S. Cerevisiae cross Flashcards

1
Q

Why use S. Cerevisiae?

A

1) Easy to grow and cross, produces many progeny
2) Lots of mutant strains available, possible to demonstrate several genetic phenomena in 1 experiment
3) Has lots of life cycle as haploid organism so no need to worry about dominance relationships

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2
Q

What is S. Cerevisiae?

A

A unicellular eukaryotic budding yeast

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3
Q

How does it get energy?

A

Using fermentation to degrade glucose and other organic nutrients

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4
Q

Draw life cycle and describe

A

(see notes for diagram)
Cells of opposite type are a and alpha and they fuse to make diploid cells that continue to proliferate by mitotic division
Diploid cells undergoing carbon/ nitrogen starvation undergo meiosis and sporulation to give rise to 4 haploid progeny enclosed in an ascus
Once spores are released they can germinate and proliferate again

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5
Q

How can S. Cerevisiae be genetically modified?

A

Targeted mutagenesis
Homologous recombination has a high efficiency so targeted insertion of exogeneous DNA fragments into the genome can be used to delete/tag ORFs (ie. RNA ORF never gets made)
(NB: ORF is also in DNA, it is the bit that gets made into the RNA ORF)

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6
Q

What are steps of making a master plate?

A

1) spread spores from the cross on a petri dish with rich medium
2) Incubate the petri dishes at 23 degrees for a week
Species will germinate and give rise to different colonies
3) Use template (see notes for diagram) to create a master plate of 48 regularly arranged colonies that can be plated onto different media to detect mutant phenotypes

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7
Q

Why were all resulting colonies white after making the master plate?
(bear in mind ade2 mutants red in limiting adenine)

A

The plate contained a large excess of adenine preventing red pigment accumulation even in ade2 mutants

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8
Q

Why are large numbers of progeny required?

A

To detect rare recombinants

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9
Q

What are the p1 and p2 bits on the master plate?

A

Controls allowing you to check phenotypes (not included in final count)

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10
Q

How do you create a random sample?

A

1) Pick randomly ie. don’t target any big/small colonies
2) Don’t pick colonies that are touching other colonies as there is a risk of mixing genotypes
3) use the flat end of a sterile toothpick and transfer it as a small streak of less than 0.5 cm onto the master plate
4) Inoculate the p1 and p2 spots with the parental strains (checks that you inoculated the samples correctly)

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11
Q

What did our experiment show about how Ras1 and Ras2 delta interact?

A

Ras 1 delta and ras2 delta are colethal and disrupt the PKA pathway that is essential in S. Cerevisiae

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12
Q

Do ras2 delta mutants grow on the URA plate?

A

No

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13
Q

Draw out Ras1 delta/+ +/Ras2 delta cross

A

(see notes for diagram)

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14
Q

What is the equation for recombination frequency?

A

Recombination frequency = (Recombination/total progeny)

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15
Q

What does it mean if a) RF=50%? b) RF is under 50% ?

A

a) genes segregate independently

b) genes are linked

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16
Q

Smaller RF means….

A

Genes are closer on chromosome

17
Q

What is epistasis?

A

A mutation “masking” the phenotype resulting from a mutation in an independent locus

18
Q

Which of the mutations we analysed was epistatic to all other mutations?

A

CDC-28-13 (temp sensitive one)

19
Q

How did plate A act as a control when we were scoring progeny?

A

Ignore strains which do not grown on plate A

All progeny should grow on this (just some are red)

20
Q

How did we get lots of data?

A

By pooling data across groups for number of colonies corresponding to each phenotype

21
Q

If a mutant trait is controlled by a single gene and you cross WT with mutant what is expected progeny ratio?

A

1 WT: 1 mutant

22
Q

If mutant trait is controlled by 2 genes and you cross 1 WT to mutant what is expected progeny ratio?

A

3 WT : 1 mutant

23
Q

How can you test of ratios differ between crosses?

A

Do a chi squared –> indicates if difference between observed and expected values is due to chance

24
Q

What does a chi squared test not show?

A

1) If the cross was carried out correctly
2) If the results are accurate
3) If we have chosen the correct genetic explanation for the results