yeast and mutant screens Flashcards

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1
Q

useful properties of yeast for genetics?

A
  1. Can exist stably as haploids (n) or diploids (2n)
  2. Are single-celled eukaryotes
  3. Forms compact colonies on plates
  4. Grown in large quantities in liquid medium
  5. Rapid life cycle (90 minutes)
  6. Small, compact genome (12 Mb)
  7. Efficient homologous recombination (handy for making knockouts)
  8. Excellent genomic resources https://yeastgenome.org/

CON - cant be used to study multi-cellular processes involving cell-cell signalling

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2
Q

explain the yeast life cycle

A

Two different haploid mating types, a and α. Determined by MATa and MATα
Two haploids of the different mating types can fuse to form a diploid…

Can enter mitosis and produce more diploids

Can undergo meiosis, forming an ‘ascus’. forms four haploid spores, two will be ‘a’, two will be α

***Haploids can also just undergo mitosis (no fusion first) and continue producing more haploids (not four tho obvi as its not mitosis)

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3
Q

how can the haploid/diploid aspect of yeast be useful?

A

You can immediately see the impact of a recessive mutation in a haploid (only one mutant copy needed)

Can look at lethal mutations in diploid heterozygotes, when this would kill haploids immediately

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4
Q

what was the goal of the Saccharomyces Genome Deletion Project?

A

generate a complete set (as far as possible) of yeast deletion strains with the overall goal of assigning function to the ORFs through phenotypic analysis of the mutant

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5
Q

how were mutants created in the Saccharomyces Genome Deletion Project?

A

Used a PCR-based deletion strategy

had a normal yeast genome…
- Designed linear DNA to perform knockout:
Ends = 45 bp primers - homologous to target gene to ensure the knockout occurs in the right place
- Use of an upstream and downstream tag to later map where the knockout occurred/which gene was knocked out
- Kanamycin resistance cassette replaced the yeast ORF to prevent replication/knock out the gene in question, and allow for selection of successful integration by homologous recombination (by applying the antibiotic kanamycin)

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6
Q

what mutant collections did the Sac. genome deletion project make?

A

Haploid (n) mating type a

Haploid (n) mating type α

Diploid (2n) homozygous for knockout (for non-essential genes)

Diploid (2n) heterozygous - necessary for knocking out essential genes (will still be able to live as it should have one functional copy)

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7
Q

give an overview of a forward genetic screen

A

Finding the gene responsible for the phenotype you are interested in

Amass mutants that can’t do what you are interested in

Cross with WT individuals to see if offspring show ratios of WT:mutant phenotype to see if it shows single-gene inheritance (has to be this way, you want to focus on one gene at a time)

Deduce gene function at molecular level

Deduce how the gene interacts with other genes to produce the trait you are interested in

Identify and test homologues in other species (yeast is a good starting point, but not entirely applicable to humans, tho high conservation between yeast and humans = likely to be a homologue)

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8
Q

in a forward genetic screen - what choice would you make for recessive mutations?

mutations in genes for essential processes?

A

recessive - immediately see the phenotype in a haploid

essential - use a conditional mutant (e.g. temperature sensitive) in order to keep your cells alive - turn up the heat - view the mutation is action

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9
Q

how are temperature sensitive mutants (or condition sensitive ones) found?

A
  1. Add mutagen to yeast to create point mutations
  2. Collect lots of colonies
  3. Use velveteen ‘stamp’ to copy colonies from one plate to another
  4. You can then grow identical colonies in two different conditions (for us it would be temperature) and play spot the difference, e.g. one of the colonies on your plate growing only at the low temperature and not at the more restrictive temperature
    These are mutations across the genome, so not all secretion obviously, just all lethal (if its growing vs not growing at all) and temperature sensitive
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10
Q

why do you need temperature sensitive mutants when looking for genes involved in secretion?

A

mutations in secretion genes would be lethal

so for haploids it would have to be temperature sensitive to allow them to survive at all (at the permissive temperature)

for diploids you could use heterozygotes but then how do you see the result of the mutation ALONE?

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11
Q

once a load of temperature sensitive lethal mutants had been identified, how did they find out which were mutants in the secretion pathway?

A

secretion assay - looked for mutants unable to uptake or secrete assayable molecules

Focused on -
- Acid phosphatase and invertase (converts sugars from one to another), two enzymes usually secreted by yeast

  • Chromate (lethal IF uptaken)
  • Sulphate permease - transport protein that allows chromate into the cell

A WT cell - the surrounding media should have acid phosphatase and invertase present (you can assay for their activity)
In a mutant - these enzymes would be stuck inside the cell

  1. Grow mutant in permissive (low) temperature and then shift to non-permissive temperature
  2. Assay the growth medium for enzyme (acid phosphatase) activity at various time points across the experiment

Select mutants that fail to secrete acid phosphatase at non-permissive temperatures - the graph shows the shift to the restrictive temperature and the SEC mutants no longer secreting acid phosphatase (less of its activity seen in the assay)

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12
Q

what was seen in the first sec mutants viewed?

How was this information used to perform a more efficient, second screen for secretion mutants?

A

Loads of vesicles - cus it cannot secrete them
Smaller - cannot grow as cannot deliver stuff to the membranes
Darker - dense as it is producing materials to grow but cannot

As the mutants appeared darker - he used a sucrose density gradient to isolate all the different mutants easily

Grows mutants at permissive temperature
Puts them at the restrictive temperature
Allows to grow
Separates on density gradient - to get all the ones with mutations affecting secretion

Allows him to screen more and more quickly

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13
Q

surprise karaoke! Sing your favourite song

A
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14
Q

what is the purpose of complementation tests?

what are they?

A

Purpose - to identify if these were mutations in different genes, or just a different mutations in the same gene

Cross two recessive mutations
Check whether the resultant F1 diploid individual(s) have wild-type or mutant phenotype…

If the F1 progeny display the mutant phenotype, then the two mutations must be recessive alleles of the same gene (in a diploid a recessive mutant only shows if both alleles are mutated)

If wild-type, then the two mutants are said to have complemented (been in different genes). This must be the case because their has to be a functional copy of each gene

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15
Q

what was identified in this screen for mutations of the secretion pathway?

A

sec1 - was for a SNAP….
The density gradient allowed him to find another 23 sec mutants

Group 1 - Ten genes that when mutated accumulate secretory vesicles (like sec1) look like an aero
2. Group 2 - Nine genes that when mutated accumulate ER-like structures (ER to golgi)
3. Group 3 - Two genes that when mutated caused strange golgi-like organelle to form (golgi to vesicle)

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16
Q

temperature sensitive mutants are typically?

A

thought to be caused by typically point mutations prone to mis-folding into an inactive form at restrictive temperatures

Permissible temperature - e.g. 25oC you’ve got a functional protein, 37oC the protein misfolds and you see the phenotype of the mutant
Remember - the mutation is present at both temperatures, it’s impact/phenotype is only visible at a certain temperature

16
Q

why was the secretion study relevant?

A

The proteins involved in secretion are conserved across eukaryotic organisms - so this study was very relevant in wider biology and medicine. Secretion is essential for delivery of membrane proteins, neurons, biotechnology (want cells producing your drug to pump it out) etc…

17
Q

chromosome instability (CIN) -

what is it?
caused by?
what does it result in?

A

Chromosome instability = hallmark of cancer
Its a change in chromosome structure/number causing chromosome gain/loss

Caused by - failures in mitotic chromosome transmission (a failure to divide chromosomes correctly in mitosis) OR defects in the mitotic spindle checkpoints

Unbalanced division of chromosomes in mitosis leads to aneuploidy offspring (as in cells) - leading to cell death OR cancer progression
CIN exacerbates tumorigenesis - accumulative large-scale genome rearrangements increases chances of disrupting expression/function of oncogene or tumour suppressor gene

18
Q

explain what is meant by variants of uncertain significance

A

Cancer cells accumulate LOTS of mutations as cell cycle checkpoints etc… are not functioning
So they have lots abnormal cDNAs - in sequence and expression level

Finding out the functional consequence of these variants is a problem - the variants are being identified faster than their function can be figured out. We need to know which variants/changes in genes are driving cancer progression

how do you identify which is the driver mutation amongst passenger mutations?

19
Q

name a few methods used to distinguish passenger and driver mutations (VUS classification)

A

biochemical functional assays

integrated algorithms

personal and family history of cancer of the carrier

severity of protein modification

loss of heterozygosity

20
Q

what’s the idea behind the CIN model in yeast (paper 1)?

A

29% of all ‘essential’ genes in yeast were known to give a CIN phenotype when mutated
They asked can the human copy of the gene - orthologue - do the job in a yeast cell for a yeast gene that has been knocked out - this is cross species complementation

21
Q

define orthologue and paralogue

A

orthologue = genes inherited from a common ancestor performing the same role in different organisms

Paralogue = genes which resulted from gene duplication events within a genome

22
Q

what did this paper - ‘Paper 1 - using yeast as a platform to study human genetic variants (Hamza, Akil et al)’ - do?

A

(with 322 genes)
Diploid yeast cell heterozygous for a mutation in an essential gene (one mutant one functional) → add a a WT functional copy of the human orthologue via a plasmid
NOTE - they chose the human orthologue, going by one with the closest sequence to the yeast version

Allow sporulation - division into four haploid cells

Possibilities -

The two cells with the mutant gene die
The other two survive thanks to the functional yeast gene

OR

All four spores survive because complementation occurred, and in the cells with the mutant yeast copy of the gene, the human homologue compensated

23
Q

Hamza, Akil et al did pool by pool screens as well - what were these and why?

A

Used ‘pools’ of yeast of heterozygous mutants for 622 essential yeast genes
Used a mixture of plasmids with a total of 1010 human cDNAs to transform the yeast

why - Unbiased strategy
To identify unknown orthologues, OR complementation between non-orthologous genes i.e. see if there were any surprises

Found -
65 human cDNAs that complement 58 yeast null mutants, 20 of which were newly discovered pairs
Looked for patterns - didn’t really know, said they were cytoplasmic proteins unlikely to have regulatory roles. But the researchers did not give up…

24
Q

Hamza, Akil et al did their one to one complementation screens, and their pool-pool screens and what else? (the important one in my opinion)

A

testing tumour specific missense mutations - can mutant human genes still complement yeast genes? if answer is no could be because it’s a driver mutation…

Made haploid strain with yeast mutant and human missense mutant to see if mutated human gene could still compensate

Tested for viability i.e. which ones survived - in 4/35 cases the mutant human gene could no longer complement the yeast gene - these 4 cases likely to be driver mutations in the cancer

Tested the ones that survived for altered growth rates (18 grew slower, 1 grew faster, 12 were unchanged)

Tested for resistance to DNA damaging agents - most showed altered sensitivity

Takeaway - complementation could be used to differentiate driver mutations from passenger mutations

25
Q

Hamza, Akil et. al, 2015 -
what question were they trying to answer?

takeaway?

Why are diploid heterozygotes used in the one-to-one screen examining the variants of unknown significance?

A

Which are the ‘driver’ mutations and which are the ‘passenger’ mutations

Because the mutations might* have been in essential genes – to maintain the line

Takeaway - complementation could be used to differentiate driver mutations from passenger mutations

26
Q

what are the two ways of measuring how old yeast is? how are they mesasured?

A

CLS - chronological life span = amount of time a yeast cell can survive in the stationary phase. Quantified by calculating viability - the fraction of the culture able to reenter the cell cycle after an extended state of quiescence/not growing

to measure it - remove cell each day/regular timepoints and see when it can no longer re enter cell cycle

RLS - replicative lifespan = number of daughter cells produced by one mother cell
Replicative viability quantified by mean number of daughters produced from mothers of a particular strain background before senescence (before it stops dividing)

to measure it - Observed cells under microscope, when it produced daughter cells they were teased away and counted. Its looking at the number of mitotic divisions

27
Q

is yeast as a model for aging reliable?

A

More evidence is being found that suggests that ageing is determined by ancestral evolutionary origins in part, suggesting some genetic basis for ageing → model organisms like yeast can therefore be used to study it

28
Q

McCormick, Mark A et al, 2015
“A Comprehensive Analysis of Replicative Lifespan in 4,698 Single-Gene Deletion Strains Uncovers Conserved Mechanisms of Aging”

what were the two main goals?

A

Goal number 1 → the study hopes to find genes in yeast that are already known to influence life span from other model organisms like worms (and therefore you don’t have to work in worms)

Goal number 2 → find some new ones as well

29
Q

what did McCormick, Mark A et al do in their screen?

A

Screened a single gene deletion strain library - this is a load of yeast with different null mutations in non-essential genes (haploids i think - remember these can undergo mitosis)

For each mutant they measured the replicative age (No. of daughter cells)

found -
238/4698 mutants showed longer lifespan than the WT
So these 238 genes normally would act to shorten the cell’s lifespan, most likely

30
Q

paper 2 - the 238/4698 mutants were classified by function.

what were the 5 classifications?

A
  1. Translation (cytosolic and mitochondrial)* (goal 1, seen in worms)
  2. TCA cycle* - A nearly universal metabolic pathway (goal 1, seen in worms)
  3. Proteasomal activity** (goal 2 - new)
  4. Mannosylation** - a bit like glycosylation (goal 2 - new)
  5. The SAGA complex*** - A SAGA-type histone acetyltransferase complex that recruits basal transcription machinery (goal 2 - new, worms don’t even have a homologue for this, but humans do have the STAGA complex)
31
Q

LOS 1 (loss of suppression 1) was one of the 238 mutants identified.

what was found when it was further investigated?

A

Deletion of this had the biggest effect on increasing yeast lifespan
Codes for a nuclear pore involved in nuclear export of pre t-RNAs and re-export of mature t-RNAs

Loss of the gene lengthened replicative lifespan, as did exclusion of the protein due to caloric restriction

***Mutants - accumulate t-RNAs in the nucleus

32
Q

paper 2 - LOS 1 - is it the tRNA accumulation in the nucleus causing an increased lifespan?

A

overexpression of MRT10 also results accumulation of tRNA in nucleus (by pumping tRNA from cytoplasm into the nucleus) AND the overexpression increased lifespan also, so answer = yes kind of maybe

33
Q

why might tRNA accumulation in the nucleus result in longevity? (hint - caloric restriction)

A

Dietary restriction extends lifespan of all organisms including yeast
Restricting the diet of los1 deletion strain of yeast did not extend lifespan any further tho so…

Dietary restriction’s effect of extending lifespan must be entirely by increasing accumulation of nuclear tRNA
Or the lifespan of the los1 deletion strain is the theoretical maximum for any yeast cell

No suggestions from authors on how to distinguish between these possibilities, but do believe los1 is a conversion point for multiple signalling pathways involved in ageing including the dietary restriction ‘master switch’ TOR, and DNA damage coordinator Gcn4

34
Q

summarise findings of paper 2 - McCormick, Mark A

A

4,698 gene deletions tested: the most comprehensive yeast data set on aging

Longevity genes found were often orthologues of known lifespan-extending C. elegans genes suggests conservation across species

Genome-wide information uncovered novel aging pathways such as tRNA transport