Plasmid cloning

Question 1

What are cloning vectors used for?

Answer 1

Cloning fragments of DNA.

Question 2

Where are cloning vectors derived from?

Answer 2

Naturally occurring plasmids or bacteriophages. They have been modified to make them more convenient and versatile.

Question 3

What are plasmids?

Answer 3

Closed-circular double stranded DNA molecules found in many prokaryotic cells. They are self-replicating.

Question 4

What are bacteriophages?

Answer 4

They are viruses that infect bacteria. They consist of a nucleic acid molecule surrounded by a protein coat, and can be DNA or RNa.

Question 5

What are lambda bacteriophages used for?

Answer 5

Used for cloning large pieces of DNA.

Question 6

What are M13 bacteriophages used for?

Answer 6

DNA sequencing.

Question 7

What are the differences between lambda and M13?

Answer 7

Lambda is double stranded whereas M13 is single stranded and is 50kb compared to 10kb.

Question 8

What are the desirable characteristics of plasmid cloning vectors?

Answer 8

Having a small size, a range of unique restriction sites, selectable markers, positive selection and a high copy number.

Question 9

What benefits does having a small size of cloning vector provide?

Answer 9

They are easier to manipulate, have a high copy number, more insert per ug of plasmid and are easier transformable (easier made to enter a bacterial cell than a larger one)

Question 10

Why is having a unique range of restriction sites useful?

Answer 10

It allows cloning fragments to be produced by several enzymes.

Question 11

Why is having a selectable marker in a cloning vector useful?

Answer 11

Plasmid containing cells can be distinguished from plasmid lacking cells. This selectable marker is often an antiobiotic resistance gene such as beta lactamase (resistance to ampicillin)

Question 12

Why is positive selection useful for cloning vectors?

Answer 12

Enables bacterial cells that contain a cloned insert to be identified.

Question 13

Why is a high copy number useful for cloning vectors?

Answer 13

Ensures as much plasmid DNA can be purified from each cell as possible.

Question 14

Why is the plasmid pBR322 cloning vector named as it is?

Answer 14

p=plasmid, BR= Bolivar and Rodriguez.

Question 15

What are the selectable markers in pBR322?

Answer 15

Ampicillin and tetracycline resistance.

Question 16

How large is pBR322?

Answer 16

4.4Kb in size.

Question 17

What are some of the restriction sites in pBR322?

Answer 17

PstI, EcoRI and HindIII.

Question 18

Does pBR322 have a high copy number or positive selection?

Answer 18

It has a moderate copy number (15 per cell) and does not have true positive selection but something similar - insertional activation.

Question 19

Where is BamHI restriction site found in pBR322?

Answer 19

Within the tetracycline resistant gene.

Question 20

What is the first step for cloning into pBR322?

Answer 20

The cloning vector is cut with BamHI and may (or may not be) dephosphorylated with alkaline phosphatase.

Question 21

What is the second step in cloning into pBR322?

Answer 21

The DNA to be cloned is cut with the same enzyme (or an enzyme that produces the same sticky ends) and is not dephosphorylated (or is if the plasmid wasn’t).

Question 22

What is the third step in cloning into pBR322?

Answer 22

The plasmid and the DNA is mixed together in a suitable buffer in the presence of DNA T4 ligase and ATP. The molecules will join together in all possible combinations.

Question 23

What are the products of ligation in cloning into pBR322?

Answer 23

Unligated vector molecules, unligated DNA fragments, self-ligated vector molecules, wrong recombinant molecules, the desired recombinant molecule, multiple vectors stuck together and linear products.

Question 24

What is the next step after mixing the DNA together in cloning into pBR322?

Answer 24

The ligated DNA must be inserted into the E.coli cells by transformation.

Question 25

How must the E.coli cells be treated to prepare them to take up the recombinant plasmids?

Answer 25

They must be made competent by treating them with cold CaCl2, enabling them to take up naked DNA.

Question 26

What are the following steps in transformation when cloning into pBR322?

Answer 26

The DNA must be incubated with the competent cells on ice (so the DNA can bind to the cell exterior) and the mixture must be given a heat shock at around 42 degrees which allows the DNA to enter. The cells are now transformed and can be plated onto agar medium in petri dishes.

Question 27

What is a problem with the transformation method in cloning into pBR322?

Answer 27

The process is inefficient and only a small minority of the cells take up the plasmid DNA (around 0.01%)

Question 28

How can the E.coli cells that contain the plasmids be selected for?

Answer 28

If ampicillin is included in the medium the cells are being grown on, only those containing the plasmids will be able to grow due to the resistant gene. The antibiotic acts as a positive selective agent.

Question 29

What is a problem with this selection process?

Answer 29

All plasmid containing cells will grow into colonies and there is no differentiation between empty plasmids, recombinant plasmids or your clone.

Question 30

How can the problem with the selection process for cells containing plasmids be overcome?

Answer 30

Using insertional inactivation.

Question 31

What is insertional inactivation?

Answer 31

This is when extra DNA can be inserted into a gene to inactivate the gene. pBR322 contain resistance to ampicillin and to tetracycline, so a gene can be inserted into the tetracycline resistant gene so that it is inactivated.

Question 32

How can insertional inactivation provide a solution to the problem of identifying the correct recombinants?

Answer 32

The growth of colonies on ampicillin and tetracycline can be compared. Only non-recombinants will grow on the tetracycline plate which allows the identification of recombinants on the ampicillin plate which can be picked for further analysis.

Question 33

What are some key features of the plasmid cloning vector pUC8?

Answer 33

It is small in size (2750 bp), it is only ampicillin resistant, its unique restriction sites are clustered together in the lacZ’ gene of the beta galacosidase from the lac operon.

Question 34

What is special about the lacZ’ gene?

Answer 34

It contains a BamHI restriction site that when new DNA is inserted, will inactivate the lacZ’ so the beta galactosidase resistance is inactivated. This forms the basis for positive selection.

Question 35

How is this feature of lacZ’ gene useful when cloning in pUC8?

Answer 35

A mutant E.coli strain can be used that only has one part of the beta galactosidase gene and the other part being containing on the pUC38 plasmid. This means that only the non-recombinant plasmids will produce the complete beta galactosidase enzyme.

Question 36

What two chemicals must the medium contain that the transformed E.coli cells (with pUC38) contain in order for the positive selection to work?

Answer 36

IPTG and X-gal.

Question 37

What do each of these chemicals (in the medium for pUC38 for positive selection) do?

Answer 37

IPTG is an artificial inducer for beta galactosidase and X-gal is an artificial substrate for beta galactosidase.

Question 38

How do these chemicals in pUC38 plating allow for positive selection to occur?

Answer 38

Beta galactosidase converts X-gal to a blue pigment, so recombinant pU38 colonies will remain white whereas non-recombinant colonies will be blue. The white colonies cna be picked for further analysis.

Question 39

What are the advantages of positive selection over replica plating?

Answer 39

It is easier.