2025-2026 Genome technology Guy Van Camp
GENOME TECHNOLOGY AND
APPLICATIONS
CHAPTER 1: DNA CLONING (VAN CAMP)
4 STEPS OF DNA CLONING:
1) In vitro construction of a recombinant DNA molecule
a) Use a vector: extract plasmid from bacteria and cut with restriction enzyme (restriction
endonucleases). Then extract DNA of interest by cutting it with the same restriction enzymes
and ‘paste’ it (ligase) into the plasmid.
à Vector must contain replicon = DNA sequence that is capable of independent replication (it
has an ORI = origin of replication)
à Replicon is specific for the host
à Vector contains a lot of features used in the cloning process
2) Transformation
a) Put recombinant DNA into a host cell
b) Often bacteria or yeast is used: it grows easily and has fast reproduction
c) Often E. coli is used: it is convenient and can grow on plates and in liquid
à not for big human genes: these need post translational modifications that won’t happen in the
bacteria
d) But, for expression studies you must use eukaryotic cells: often insect cells are used
à You must make cell cultures, this takes longer
à Usually you first clone/amplify your recombinant DNA in
bacteria before you put it in eukaryotic cells
3) Selective propagation of clones
a) You must make sure that only the cells with the recombinant DNA
can grow/ that you are able to recognize them. One cell will form 1
colony and all cells in the same colony are identical/clones. (pick 1 colony using “enting” and
grow it in liquid)
b) E.g. antibiotics are used: in the vector you express an antibiotic-resistant gene. All cells that have
successfully taken up the vector will be resistant to antibiotic and will be able to grow.
4) Isolation of recombinant DNA clones
a) You lyse the cell and purify the plasmid
1
,2025-2026 Genome technology Guy Van Camp
RESTRICTION ENDONUCLEASES
• Nomenclature: 1 letter genus – 2 letter species – number
E.g. HaeIII = Haemophilus Aegypticus third
• Normally restriction enzymes are used as a defense mechanism of bacteria. They cleave the DNA of
bacteriophages that enter the bacterium. It cuts at specific DNA-sequences of bacteriophage.
o So how does restriction endonuclease know that it is not cutting its own bacterial DNA? à
with methylation: the bacterium will methylate its own DNA so restriction endonuclease
cannot cut. Usually, the bacterium does not only contain a restriction endonuclease, but
also a matching DNA-methylase.
• Restriction endonucleases are highly specific for 1 sequence in DNA and will only cut at that
sequence = recognition sequence. This sequence is usually a palindrome of 4-8 bp (both strands
read the same from 5’-3’ e.g. 5’-GGATCC-3’, 3’-CCTAGG-5’)
• Restriction endonucleases can cut in different ways:
o Blunt ends: the endonuclease cuts in a straight
line on the symmetry axis resulting in fragments
with blunt ends
o Sticky ends: the endonuclease does not cut in a straight line. It will create overhangs. If the
overhang has a 3’ end = 3’ overhang. When it is at the 5’ end = 5’ overhang (5’ overhangs are
more common than 3’ overhangs). These overhangs are ‘sticky ends’ since they can bind with
each other if they are complementary. To form covalent bonds, ligase is needed.
à Before using the ligase to hybridize bonds you must
inactivate restriction endonuclease, this can be done using
heat.
o In this example you see that MboI cuts after the G of GATC-
CTAG and that BamHI cuts after the second G of GGATCC-
CCTGG. This results in the same overhangs: GATC-CTAG,
so MboI and BamHI are compatible (their fragments can
hybridize). The difference between the 2 is that MboI cuts a
sequence of 4 bases and BamHI of 6 bases. Restriction enzymes
o Isoschizomeres = different restriction enzymes that have the same Enzyme Recognition Expected size*
sequence (human genome)
recognition pattern. The enzymes come from different bacteria à some are AluI AGTC 300 bp
cheaper, easier to use or more reliable for your experiment, but you can use HaeIII GGCC 600 bp
TaqI TCGA 1400 bp
both EcoRI GAATTC 3100 bp
o Depending on the recognition sequence, restriction enzymes will cut less or BamHI GGATCC 7000 bp
more frequently in the DNA à this depends on the size of the recognition
SmaI CCCGGG 78.000 bp
BssHII GCGCGC 390.000 bp
sequence (4-8bp) and if there are CpG sites (more CpG = less cutting) NotI GCGGCCGC 9.766.000 bp
• CpG-sequence/ CpG-islands: *40% G+C, CpG 20% of expected frequency
This is a special sequence since it allows methylation (methylations can only occur in CG-regions,
regions that are rich in CG = CpG-island)
o Consequence: CpG-islands are rare in human genome. So, if a restriction endonuclease
recognizes a CpG-site, it will cut less frequently resulting in larger fragments.
2
,2025-2026 Genome technology Guy Van Camp
DNA LIGASE
• It can restore a covalent bond between two DNA-fragments. This is
important for sticky ends: even if they ‘stick together’ this is not stable
since these are no covalent bonds.
• For Blunt ends the forming of covalent bonds is more difficult
• DNA ligase uses the -OH at 3’ of nucleotide and the PO42- at 5’ of
nucleotide: see figure
• Once you have cut your vector and your target DNA with your
endonucleases there are 3 possible ways in which they can ligate, but only
1 of these 3 ways is correct:
o The target DNA will bind to itself (multiple DNA fragments are
produced with the same sticky ends that can bind together) =
intermolecular concatemer
o The vector can bind to itself, since the fragment that we cut out of
vector can anneal to vector again or the two sticky ends of vector
might be complementary and bind each other = cyclization/
intramolecular vector
o The target DNA binds in vector DNA and recombinant DNA is
formed = only good way of ligation.
• Remember that we needed to select for the host cells that contained the recombinant DNA, this is
partly why (also because sometimes the host cell doesn’t take up vector/ recombinant DNA at all)
AVOID RECIRCULARIZATION
• 2 ways to avoid the recircularization of vector:
1. Use two different restriction enzymes. Instead of a single cut with the same sticky ends, a piece
with different sticky ends will be cut out of the vector. The insert will be cut using the same 2 RE’s.
These sticky ends are not complementary they cannot bind to each other. Only a DNA-fragment that
has the same sticky ends can be inserted, in the right direction!!! (you first need to remove the pieces
that are cut out because otherwise these could be ligated in the vector again)
2. Use dephosphorylation: ligase will hybridize the sticky ends using the OH at 3’ and the PO42- at the 5’
end of nucleotide. The endonucleases will cut the vector this way, leaving the OH and PO42-
accessible. By using alkaline phosphatase, you can remove the PO42- from the 5’ resulting in an OH.
Now there cannot be hybridization between the two sticky ends. However, to ligate the insert, you
must use an insert that still has the PO42- to make sure it can bind.
3
, 2025-2026 Genome technology Guy Van Camp
ORI = ORIGIN OF REPLICATION
• Rember that the vector must be a replicon = DNA sequence that is able to replicate on its own (so it
must contain an ORI = origin of replication = allows replication independent of the host
chromosome)
• The ORI is very specific for the host: for yeast you need a certain ORI and for E. coli you will need
another ORI
• The bacterial chromosome is circular and has 1 ORI. The human chromosome has multiple ORI’s so
replication can start at different points simultaneously, otherwise replication of all the chromosomes
would take too much time.
• ORI will determine the copy number in bacterium:
o Copy number of plasmids is high, so the ORI is highly active, the bacterium will have more
plasmids
o Copy number of chromosomes in bacterium is 1. The ORI is inhibited when there is no
replication of bacteria, so we only have 1 chromosome per cell
VECTORS
• The most used vectors are plasmids or bacteriophages.
o Plasmids:
= small circular DNA molecules in bacteria
= high copy number (multiple copy number ORI)
= are found in nature
= contain only a few genes
= transmitted vertically (mother-cel to daughter-cel)
= transmitted horizontally (This can cause problems in hospitals where antibiotic resistant
genes in plasmids of bacteria can be transmitted easily to next patient this way = super-
bacterium = resistant to antibiotics)
= have a supercoil structure. This way they can easily be purified from rest of cell-debris
after lysis of cell (note: the chromosomes of bacterium can also have a supercoil structure,
but they will lose this structure when cell is lysed)
o Bacteriophages
= viruses that infect bacteria
= linear or circular genome
= can be found outside the cell in a protein coat
PUC19 = CLASSICAL PLASMID
This is an example of a plasmid for cloning in bacteria:
• The vector has an ampicillin resistance gene
• It has an ORI with a high copy number
• It has multiple cloning sites (MCS): it has multiple
sites where only 1 restriction enzyme can cut (the
restriction enzyme can only cut in that place)
• It has a LacZ’ and lacl (for recombinant screening à see next page)
4
GENOME TECHNOLOGY AND
APPLICATIONS
CHAPTER 1: DNA CLONING (VAN CAMP)
4 STEPS OF DNA CLONING:
1) In vitro construction of a recombinant DNA molecule
a) Use a vector: extract plasmid from bacteria and cut with restriction enzyme (restriction
endonucleases). Then extract DNA of interest by cutting it with the same restriction enzymes
and ‘paste’ it (ligase) into the plasmid.
à Vector must contain replicon = DNA sequence that is capable of independent replication (it
has an ORI = origin of replication)
à Replicon is specific for the host
à Vector contains a lot of features used in the cloning process
2) Transformation
a) Put recombinant DNA into a host cell
b) Often bacteria or yeast is used: it grows easily and has fast reproduction
c) Often E. coli is used: it is convenient and can grow on plates and in liquid
à not for big human genes: these need post translational modifications that won’t happen in the
bacteria
d) But, for expression studies you must use eukaryotic cells: often insect cells are used
à You must make cell cultures, this takes longer
à Usually you first clone/amplify your recombinant DNA in
bacteria before you put it in eukaryotic cells
3) Selective propagation of clones
a) You must make sure that only the cells with the recombinant DNA
can grow/ that you are able to recognize them. One cell will form 1
colony and all cells in the same colony are identical/clones. (pick 1 colony using “enting” and
grow it in liquid)
b) E.g. antibiotics are used: in the vector you express an antibiotic-resistant gene. All cells that have
successfully taken up the vector will be resistant to antibiotic and will be able to grow.
4) Isolation of recombinant DNA clones
a) You lyse the cell and purify the plasmid
1
,2025-2026 Genome technology Guy Van Camp
RESTRICTION ENDONUCLEASES
• Nomenclature: 1 letter genus – 2 letter species – number
E.g. HaeIII = Haemophilus Aegypticus third
• Normally restriction enzymes are used as a defense mechanism of bacteria. They cleave the DNA of
bacteriophages that enter the bacterium. It cuts at specific DNA-sequences of bacteriophage.
o So how does restriction endonuclease know that it is not cutting its own bacterial DNA? à
with methylation: the bacterium will methylate its own DNA so restriction endonuclease
cannot cut. Usually, the bacterium does not only contain a restriction endonuclease, but
also a matching DNA-methylase.
• Restriction endonucleases are highly specific for 1 sequence in DNA and will only cut at that
sequence = recognition sequence. This sequence is usually a palindrome of 4-8 bp (both strands
read the same from 5’-3’ e.g. 5’-GGATCC-3’, 3’-CCTAGG-5’)
• Restriction endonucleases can cut in different ways:
o Blunt ends: the endonuclease cuts in a straight
line on the symmetry axis resulting in fragments
with blunt ends
o Sticky ends: the endonuclease does not cut in a straight line. It will create overhangs. If the
overhang has a 3’ end = 3’ overhang. When it is at the 5’ end = 5’ overhang (5’ overhangs are
more common than 3’ overhangs). These overhangs are ‘sticky ends’ since they can bind with
each other if they are complementary. To form covalent bonds, ligase is needed.
à Before using the ligase to hybridize bonds you must
inactivate restriction endonuclease, this can be done using
heat.
o In this example you see that MboI cuts after the G of GATC-
CTAG and that BamHI cuts after the second G of GGATCC-
CCTGG. This results in the same overhangs: GATC-CTAG,
so MboI and BamHI are compatible (their fragments can
hybridize). The difference between the 2 is that MboI cuts a
sequence of 4 bases and BamHI of 6 bases. Restriction enzymes
o Isoschizomeres = different restriction enzymes that have the same Enzyme Recognition Expected size*
sequence (human genome)
recognition pattern. The enzymes come from different bacteria à some are AluI AGTC 300 bp
cheaper, easier to use or more reliable for your experiment, but you can use HaeIII GGCC 600 bp
TaqI TCGA 1400 bp
both EcoRI GAATTC 3100 bp
o Depending on the recognition sequence, restriction enzymes will cut less or BamHI GGATCC 7000 bp
more frequently in the DNA à this depends on the size of the recognition
SmaI CCCGGG 78.000 bp
BssHII GCGCGC 390.000 bp
sequence (4-8bp) and if there are CpG sites (more CpG = less cutting) NotI GCGGCCGC 9.766.000 bp
• CpG-sequence/ CpG-islands: *40% G+C, CpG 20% of expected frequency
This is a special sequence since it allows methylation (methylations can only occur in CG-regions,
regions that are rich in CG = CpG-island)
o Consequence: CpG-islands are rare in human genome. So, if a restriction endonuclease
recognizes a CpG-site, it will cut less frequently resulting in larger fragments.
2
,2025-2026 Genome technology Guy Van Camp
DNA LIGASE
• It can restore a covalent bond between two DNA-fragments. This is
important for sticky ends: even if they ‘stick together’ this is not stable
since these are no covalent bonds.
• For Blunt ends the forming of covalent bonds is more difficult
• DNA ligase uses the -OH at 3’ of nucleotide and the PO42- at 5’ of
nucleotide: see figure
• Once you have cut your vector and your target DNA with your
endonucleases there are 3 possible ways in which they can ligate, but only
1 of these 3 ways is correct:
o The target DNA will bind to itself (multiple DNA fragments are
produced with the same sticky ends that can bind together) =
intermolecular concatemer
o The vector can bind to itself, since the fragment that we cut out of
vector can anneal to vector again or the two sticky ends of vector
might be complementary and bind each other = cyclization/
intramolecular vector
o The target DNA binds in vector DNA and recombinant DNA is
formed = only good way of ligation.
• Remember that we needed to select for the host cells that contained the recombinant DNA, this is
partly why (also because sometimes the host cell doesn’t take up vector/ recombinant DNA at all)
AVOID RECIRCULARIZATION
• 2 ways to avoid the recircularization of vector:
1. Use two different restriction enzymes. Instead of a single cut with the same sticky ends, a piece
with different sticky ends will be cut out of the vector. The insert will be cut using the same 2 RE’s.
These sticky ends are not complementary they cannot bind to each other. Only a DNA-fragment that
has the same sticky ends can be inserted, in the right direction!!! (you first need to remove the pieces
that are cut out because otherwise these could be ligated in the vector again)
2. Use dephosphorylation: ligase will hybridize the sticky ends using the OH at 3’ and the PO42- at the 5’
end of nucleotide. The endonucleases will cut the vector this way, leaving the OH and PO42-
accessible. By using alkaline phosphatase, you can remove the PO42- from the 5’ resulting in an OH.
Now there cannot be hybridization between the two sticky ends. However, to ligate the insert, you
must use an insert that still has the PO42- to make sure it can bind.
3
, 2025-2026 Genome technology Guy Van Camp
ORI = ORIGIN OF REPLICATION
• Rember that the vector must be a replicon = DNA sequence that is able to replicate on its own (so it
must contain an ORI = origin of replication = allows replication independent of the host
chromosome)
• The ORI is very specific for the host: for yeast you need a certain ORI and for E. coli you will need
another ORI
• The bacterial chromosome is circular and has 1 ORI. The human chromosome has multiple ORI’s so
replication can start at different points simultaneously, otherwise replication of all the chromosomes
would take too much time.
• ORI will determine the copy number in bacterium:
o Copy number of plasmids is high, so the ORI is highly active, the bacterium will have more
plasmids
o Copy number of chromosomes in bacterium is 1. The ORI is inhibited when there is no
replication of bacteria, so we only have 1 chromosome per cell
VECTORS
• The most used vectors are plasmids or bacteriophages.
o Plasmids:
= small circular DNA molecules in bacteria
= high copy number (multiple copy number ORI)
= are found in nature
= contain only a few genes
= transmitted vertically (mother-cel to daughter-cel)
= transmitted horizontally (This can cause problems in hospitals where antibiotic resistant
genes in plasmids of bacteria can be transmitted easily to next patient this way = super-
bacterium = resistant to antibiotics)
= have a supercoil structure. This way they can easily be purified from rest of cell-debris
after lysis of cell (note: the chromosomes of bacterium can also have a supercoil structure,
but they will lose this structure when cell is lysed)
o Bacteriophages
= viruses that infect bacteria
= linear or circular genome
= can be found outside the cell in a protein coat
PUC19 = CLASSICAL PLASMID
This is an example of a plasmid for cloning in bacteria:
• The vector has an ampicillin resistance gene
• It has an ORI with a high copy number
• It has multiple cloning sites (MCS): it has multiple
sites where only 1 restriction enzyme can cut (the
restriction enzyme can only cut in that place)
• It has a LacZ’ and lacl (for recombinant screening à see next page)
4
