place a GFP tag at the N-terminus of a gene.
Could you use the same technology??
Since this technology requires that you insert a marker (e.g. URA3), you can’t make subtle
changes.
You could generate a plasmid that allows you to put a tag at the 5’ end of gene
and express the gene off of a heterologous promoter, such as the Gal promoter.
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To get around this, one can the following trick:
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VI) Doing a screen in yeast.
A) Example of a simple screen: a screen for Ade- strains, that is, strains which cannot
make adenine and require that you add it to the agar plate. We want to find all of the genes
that are required to make adenine. There may be many genes in this pathway.
1) Mutagenesis-getting our mutants. Since yeast is haploid, this is easy!
2) we will spread our mutagenized cells on a plate containing adenine. Why with
adenine? Because we want mutants that require the addition of adenine to grow!
3) we will replica this plate to a control plate that has adenine and an experimental
plate that does not
4) we will identify colonies that don’t grow on the –ade plate and pick them from the
+ade plate
Mutagenize (e.g. EMS)
Wow, we isolated 117 mutants. This would have taken about a month. Now we want to
analyze our mutants. What do we want to know?
1) Are the mutants recessive?
2) How many genes (complementation groups) did we get. Note that some of the
mutants may be affected in the same gene.
3) Is only one gene affected for any given mutant? In some cases the phenotype
actually requires mutations in two separate genes. This is more common with more
complex phenotypes (like temperature sensitivity), and less common with phenotypes like
requiring adenine.
4) What gene is affected in each mutant. That is, we want to clone our genes.
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Let’s look at these one at a time:
1) Are the mutants recessive.
a) Mate each of our mutants to a wildtype strain. Select the resulting diploid. If this
diploid is wildtype, then the mutant is recessive.
b) What does this mean? Assuming that the gene is not dosage sensitive (i.e. that the
cell doesn’t care if it has one or two copies of the protein produced), then this suggests that
the gene has lost a function. The vast majority of yeast genes are NOT dosage sensitive, at
least using the rather non-quantitative assays for phenotype commonly performed.
c) If it is dominant this can be for several reasons:
i) Dominant negative. This is an allele that somehow interferes with the
wildtype allele. An example. Imagine that there is a step in the adenine biosynthesis
pathway that uses a large complex made of 10 molecules of the same protein. Lets say there
is an allele of a gene, lets call it YFG1-5, that can enter into this complex, but it doesn’t
work. This molecule will poison almost all of these complexes in the cell.
- Does YFG1-5 / YFG1 = yfg1! / yfg1!? (and does it have the same
phenotype as the yfg1! or the YFG1-5 haploids)?
-How could you tell a dominant negative from a haplo-insufficiency?
ii) Dominant gain of function. This is an allele that is dominant because it is
over-active. Imagine that the adenine pathway is turned off when there is adenine in the
media because a transcriptional inhibitor is made. We will call this transcriptional inhibitor
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Yfg2 and say that it is normally only turned on when cells have adenine. Now imagine that
we have an allele (YFG2-5) that is always on, no matter how much adenine is in the media.
This is going to be dominant.
-Does YFG2-5 / YFG2 = yfg2! / yfg2!?
(will yfg2! = yfg2-5)?
iii) Neomorphic alleles are those that gain novel functions not had by the
starting allele. These rarely occur in screens, but effectively are evolution.
2) How many genes are represented in our 117 mutants. Many of these are mutants affected
in the same gene, we want to know how many genes we have identified.
116 of your mutants are recessive, and one is dominant.
a) Let’s mate our mutants. Imagine that we start with three mutants. Since we don’t
know how many genes are involved, we will just call them 1, 2 and 3.
We were smart when we set up our screen and we did half the screen in a MATa strain and
half in a MAT" strain. Mutant 1 is MATa, mutants 2 and 3 are MAT" strains.
We mate mutant 1 and 2 and select the diploid.
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It cannot grow on adenine.
We mate mutant 1 and 3 and select the diploid.
It can grow on adenine.
-which mutants do NOT “complement” ? Are they affected in the same gene?
-which mutants do “complement”? Are they affected in the same gene?
-would 2 and 3 complement is we could mate them?
b) You have 11 complementation groups for the recessive mutants.
So what do we do about mutant #117, which is dominant. This could be a dominant allele
of a gene that is represented elsewhere in your complementation groups by recessive alleles.
How do you figure this out? If you mate it to the ade1 mutant, the resultant diploid will be
Ade-, but that doesn’t tell you that #116 is an allele of ADE1
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You isolated ADE1 30 times, ADE2 20 times, ADE3 15 times, ADE4 15 times, ADE5 10 times,
ADE6 10 times, ADE7 5 times, ADE8 5 times, ADE9 5 times, ADE10 1 time, ADE11 1 time.
Is your screen saturating???
3) However, it was a little premature naming your genes, because you haven’t yet answered
question number 3: Is each mutant affected in only one gene? What is the alternative?:
Imagine that there are two genes in a linear pathway and each allele is very weak. You may
see no phenotype, or a very weak phenotype, with the separate alleles, but the phenotype
may be more obvious when they are in combination. Alternatively, there may be two
entirely redundant genes and you may have null alleles (entirely dead) of both. You need to
be especially suspicious of the mutants in their own complementation group. Why is that??
So what do you do? Cross your strain to wildtype, sporulate, and examine the segregants
from the cross (the strains from each spore). If you really need both genes to get the
phenotype, what will you see?
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4) Cloning your gene.
I won’t teach mapping, since no-one does this anymore. In mapping, you simply examine a
large number of available “markers” and determine if your gene is linked to any of these.
Again, we will assume that we are trying to clone a recessive mutant. Let’s assume that we
started our experiments with a strain that was MATa ura3 trp1
a) We will transform this strain with a library. This library will be random
insertions of genomic DNA from a wildtype strain that have been cloned into a vector.
You could actually either select or screen for the cells that are rescued because they contain
the plasmid containing a wildtype version of the mutated gene. We are going to do this as a
screen only because this is the way that it normally has to be done. (imagine that we were
cloning based on the color phenotype). If your original screen was done in a ura3 strain,
then the library should be made in vector with URA3 in it. You would transform one of
your mutants with this library and spread on plates lacking uracil. You would then replica
plate this to plates lacking adenine and look for those that can grow.
i) The vector needs to have a replication origin. You also want the plasmid
to be a CEN based plasmid. Why is this? What is the danger of using a high copy plasmidWe’ll see in a later lecture.
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B. How to study essential genes: conditional alleles.
1) Traditionally use temperature sensitive mutants.
a) mutagenize and grow at 23°C ! test at 36°C
b) if you know that you want a temperature sensitive allele, you may want to start
with a general screen for temperature sensitives.
2) Example. What are the genes required for budding/entering the cell cycle? Cells
bud at G1.
Mutants that can’t bud are dead, therefore these will be essential genes.
a) The CDC Screen:
i) Mutagenize cells and plate at 23°C --> replica to 36°C
ii) Pick the ts mutants from the 23°C plate and put in liquid (save a little)
iii) Bring to 36°C wait about three hours (one cell cycle takes about 2 hours)
iv) examine on a microscope slide.
This is very similar to the CDC (cell division cycle) screen that was done by Hartwell in
the 70’s, except his screen looked for arrests at all stages of the cycle and
categorized them.
Subsequently, Paul Nurse did a similar screen in pombe
This wasn’t as redundant as you might guess. Many novel genes and several key
experiments were performed in pombe that capitalized on the different biologies
of the two systems.
!
37°C
23°C
!
37°C
asynchronous !
G2/M
asynchronous
!
G2
23°C
S. cerevisiae cdc mutant
S. pombe cdc mutant
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Important point: For complex phenotypes, it is often useful to start
with a very easy, but not very specific screen, and follow up the
positives with a more labor-intensive, but more specific screen.
3) Start --->
a) After this point in G1 cells can no longer mate.
b) Onces diploids enter a normal S phase, they can no longer go through meiosis
c) Cells will grow in size in G1 until start, after which they are large enough enter S
phase and complete a cell cycle.
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d) CDC28 shown to control entry at start
i) Start is a point in G1 from which cells can mate. If they pass this point they can no
longer mate and are committed to a cell cycle. Cells will grow in size in G1 until they are large
enough to continue with no growth.
ii) CDC28 encodes a kinase that is the central regulator of the cell cycle --> well conserved,
central cell cycle player in all eukaryotes- often called cdc2 or cdk#
iii) CDC28 acts with a second protein called a cyclin
Why aren’t the cyclin genes CDC mutants?
They are redundant. Some cyclin mutants do have CDC phenotypes in another yeast, pombe
How were they found in yeast? High copy suppression screen with a temperature
sensitive allele of CDC28, called cdc28-4 (We’ll get to this later)
Cdc28 associates with different cyclins at different parts of the cell cycle.
! Cln1, Cln2, and Cln3 function in G1-S transition
! Clb1, Clb2, Clb3, Clb4, Clb5, & Clb6 function in S phase and mitosis
! One important way that it is regulated is by degradation of those cyclins
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