L9 Introduction and maintenance of exogenous DNA into host cells

Question 1

Saccharomyces cerevisiae

Answer 1

single cells

yeast

Question 2

Aspergillus nidulans

Answer 2

multinucleate hyphae

fungi

Question 3

Drosophila melanogaster

Answer 3

P-elements

Question 4

Arabidopsis thaliana

Answer 4

Ti Plasmid

Question 5

Mus musculus

Answer 5

Embryonic stem cells

Question 6

Yeast transformation

Answer 6

1. uninucleate cells divide by budding
2. cell wall degrading enzymes
3. sphaeroplasts - osmotically sensitive
4. DNA + Ca2+ + PEG => transformations (creates small pores in cell membrane)

Question 7

YIp

Answer 7

See OneNote diagram
Bacterial plasmid backbone + yeast selectable marker
- YIp integrates into the genome by homology

Question 8

YEp/YRp

Answer 8

YIp + yeast origin of replication

• high transformation freq.
• not integrated
• high copy number (~30 per cell) but poor segregation during cell division as it does not have a centromere
• maintained and replicates independently, recombination not required
• shuttle vector

YEP - high copy number, natural yeast plasmid
YRp - yeast chromosome, low copy number BUT the SAME with regards to TRANSFORMATION

Question 9

YCp

Answer 9

YEp/YRp + centromere

• high transformation freq.
• good segregation
• single copy/cell

Question 10

YAC

Answer 10

yeast artificial chromosome

• can be used to clone very large DNA fragments
• stability increases and copy number decreases with size

Question 11

Yeast selectable markers

Answer 11

See OneNote diagram

- mainly WT amino acid auxotrophic markers

Question 12

Yeast transformants

Answer 12

• Transformants are due to HOMOLOGOUS recombination either by single or double cross overs

single => left with both WT, mutant and part of the plasmid, addition transformants

double => WT replaces mutant copy, less common as it requires double recombination, substitution transformants

Question 13

Plasmids with yeast origin of replication

Answer 13

maintained in the nucleus but are replicated independently of the yeast genome, integration is limiting for YIp vectors

Question 14

Using transformation to clone a gene by complementation

Answer 14

See OneNote

- figure out what the mutant is

Question 15

Genomic library

Answer 15

See OneNote

Collection of shuttle vectors that contain different sections of that genome => shuttle vector then inserted into the yeast mutant

Can isolate the shuttle vectors as they are replicating independently and put it back into E.coli in order to make more copies that we are able to sequence and find out the exact sequence that complemented our mutant - subclone to define minimal complementing region

Question 16

Do you need to know if your mutation is dominant or recessive?

Answer 16

Yes

Question 17

Using transformation to investigate gene expression with a reporter

Answer 17

See OneNote

• vary deletion region and assay for expression
• Region of interest can be transferred into a vector and analysed

Question 18

Using transformation for overexpression/gene dosage experiments

Answer 18

• use different plasmids to vary copy number of your gene
• Different vectors have different copy numbers so would produce different amount of product

OR
- use highly expressed promoter to drive expression of your gene

Question 19

Using transformation to characterise gene function in yeast

Answer 19

gene inactivation and gene replacement - specific mutations by homologous recombination

Question 20

Gene inactivation - strategy 1

Answer 20

See OneNote

• insert selectable marker into the gene to be inactivated on a YIp plasmid => gene cut in half => inactivated
• double X over

Question 21

Gene inactivation - strategy 2

Answer 21

See OneNote

- insert internal region only of gene into YIp vector

Question 22

Gene inactivation constructs

Answer 22

See OneNote

Question 23

What if the inactivation is lethal?

Answer 23

Use a diploid recipient e.g. Aspergillus => heterozygous gene knock-out

Shift to poor nutritional conditions to induce sporulation

Question 24

Genotypic ratio of heterozygous knock out

Answer 24

2 ura3- : 2URA3+ = not essential

2 ura3- : 0URA3+ = eseential

Question 25

Gene inactivation

Answer 25

to create loss of function mutant

Question 26

Gene replacement

Answer 26

See OneNote for steps | to introduce more subtle mutations into the genome

Question 27

A genetic way of ensuring integration at the gene of interest?

Answer 27

- use Aspergillus transformation

Question 28

Aspergillus transformation

Answer 28

See OneNote - Rapid, generates protoplasts, add DNA, isolate transformants - Auxotrophic and drug resistant selectable markers

Question 29

Aspergillus nidulans

Answer 29

- Ascomycete - Haploid or diploid - Asexual and sexual cycles BUT multinucleate hyphae

Question 30

A.nidulans integration

Answer 30

non-homologous integration occurs a lot more often than homologous integration (unlike in yeast)

Question 31

Non-homologous >> Homologous ad. and disad.

Answer 31

Advantages 1. can introduce any gene (no homology) 2. can leave native gene intact 3. can introduce multiple copies of gene Disadvantages 1. possible positions effects on expression 2. homologous events more difficult to isolate

Question 32

Gene targeting in Aspergillus

Answer 32

See OneNote | - targeting DNA to the argB locus

Question 33

Aspergillus conidiation

Answer 33

- asexual life cycle - ordered developmental program - requires the product of the brlA gene - brlA mRNA produces only with the onset of conidiation (encodes a TF)

Question 34

Does brlA promote conidiation or does conidiation promote brlA?

Answer 34

See OneNote

Question 35

Aspergillus transformation

Answer 35

- high freq. transformation => clone by complementation - DNA must be integrated, most integrations are non-homologous - homologous integrations do occurs => gene targeting and inactivation are possible - reporter genes can be easily introduced

Question 36

Ku70/Ku80

Answer 36

- required for DNA repair by NHEJ - identified and inactivated A.nidulans nkuA (A.nidulans homologue of the human Ku70) - in inactivated nl=kuA, integrations were essentially 100% homologous => gene inactivation/replacement now east