1.1. Cell-based DNA cloning
1.1.A. Principles of DNA cloning
Cell-based DNA cloning comprises of four steps:
1. In vitro construction of a recombinant DNA molecule
2. Transformation to a host (usually E. coli)
3. Selective propagation of clones
4. Isolation of recombinant DNA clones
① In vitro construction of a recombinant DNA molecule
❥ Requires the cutting and pasting of DNA
❣ This is done using restriction endonucleases (RE), these are restriction enzymes that will cut a DNA
molecule at a specific site
❣ DNA ligase that sticks the cut ends back together
❥ Requires a replicon
❣ This is a piece of DNA that can replicate independently
❣ This is host specific
❣ Usually a “vector” is used that contains many features that are
used in the cloning process
⤷ Is usually a plasmid: short + many copies present + they replicate independently
② Transformation
❥ The recombinant DNA molecule is introduced in a host cell
❣ Usually a bacterium or yeast: easy to grow and fast to reproduce
❥ In bacteria such as E. coli, large genomes can be inserted but the
expression of such genomes (that will make large proteins) is not
possible
❣ Bacteria can’t synthesize such big proteins
❣ Bacteria also can’t modify such bit proteins
❣ So for expression studies, cloning is done in eukaryotic cells
③ Selective propagation of clones
❥ Cells are plated on agar and each individual cell
forms a colony
❥ Each colony is a clone: all cells in that colony are
identical and have the same ancestor cell
❥ One colony can then be grown in liquid medium
to obtain more cells
④ Isolation of recombinant DNA clones
❥ The recombinant DNA is purified from the cells
❣ The cell are lysed and the plasmids are isolated and taken out
❥ Results: recombinant DNA clones
,⑤ Overview of the process
1.1.B. Toolkit of cell-based cloning
① Restriction endonucleases (RE)
❥ Nomenclature of RE...
❣ 1st letter: genus ⇒ HaeIII ↠Hemophilus aegypticus
❣ 2nd and 3rd letters: species ⇒ HaeIII ↠ Hemophilus aegypticus
❣ Followed by a number: HaeIII
❥ RE naturally occur in bacteria as a defence mechanism against bacteriophages
❣ When a bacteriophage inserts its viral genome into a bacterium, the RE will cleave that genome
❣ The bacterial genome also has recognition sites but these are ‘hidden’ via methylation which inhibits
the RE from cutting into the bacterial genome
❥ Type II RE will cut a specific recognition sequence
❣ Usually 4-8 bp long
❣ Usually a palindrome
❥ RE can cleave in two ways...
❣ On the symmetry axis
⤷ Here it cuts straight through both strands creating two
blunt ends
❣ Non-symmetrical
⤷ Here it cuts the DNA at two locations that don’t align
⤷ This creates overhangs or sticky ends
⤷ We can distinguish a 3’ and 5’ prime overhang
⤷ This reaction can be reversed by DNA ligase
,❥ RE can have many different recognition sequences
❣ In general the human genome has more AT than GC, this
causes RE with AT in their sequence to be shorter fragments
(they find the next recognition site quicker)
❣ Recognition sequences that contain GC or are long usually
result in large fragments because the odds of finding that
same sequence in the genome is smaller and so it takes a
lot longer finding a similar site
❥ Different RE with the same recognition sequence are called
isoschizomeres, they won’t necessarily have the same sticky ends sequences
❥ Some RE can have compatible sticky ends despite having different recognition sites (so they aren’t
isoschizomeres)
BamHI:
MboI:
② DNA ligase
❥ The ligation reaction from DNA ligase
with the target DNA fragment and vector
DNA won’t necessarily cause the ideal
intermolecular vector-target
recombinant DNA result instead the
ligation reaction can have many results
❥ DNA ligase can restore a covalent bond
in a DNA molecule; this is easier for
sticky ends than blunt ends
❥ Several different fragments that are ligated together are called a
concatemer
❥ Intramolecular ligation is called cyclisation
③ Origin of replication (ORI)
❥ An ORI allows replication independent of the host chromosome
❥ Independent replication facilitates purification of recombinant molecule
❥ Bacterial chromosome contains...
❣ A circular chromosome with one ORI that limits the number of chromosomes to one per cell
❣ Plasmids with specific types of ORI
, ④ Vector
1. Plasmids
❣ Small, circular DNA molecules in bacteria
❣ Usually contain multiple copy number ORI and a few genes
❣ Can be transmitted...
⤷ Vertically from parent to daughter cell during binary fission
⤷ Horizontally from one bacterium to another via conjugation
⧙ This is what facilitates antibiotic resistance
⧙ If one bacterium has an antibiotic resistance gene it can pass it to another horizontally via
conjugation thus rendering that other bacterium resistant too
❣ Has a supercoil structure used for replication (like bacterial chromosomes)
2. Bacteriophages
❣ Are viruses that infect bacteria
⤷ This means that these organisms are naturally skilled at injecting a genome into a bacteria
⤷ So very useful for transformation (where we enter the recombinant DNA into the bacterium)
⤷ Can have linear or circular genomes
⤷ Can be found outside of cells, in a protein coat
❥ How can we avoid recirculation in a plasmid vector?
1. Using two different RE
⤷ This can be achieved by using two different RE to cut the vector
⤷ If for example you cut your plasmid using EcoRI en HindIII than you cut out a fragment, and the
remaining vector has two overhangs that are not
compatible and thus won’t rehybridize
⤷ This allows you to then easily insert your fragment
⧙ If you cut your fragment using the same RE as the
vector, then your fragment will be compatible
⧙ Then you insert and ligate your fragment with the
vector fragment
2. Using dephosphorylation
⤷ The ligation reaction hinges on the presence of...
⧙ An OH--group at the 3’ end
⧙ A PO43--group at the 5’ end
⤷ When both are present, DNA ligase can ligate these together
⤷ So if you dephosphorylate the 5’ end so that an OH-group is present instead, you will block DNA
ligase from ligating the two overhangs together
⤷ This can be achieved using alkaline phosphatase
⧙ It removes the 5’ phosphate group
⧙ It leaves a sugar and base
⤷ Result: vector can’t recircularize
⤷ If your insert then contains a 5’ phosphate group then it can be
included in the vector
❥ A vector usually contains multiple cloning site; a region with
multiple RE-sites