Lecture 22

Question 1

Creating a complete loss of function in yeast - double cross over:

Answer 1

• Insert a selectable marker (eg. LEU2+) in the middle of the gene, destroying the function of the gene
• Transform into a recipient
• A double recombination event will cause the construct of interest to be inserted and this can be selected for by selecting for the selectable marker
• The LEU2+, Gene X- is generated

Question 2

Creating a complete loss of function in yeast - single cross over:

Answer 2

• Insert a small portion - internal regions only of gene into YIp vector, by delete the 5’ and 3’ end.
• A single recombination event will integrate (by homology) the entire construct
• The genomic copy of our gene will lack a functional copy, one copy will lack the 5’ end the other will lack the 3’ end

Question 3

What will happen if your gene is essential?

Answer 3

• The inactivation will be lethal
• It can be hard to know WHY you are not getting transformation (is this lethal, for example?)
• Yeast is haploid but can be diploid
• a type cells and alpha type cells can mate to form an a/alpha haploid, which replicate.
• Under stress (nitrogen limitations) meiosis of diploid yeast is induced, and a, a, alpha and alpha will be produced, inside an ascus.

Question 4

Gene inactivation in a diploid recipient:

Answer 4

• Create a ura3- which is viable, and transform the inactivation construct into a diploid containing a WT copy Y+ and an inactive copy Y-.
• Conditions can be shifted to turn it back into a haploid state.
• Two haploids will be derived from the WT version of the gene and two will be inactive, so we can expect 2 Ura+ and 2 Ura-.
• If you see this ratio you know that this gene is not essential as everything is retrieved
• If it is an essential gene you will only get WT (Ura+) haploid yeast

Question 5

Two step gene replacement strategy uses:

Answer 5

• Introduce small changes in the DNA, such as a couple of bases changed

Question 6

Two step gene replacement strategy using a YIp: Step 1:

Answer 6

• Into the YIp, introduce a YIp with a selectable marker and an altered version of your gene
• Select for the selectable marker and introduce the whole plasmid in yeast
• Finish with the (new) altered version of your gene present in the genome and the old gene

Question 7

Two step gene replacement strategy: Step 2:

Answer 7

• Expose the effect of this mutation
• Looking for intra-chromosomal duplication event between the sequences ‘old’ and ‘new’
• A recombination event could occur and create copies with ONLY the old copy and the other ONLY the new copy
• This gives us a yeast genome with only the new copy of the gene (no selectable marker or WT copy), determining through RE analysis

Question 8

Uses of two step gene replacement in HIS4 mutants:

Answer 8

• HIS4 gene is regulated by GCN4, but there are no mutants with altered control regions (cis) isolated
• Create a set of deletions for HIS4, taking out different parts of the promoter
• These were individually introduce back into yeast using 2 step gene replacement

Question 9

List of steps:

Answer 9

1. Take a YIp with selectable marker and the construct containing a deletion in the 5’ region of HIS4
2. Introduce into yeast
3. Select for URA3+
4. Finish up with a yeast transformatnt containing the old and new version and the selectable marker

Question 10

Second half of steps:

Answer 10

5. Remove the selectable marker and the old copy

6. Regions of recombination with the 2 regions of homology produces the WT and the mutated copy of HIS4

Question 11

HIS4 two step gene replacement results:

Answer 11

4 distinct classes:

• 1. No effect
• 2. Expression of HIS4- dropped, but was still regulated
• 3. Regulation of HIS4 was lost, even though the gene was still expressed
• 4. Loss of expression altogether

Question 12

Loss of regulation of HIS4:

Answer 12

• Regulation by GCN4 must occur in the -138 region of the DNA. This is the promoter
• It was found that there were two other copies of the same sequence, but it was only when you mutate the last site
• There are three copies, so if only one is mutated there is no effect, but if you delete all other copies it is detectable

Question 13

Further steps:

Answer 13

• Took the -136 strain which has lost regulation and selected for revertants
• The -136 cuts through the sequence and introduces new sequences
• The revertants convert the base pair change back to the original nucleotide

Question 14

Different types of yeast plasmids and their uses:

Answer 14

• Clone by complementation (YRp/YEp)
• Regulation of expression (YRp/YEp or YCp)
• Over expression (YRp/Yep or YIp)
• Gene inactivation/replacement (YIp)

Question 15

Aspergillus nidulans:

Answer 15

• Can be haploid or diploid
• Have asexual and sexual cycles
• Has multinucleate hyphae
• Easy to quickly generate protoplasts

Question 16

Generating protoplasts/recombinants:

Answer 16

• Cell wall degrading enzyme in the presence of CA2+ (facilitate uptake) and PEG (make cell membrane penetrable)
• Good for drug resistance selectable markers:

Question 17

Difference between asp and yeast:

Answer 17

• Integration in yeast is exclusively by homology

- In Asp there are more non-homolgous events than homologous

Question 18

Advantages of usually being non-homologous

Answer 18

• Can introduce almost any gene (there is no homology)
• Can leave the native gene intact
• Can generate multiple copies of the gene

Question 19

Disadvantages of usually being non-homologous:

Answer 19

• Possible positions effects on expression

- Homologous events are more difficult to isolate

Question 20

Strategies developed to identify homologous integrations to be gene specific include:

Answer 20

• Targeting to the argB locus
• Transform an argB- mutant recipient with an argB- plasmid (different mutations to produce argB-) select for argB+, which will occur when the construct enters the argB- locus

Question 21

Aspergillus conidiation:

Answer 21

• Ordered developmental program leading to the production of conidia
• Conidia are asexual spores
• This requires the brlA gene. The bristleA mutant cannot make the transition to producing conidia
• brlA is expressed at the time of conidiation in a WT asp.

Question 22

Does brlA cause conidiation, or does conidiation cause the production of brlA?

Answer 22

• Expressing brlA in the mycelia (not during conidiation)..
• brlA coding region under the control of alcA promoter, intorduced into brlA- protplasts
• Transformants that are argB+ with an alcA::brlA fusion showed that in the presence of ethanol (in response to the expression of brlA) results in conidiation

Question 23

Aspergillus transformation:

Answer 23

• High frequency transformation can be cloned by complementation
• DNA must be integrated

Question 24

nkuA gene in asp is the equivalent to Ku70 in most eukaryotes:

Answer 24

• Non-homologous end joining DNA repair occurs with the use of Ku70/Ku80
• In asp, nku mutants of asp will behave like yeast, in that plasmids are targeted to a region of homology

Question 25

Differences between Asp and yeast:

Answer 25

• Simple, high efficiency transformation system
• Cloning by complementation
• Gene inactivation/replacement
• Difference in DNA repair affects the way these two fungi handle exogenous DNA