Lecture 9 - Yeast as Model Eukaryotes Flashcards

(30 cards)

1
Q

(self-assessment question) Outline the main steps in making targeted gene knockouts in mice?

A

Culture ES cells, make targeting vectors with homologous arms, transfect ES cells with correct modification, proliferate modified ES cells, add ES cells into blastocysts & implant into foster mother, select mosaic mice with targeted gene modification in the germline, back-cross to normal mice to maintain line

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2
Q

(self-assessment question) List 2 features that make mice a good model for human disease?

A
  • High levels of homology with human genome/genes
  • (relatively) rapid generation time
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3
Q

(self-assessment question) Identify a limitation of using mice as a model for human disease?

A

Not all processes are conserved between mice & humans (e.g. HIV infection)

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4
Q

(self-assessment question) Suggest a disease that a mice could be a good model for

A
  • Most diseases with a strong genetic basis have a mouse model
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5
Q

What are convenient features of S. cerevisiae?

A
  • can exist stably as haploids (n) or diploids (2n)
  • are single-celled eukaryotes
  • forms compact colonies on plates
  • grown in large quantities in liquid medium
  • rapid life cycle (90 minutes)
  • small, compact genome (12Mb)
  • efficient homologous recombination
  • excellent genomic resources
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6
Q

Who is Louis Pasteur?

A
  • grandad of microbiology
  • interested in wine & beer
  • “alcoholic” fermentation is a process correlated with the life & organization of yeast cells, not with the death or putrefaction of the cells”.
  • Denmark’s Carlsberg brewery established one of the world’s first yeast-biology labs in 1875.
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7
Q

Describe the life cycle of yeast

A
  • single-celled
  • sexual & asexual phases
  • cell divides by budding following mitosis
  • 2 mating types - determined by MATa & MATalpha
  • When 2 haploid cells of opposite mating types unite, then can undergo mitosis (remain in diploid) or meiosis (return to haploid)
  • can be stability maintained as a haploid or diploid
  • allows for exploitation
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8
Q

How does the yeast life cycle lend itself to genetic experiments?

A
  • A particular advantage of the yeasts as genetic models is their HAPLOID LIFE CYCLE.
  • recovery of recessive mutations, these are usually loss-of-functions alleles and are particularly useful in determining the normal function of the gene.
  • Even lethal mutations can be maintained if the wild-type copy is also present (using diploid heterozygotes).
  • The Yeast Knock-Out (YKO) collection was generated using both diploids & haploids
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9
Q

What is Yeast Knockout deletion collection (targeted gene knockouts)?

A
  • PCR-based deletion strategy, similar to gene targeting in mice
  • Primers were designed to add 45bp complementary sequence to a kanamycin resistance cassette.
  • Aimed to systematically knockout every open reading frame (ORF) in the genome
  • Chromosomal integration by homologous recombination

Make a linear piece of DNA that has homology upstream & downstream from where we are interested. Up-tag region (upstream tag) & down-tap (downstream tag), reference to transcription

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10
Q

What occurred during targeted gene knockouts (Saccharomyces genome deletion project)

A

Four different mutant collections were generated:
- Haploid (n) - mating type a
- Haploid (n) - mating type alpha
- Diploid (2n) - homozygous for knockout alleles (non-essential genes)
- Diploid (2n) - heterozygous for knockout alleles (for non-essential & essential genes)

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11
Q

What is Yeast’s main contributions to science?

A

Yeast is the model organism for eukaryotic cell biology:

  • Several Nobel Prizes
  • Cell cycles
  • Trafficking
  • Recombination
  • Gene interactions
  • Mitochondrial genetics
  • Genetics of mating type (and switching)
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12
Q

What are the general steps in analyzing a trait through forward genetic screening?

A
  1. Amass mutant affecting the trait of interest
  2. Cross (mate) mutant individuals to wild-type individuals to see if their descendants show ratios of wild-type-to-mutant phenotypes that are characteristic of single-gene inheritance. Do this to make sure that the interesting trait is only 1 gene.
  3. Deduce the functions of the gene at a molecular level
  4. Deduce how the gene interacts with other genes to produce the trait in question
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13
Q

What are the main steps in forward genetic screens in yeast?

A
  1. identify a process of interest
  2. predict the likely phenotype of a mutant unable to carry out that process
  3. devise a method of identifying such mutants
  4. Haploid life cycle facilitates the recovery of recessive mutations
  5. Use conditional mutants to identify genes in essential processes
  6. Identify & test homologues in other species.

Conditional mutants are where the mutant phenotypes are only visible in some conditions - e.g. temperature

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14
Q

How do you design a powerful screen?

A
  1. Identify a process of interest
  2. Predict the likely phenotype of a mutant unable to carry out that process
  3. To devise a method of identifying mutants with that phenotype
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15
Q

What is secretion?

A

Secretion is getting (usually always) proteins from inside a cell to outside a cell. Also involves intracellular trafficking. Nucleus pumps out mRNA, which is translated, before being transported through ER, Golgi etc.

When it gets to the cell membrane, it fuses then it is released.

For cell growth, it requires more material at the edge of the cell. Lack of secretion means lack of growth

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16
Q

Why is secretion important?

A
  • Essential for growth (delivery to the cell surface)
  • Essential for secretion of substances/protein/enzymes outside of cell
  • Essential for message delivery at nerve cells (and nerve-related illnesses)
  • Essential for delivery of membrane proteins to the cell membrane
  • Structurally & functionally conserved across eukaryotic organisms - essential for making these studies BIOLOGICALLY & MEDICALLY relevant.
17
Q

How are genetic screens used to solve a physiological problem?

A

The genes lead to the proteins, which leads to the biochemistry, which leads to physiology.

First, he developed an assay for secretion. He obtained a temperature-sensitive mutant collection to screen for secretion (sec) mutants. They had to be temperature-sensitive as, he predicted (correctly) that mutations that disrupt the secretion system would be lethal.

18
Q

How was a quantitative assay for secretion created?

A

Cells with chromate (lethal if uptaken) on the outside. Sulphate Permease allows the chromate to enter cell.

Then look for cell survival, which would indicate a sec (secretion) mutant, as chromate had not been uptaken - therefore problems with secretion,

19
Q

What occurred to the temperature-sensitive mutant collections for secretion?

A

Master plate with colonies and stamped the temperature-sensitive mutants. 2 different types of medium (or temperature). Stamp the same colonies twice - one on each condition.

Only interested in yeast that grows in permissive, but not in restrictive conditions.

20
Q

What was learnt about Temperature-sensitive (Ts) mutants?

A
  • mutations in essential genes cause death
  • in diploids, these are maintained in the heterozygous state
  • in haploids, heat-sensitive recessive lethal alleles are used
  • Ts alleles are thought to be caused by mutations that make the protein prone to mis-folding into an inactivate form at the restrictive temperature
  • Mutant stocks can be maintained in permissive conditions (e.g. 25 degrees Celsius)
  • The mutant phenotype can be observed by growing the mutants in restrictive (37 degrees Celsius) conditions
21
Q

What is the process for creating an assay for secretion & cell-surface growth?

A
  1. Grow mutant in permissive (low) temperature
  2. Shift to non-permissive temperature
  3. Assay the growth medium for enzyme (acid phosphatase) activity at various time points
  4. Select mutants that fail to secrete acid phosphatase at non-permissive temperature.
22
Q

At what temperature can wild-type cells secrete acid phosphatase?

A

At both temperatures:
- permissive (25 degrees Celsius)
- restrictive (37 degrees Celsius

23
Q

What is the process of isolation of the first sec mutants?

A
  • Grow yeast & assay the media for acid phosphatase activity
  • Sec1 & Sec2 cells stop secreting acid phosphatase when shifted to 37C (the restrictive temperature)
  • Wild type cells continue to secrete acid phosphatase after the first to 37C (the restrictive temperature)
24
Q

How does a sec mutant look under a microscope?

A

Smaller cell, darker, packed with vesicles. Suggested that it looked denser (mass per volume). This means that this mutant means that vesicle production continues, but they can’t fuse with the outer membrane & therefore can’t release their contents.

It was clear that there was a genetic basis of membrane trafficking. Concluded that sec1&2 are more dense than wild-type cells when they are grown at restrictive temperatures. This is probably because the vesicles are blocked at delivery to the plasma membrane, so cells expand as they grow - become denser.

25
Describe the characterisation of sec1
- sec1 was cloned and it encoded a SNAP receptor protein which is essential for membrane fusions - hence sec1 mutants are blocked at secretion & cell-surface assembly
26
How might you be able to find more sec mutants that are defective at different points in the pathway?
They can be enriched using a density gradient
27
Describe the second (improved) mutant screen
- used density-gradient separation to enrich for secretion mutants from a population of temperature-sensitive (Ts) mutants] - shift cells to a non-permissive temperature - allow to grow - separate on a density gradient - collect the dense ones - screen those dense ones for acid phosphatase secretion - can screen a much bigger population this way - identified a further 23 sec mutants - used COMPLEMENTATION & phenotypic analysis to organise his mutant collection into groups
28
How do complementation tests occur?
- inter-crossing 2 independent individuals that are recessive mutations - check whether F1 diploid individual(s) have wild-type or mutant phenotype - if the F1 progeny display the mutant phenotype, then the 2 mutations must be recessive alleles of the same gene. - if wild-type, then the 2 mutants are said to have complemented (been in different genes) This is easy in yeast because of the haploid/diploid phases
29
What groups were the 32 SEC genes separated into?
Group 1 - 10 genes that when mutated accumulate secretory vesicles (like sec1) Group 2 - 9 genes that when mutated accumulate ER-like structures Group 3 - 2 genes that when mutated caused strange golgi-like organelle to form
30
What does it signify if, when 2 independently-derived recessive mutant alleles producing similar recessive phenotypes fail to complement?
They must be alleles of the same gene.