Institute of Molecular Life Sciences (IMLS)
Prof. Michael Hengartner
Dr. Monika Hediger
Practical course
BIO 111
Part 2: Molecular Genetics
Course-4: 08. Nov. / 09. Nov. / 15. Nov. / 16. Nov. 2018
Course-5: 22. Nov. / 23. Nov. / 29. Nov. / 30. Nov. 2018
HS 2018
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
1
Table of contents
Table of contents
A) Human genetic diversity
1. Human trait: “Bitter” taste perception_the PTC gene
1.1. Introduction
1.1.1. “Bitter” taste perception: the TAS2R38 gene (PTC taste receptor)
p. 3
1.2. Goal of the Experiment
p. 8
1.3. Polymerase Chain Reaction (PCR): Theory
p. 8
1.4. Protocol
1.4.1. DNA Isolation of cheek cells
1.4.2. Setting up PCR reaction
1.4.3. Setting up sequencing reaction
p. 10
1.5. Analysis of PTC genotype and phenotype
1.5.1. Determining your PTC genotype
1.5.2. Testing your PTC phenotype
p. 13
1.6. Questions
p. 14
B) Recombinant DNA Technology
2. Isolation of pB30 plasmid DNA (Miniprep-Protocol)
2.1. Introduction
2.1.1. Principles of Plasmid DNA Isolation
p. 15
2.2. Goal of the Experiment
p. 16
2.3. Protocol
p. 17
2.4. Quantification of the isolated DNA
2.4.1. DNA concentration
2.4.2. Purity of DNA
2.4.3. Concentration and purity of “your” DNA sample
p. 19
2.5. Questions
p. 20
3. Restriction analysis of plasmid pB30
4.
3.1. Introduction
3.1.1. Plasmid vectors
3.1.2. Restriction endonucleases
3.1.3. Agarose gel electrophoresis
p. 22
3.2. Goal of the Experiment
p. 27
3.3. Protocol
3.3.1. Restriction digest
3.3.2. Separating DNA fragments by agarose gel electrophoresis
p. 28
3.4. Analysis of agarose gel
3.4.1. Fragment size expectations of restriction digest
3.4.2. Actual fragment sizes
p. 30
3.5. Questions
p. 31
References
p. 33
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
Time table
Time table: Part-2_Molecular Genetics
Room: Practical room 14-F-21
Time: 13h00 – 17h00
Course-4: Thu_08.11.2018 ; Fri_09.11.2018 ; Thu_15.11.2018 ; Fri_16.11.2018
Time
Experiments
13:00 – 15:00
1. Human trait: “Bitter” taste perception_the PTC gene
1.1. Introduction
1.4. Protocol: 1) DNA-Isolation and 2) setting up PCR reactions
1.3. PCR: Theory
short break
15:00 – 16:30
2. Isolation of Plasmid DNA
2.1. Introduction
2.3. Protocol: Isolation of Plasmid-DNA
2.4. Quantification of isolated DNA
2.5. Solving and discussion of questions
16:30 – 17:00
1. Human trait: “Bitter” taste perception_the PTC gene
1.4. Protocol: 3) Setting up sequencing reactions
Course-5: Thu_22.11.2018 ; Fri_23.11.2018 ; Thu_29.11.2018 ; Fri_30.11.2018
Time
Experiments
13:00 – 15:30
3. Restriction analysis of Plasmid DNA
3.1. Introduction
3.3. Protocol: 1) Setting-up of restriction digest
3.4. Agarose gel electrophoresis: 1) expected fragment sizes
3.3. Protocol: 2) Loading digest on agarose gel
short break
15:30 – 16:30
1. Human trait: “Bitter” taste perception_the PTC gene
1.1. Introduction/Recapitulation
1.5. Analysis of PTC genotype and phenotype
1.6. Solving and discussion of questions
16:30 – 17:00
3. Restriction analysis of Plasmid DNA
3.4 Agarose gel electrophoresis: 2) Analysis of results
3.5. Solving and discussion of questions
2
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
A) Human genetic diversity → Experiment 1: “Bitter” taste perception_the PTC gene
A) Human genetic diversity
With the development of DNA-based technologies, geneticists now have the ability to
observe directly differences between the DNA sequences of individuals throughout their
genomes, and they can measure theses differences in large samples of individuals in many
species. The result has been a revolution in our understanding of genetic variation in
populations.
Single nucleotide polymorphisms (SNPs) are the most prevalent types of polymorphism in
most genomes. SNPs occur within genes (including exons, introns, and regulatory regions)
as well as outside of coding regions. SNPs within protein-coding regions can be classified
into one of three groups: synonymous (if the different alleles encode the same amino acid),
nonsynonymous (if the two alleles encode different amino acids), and nonsense (if one allele
encodes a stop codon). Thus, it is sometimes possible to associate a SNP with functional
variation in proteins and an associated change in phenotype.
1. Human trait: “Bitter” taste perception
1.1. Introduction
Taste perception is triggered by chemicals when they come in contact with taste-receptor
cells (TRCs) of the tongue. Five basic tastes are recognized by most animals: sweet, sour,
bitter, salty, and umami (the taste of monosodium glutamate).
TRCs are assembled into taste buds, which are distributed across different papillae of the
tongue. There are about 5000 taste buds in the oral cavity.
There are three types of papillae: Circumvallate papillae are found at the very back of the
tongue. Foliate papillae are present at the posterior lateral edge of the tongue and fungiform
papillae are found in the anterior two-thirds of the tongue (Fig. 1)
Each taste bud contains ~100 TRCs. TRC’s are classified into three subtypes and all taste
buds contain cells of all three subtypes (Fig. 2).
Figure 1
Three types of taste
papillae located on the
human tongue
3
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
A) Human genetic diversity → Experiment 1: “Bitter” taste perception_the PTC gene
Figure 2: Three different taste bud cell types
Figure 2
1.1.1. “Bitter” taste perception: the TAS2R38 gene (PTC taste receptor)
The sense of bitter taste is mediated by a group of bitter taste receptor proteins. There are ca.
30 genes for different bitter taste receptors in mammals.
Taste bud cell type II – bitter recognition
Figure 3
The gene encoding the taste receptor TAS2R38 (taste receptor type 2, member 38), which
enables us to taste the bitter compound phenylthiocarbamide (PTC) or the related
compound propylthiouracil (PROP), was identified in 2003 [Kim, 2003]. It resides on
chromosome 7, position q35. Within this gene, several single nucleotide polymorphism
(SNPs) have been identified, all of which result in amino acid changes in the protein
(nonsynonymous SNPs). Three of these SNPs are the most common in the human population.
They reside on nucleotide position 145 (SNP-1: C ® G), nucleotide position 785 (SNP-2: C ® T)
and nucleotide position 886 (SNP-3: G ® A), and alter the amino acid sequence (see Fig. 4, 5).
4
5
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
A) Human genetic diversity → Experiment 1: “Bitter” taste perception_the PTC gene
Figure 4
These three most common SNPs are inherited as a unit (also called haplotype). There are two
common haplotypes in the human population: One haplotype correlates most strongly with
bitter-taste ability. This “wildtype” protein contains the three amino acids Proline (P),
Alanine (A) and Valine (V) at the SNP positions (SNP-1, -2, -3) and is called the dominant
“PAV” taster haplotype (= dominant taster “T” allele).
The second specific combination of the three SNPs correlates with the inability of PTC
tasting. This “mutant” protein contains the three amino acids Alanine (A), Valine (V) and
Isoleucine (I) at the SNP positions (SNP-1, -2, -3) and is called the recessive “AVI” non-taster
haplotype (= recessive non-taster “t” allele).
Thus, the hTAS2R38 polymorphisms that differ on chromosomes within an individual and
between individuals code for functionally distinct receptor types that directly affect
bitterness perception of N-C=S-containing compounds (Bufe et al., 2005)
Figure 5
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
A) Human genetic diversity → Experiment 1: “Bitter” taste perception_the PTC gene
Frequencies of tasters and non-tasters in different human populations
It should be noted that PTC taste sensitivity is not an all or nothing trait. However, people
homozygous for the PAV/PAV genotype have the highest PTC scores (most find PTC
intensely bitter), PAV heterozygotes (genotype PAV/AVI) generally have a mean PTC score
(find PTC somewhat bitter), and AVI/AVI homozygote persons have the lowest PTC score
(for most of them the PTC compound has no taste at all).
In addition to the high-frequency PAV and AVI haplotypes, also two rare (AAV and AAI)
and four extremely rare (PVI, PAI, AVV, PVV) haplotypes are found in the human
population (Table 1 and Figure 1, Risso et al, 2016). The rare AVI/AAV heterozygotes found
in European populations had a mean PTC score slightly, but significantly, higher than the
AVI/AVI homozygotes.
Table 1
There is a global predominance of the PAV and AVI haplotypes of TAS2R38. The AVI form
is less common in Africa and Asians. The AAI haplotype is primarily present in Africans and
much less common in all the other populations.
Evolution of Taster and non-tasters in human populations
Sequencing the PTC gene from several non-human primates determined that all were homozygous for
the PAV form. Thus, the AVI nontaster haplotype arose after humans diverged from the most recent
common primate ancestor. There are cases of PTC nontaster phenotype in chimpanzees. However the
non-taster phenotype appears to be due to a mutation of the initiation codon of the PTC gene, such that a
downstream ATG is used as the start codon for translation, resulting in a truncated protein that does not
respond to PTC.
The persistence of common haplotypes over time and their widespread distribution across globally
diverse populations suggest that these variants are functionally important. The dominant PAV taster
haplotype is hypothesized to have played a role in detecting potentially toxic substances. Bitter taste is
innate and triggers stereotypical behavioral outputs leading to rejection. Although a clear correlation
between bitterness and toxicity has not been established, it is generally believed that this taste quality
prevents mammals from intoxication by avoiding ingestion of potentially harmful food constituents
(Meyerhofer, 2009). However, the selective force maintaining other common AVI (AAV and AAI)
haplotypes for extraordinarily long periods of time remains yet unclear.
6
7
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
A) Human genetic diversity → Experiment 1: “Bitter” taste perception_the PTC gene
Human TAS2R38, coding region: 1002nt
The human TAS2R3 gene consists of a single coding exon 1002 bp long, encoding a 333
amino acid, 7-transmembrane domain G-protein-coupled receptor.
Shown is the coding DNA strand of the “wildtype” = taster TAS2R38 gene and below the
amino acid sequence of the TAS2R38 “wildtype” protein.
Non-Taster:
Taster: 5’-ATGTTGACTCTAACTCGCATCCGCACTGTGTCCTATGAAGTCAGGAGTACATTTCTGTTCATTTCAGTC -069
Protein
M L T L T R I R T V S Y E V R S T F L F I S V
Non-Taster:
Taster:
CTGGAGTTTGCAGTGGGGTTTCTGACCAATGCCTTCGTTTTCTTGGTGAATTTTTGGGATGTAGTGAAG -138
Protein
L E F A V G F L T N A F V F L V N F W D V V K
Non-Taster:
Taster:
AGGCAGCCACTGAGCAACAGTGATTGTGTGCTGCTGTGTCTCAGCATCAGCCGGCTTTTCCTGCATGGA -207
Protein
R Q P L S N S D C V L L C L S I S R L F L H G
Non-Taster:
Taster:
CTGCTGTTCCTGAGTGCTATCCAGCTTACCCACTTCCAGAAGTTGAGTGAACCACTGAACCACAGCTAC -276
Protein
L L F L S A I Q L T H F Q K L S E P L N H S Y
Non-Taster:
Taster:
CAAGCCATCATCATGCTATGGATGATTGCAAACCAAGCCAACCTCTGGCTTGCTGCCTGCCTCAGCCTG -345
Protein
Q A I I M L W M I A N Q A N L W L A A C L S L
Non-Taster:
Taster:
CTTTACTGCTCCAAGCTCATCCGTTTCTCTCACACCTTCCTGATCTGCTTGGCAAGCTGGGTCTCCAGG -414
Protein
L Y C S K L I R F S H T F L I C L A S W V S R
Non-Taster:
Taster:
AAGATCTCCCAGATGCTCCTGGGTATTATTCTTTGCTCCTGCATCTGCACTGTCCTCTGTGTTTGGTGC -483
Protein
K I S Q M L L G I I L C S C I C T V L C V W C
Non-Taster:
Taster:
TTTTTTAGCAGACCTCACTTCACAGTCACAACTGTGCTATTCATGAATAACAATACAAGGCTCAACTGG -552
Protein
F F S R P H F T V T T V L F M N N N T R L N W
Non-Taster:
Taster:
CAGATTAAAGATCTCAATTTATTTTATTCCTTTCTCTTCTGCTATCTGTGGTCTGTGCCTCCTTTCCTA -621
Protein
Q I K D L N L F Y S F L F C Y L W S V P P F L
Non-Taster:
Taster:
TTGTTTCTGGTTTCTTCTGGGATGCTGACTGTCTCCCTGGGAAGGCACATGAGGACAATGAAGGTCTAT -690
Protein
L F L V S S G M L T V S L G R H M R T M K V Y
Non-Taster:
Taster:
ACCAGAAACTCTCGTGACCCCAGCCTGGAGGCCCACATTAAAGCCCTCAAGTCTCTTGTCTCCTTTTTC -759
Protein
T R N S R D P S L E A H I K A L K S L V S F F
Non-Taster:
Taster:
TGCTTCTTTGTGATATCATCCTGTGCTGCCTTCATCTCTGTGCCCCTACTGATTCTGTGGCGCGACAAA -828
Protein
C F F V I S S C A A F I S V P L L I L W R D K
Non-Taster:
Taster:
ATAGGGGTGATGGTTTGTGTTGGGATAATGGCAGCTTGTCCCTCTGGGCATGCAGCCGTCCTGATCTCA -897
Protein
I G V M V C V G I M A A C P S G H A A V L I S
Non-Taster:
Taster:
GGCAATGCCAAGTTGAGGAGAGCTGTGATGACCATTCTGCTCTGGGCTCAGAGCAGCCTGAAGGTAAGA -966
Protein
G N A K L R R A V M T I L L W A Q S S L K V R
Non-Taster:
Taster:
GCCGACCACAAGGCAGATTCCCGGACACTGTGCTGA-3’ -1002nt
Protein
A D H K A D S R T L C
333aa
Figure 3
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
A) Human genetic diversity → Experiment 1: “Bitter” taste perception_the PTC gene
1.2. Goal of the experiment
In this experiment, a sample of human cells is obtained by saline mouthwash. DNA is
extracted by boiling with Chelex resin, which binds contaminating metal ions. Polymerase
chain reaction (PCR) is then used to amplify a short region of the TAS2R38 gene. The
amplified PCR product is then sequenced to identify two of the three most common SNPs
within the TAS2R38 gene.
Each student scores his or her genotype, predicts his or her tasting ability, and then tastes
PTC paper. Class results show how well PTC tasting actually conforms to classical
Mendelian inheritance, and illustrates the modern concept of pharmacogenetics—where a
SNP genotype is used to predict drug response.
1.3. Polymerase chain reaction (PCR): Theory
PCR is an alternative to cloning for generating essentially unlimited amounts of a sequence
of interest. PCR is an enzymatic amplification of a fragment of DNA (the target) located
between two oligonucleotides (= primers). These primers are designed so that one is
complementary to one strand of a DNA molecule on one side of the target sequence and the
other primer is complementary to the other strand of the DNA molecule on the opposite side
of the target sequence. Because the 3’ end of each primer points toward the target sequence
to be amplified, the primers are extended by the synthesis by DNA polymerase of the
sequence between them.
After denaturation of the DNA duplex, the primers anneal to the single stranded DNA
(ssDNA), giving rise to short double-stranded stretches of DNA, to which the polymerase
adds nucleotides complementary to the template strand, thereby extending the doublestranded region in a 5’ > 3’ direction.
First, a pair of oligonucleotides of about 20 nucleotides (nt) in length has to be designed and
synthesized (done by numerous companies). The primers are designed to anneal to
complementary DNA-sequences flanking the DNA region of interest, thereby bracketing the
region to be amplified. An excess of the two primers are mixed with a DNA sample
containing the target sequence, along with a DNA polymerase. The cofactor Magnesium
(Mg++) and the four deoxyribonucleoside triphosphates (dNTPs) are also provided. The
reaction mixture is then taken through multiple synthesis cycles consisting of the following:
1. Denaturation:
Heating to near-boiling temperatures (95°C) denatures the target DNA and creates a
set of single-stranded templates. Heating increases the kinetic energy of the DNA
molecule to a point that is greater than the energy needed to break hydrogen bonds
between base pairs, and the double-stranded DNA separates into single strands.
2. Annealing:
Cooling to approximately 60°C (depending on the length and the sequence of the
primer) encourages primers to anneal to their complementary sequences on the
single-stranded templates. Because the primers are added in excess and are short,
they will anneal to their long target sequences before the two original strands can
come back together.
3. Extension:
Heating to 72°C provides the optimum temperature for the DNA polymerase to
extend from the primer. The polymerase synthesizes a second strand complementary
to the original template, at a rate of up to 1000nt/minute.
8
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
A) Human genetic diversity → Experiment 1: “Bitter” taste perception_the PTC gene
During each synthesis cycle of approximately 30-60sec, the number of copies of the DNA
molecule is doubled; thus, 30 rounds of synthesis theoretically can produce a 1’000’000'000fold amplification of the target sequence in as little as 1 hour. In practice, the amplification is
rarely perfect, and a slightly lower yield will be achieved.
The high temperatures used in PCR require heat-stable polymerases. In the late 1980s, the
first such polymerase (known as Taq-polymerase) was isolated from Thermus aquaticus,
obtained from the Mushroom Pool in Yellowstone National Park. Volcanic ocean vents
became another prominent source for organisms adapted to extreme conditions with heatstable polymerases.
Temperature cycling was initially achieved by moving reaction tubes back and forth between
different water baths. This tedium was relieved by the advent of automated DNA thermal
cyclers in which the temperature of a heating block and the incubation times are controlled
by a microcomputer.
Primer
→ Primer extension
→ desired fragment (only target DNA)
→ desired fragment (only target DNA)
→ variable length strands
Figure 10-3
Introduction to Genetic Analysis,
11ed, Griffiths et al.
9
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
A) Human genetic diversity → Experiment 1: “Bitter” taste perception_the PTC gene
1.4. Protocol
The protocol we use is a modified version from the DNA Learning Center Kit, Carolina, Cold
Spring Harbor Laboratory, 2006.
1.4.1. DNA isolation of cheek cells by saline mouthwash
Each student will isolate his/her own genomic DNA from cheek cells.
Reagents
• 0.9% saline solution (in 50ml Falcon tubes).
Ingredients suited for human consumption: salt from the supermarket, water from the kitchen
faucet. Aliquots of 10ml are prepared.
• 10% Chelex solution (BT Chelex 100 Resin, BioRad, Cat.#143-2832).
Aliquots of 100µl are prepared. Solution has to be shaken frequently as the beads tend to settle
quickly.
Chelex 100 resin is a styrene divinylbenzene copolymer containing paired iminodiacetate ions,
which act as chelating groups in binding polyvalent metal ions --> Chelex sequesters heavy metals.
If they remain in the solution, they would activate DNases, which would then cleave DNA.
!! During all centrifugation steps: a) Balance tubes and b) put hinge of the tube towards
top/outside of the centrifuge, so that you know where the pellet will be !!
1.
Pour 10ml saline solution (0.9%) in your mouth and rinse your mouth vigorously for
30 – 40 sec. → Spit the solution back into the tube.
2.
Swirl the tube to keep the cells in solution and pipet 1ml into a 1.5ml Eppendorf tube.
→ Label the Eppendorf tube with your place number.
3.
Centrifuge 2 min at full speed in Eppendorf (table top) centrifuge.
4.
Pour off supernatant carefully. NDo NOT disturb the (white) pellet of cheek cells at the
bottom of the tube!
5.
Resuspend the cells in the residual volume (about 100µl), by vortexing the tube.
→ The liquid now appears opaque.
6.
Pipet 30µl of the cell suspension into one of the small PCR tubes containing 100µl of the
aliquoted Chelex solution.
7.
Close the lid and mix the content by tapping the tube with your finger.
→ Label PCR tube with your place number.
8.
Place labeled PCR tube in PCR machine and heat for 10 min at 99°C
(Ê disruption of cell membranes and release of genomic DNA into solution)
9.
After heating, vigorously shake the PCR tube by hand for 5 sec.
10. Place the PCR tube into an adaptor and centrifuge at full speed for 2 min.
(Ê This will pellet the Chelex beads with bound metal ions (e.g. Ca2+, Mg2+, Al3+, etc.) and cell
debris, leaving the genomic DNA suspended in the supernatant above the beads.)
11. Store the PCR tube on ice.
You will later directly use 2µl of the clear supernatant containing YOUR genomic DNA
to set up the PCR reaction (Nbe careful not to disturb the pellet of cell debris and
Chelex beads at the bottom of the tube !).
10
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
A) Human genetic diversity → Experiment 1: “Bitter” taste perception_the PTC gene
1.4.2. Setting up PCR reaction
Each student (and each assistant) will set up 1 PCR reaction using his/her own genomic
cheek cell DNA preparation as template to amplify part of the TAS2R38 (= PTC) gene.
Reagents
• Cheek cell DNA from part 1 (protocol 1.4.1.)
• PTC gene-specific DNA primers (10 µM stock solution):
TasterSNP_fwd (pos 669 – 688): 5’-CATGAGGACAATGAAGGTCT-3’
TasterSNP_rev (pos 955 – 974): 5’-TGGTCGGCTCTTACCTTCAG-3’
• GoTaq G2 DNA Polymerase, Promega, Cat.# M7845
• PCR Buffer (green color due to loading dye; contains Mg2+that is needed as cofactor for Taq DNA polymerase)
• dNTP mix (10 mM of each deoxyribonucleotide: dATP, dCTP, dTTP, dGTP)
The PCR reaction is done in a final volume of 25 µl. Keep the tubes always on ice!
1. Assistant:
• Spin down all tubes with the reagents for a few seconds before use ‼!
• Set up a mastermix for the PCR reaction in an Eppendorf tube containing all components except the DNA
template: It is important to add the components in the same order as indicated below and to use new tips for every
component. Carefully/slowly soak up and release the PCR buffer (the solution is viscous (contains glycerol) and
therefore less easy to handle than a completely watery solution)
• If you have less than 8 students → calculate the correct amount for the mastermix!
Mastermix for PCR reaction (keep on ice):
Reagent
Amount for 1 or 10,
1x
10x
ddH2O
5x PCR Buffer
dNTP mix (10mM)
Primer_fwd (10µM)
Primer_rev (10µM)
Taq polymerase
16.65 µl
5.00 µl
0.25 µl
0.50 µl
0.50 µl
0.10 µl
166.5 µl
50.0 µl
2.5 µl
5.0 µl
5.0 µl
1.0 µl
Total volume
23.00 µl
230.00 µl
or different number of tubes:
........ students + 1 (assistant) + 1 “reserve” = ........ x
Assistant: mix well mastermix (N Do not vortex!) and spin down for a few seconds.
2.
Assistant: label a stripe of 8 PCR tubes (with numbers 1 – 8 and the assistants name) + 1
additional PCR tube for yourself
3.
Assistant: add 23µl of mastermix into each PCR tube
4.
Each student: pipet 2µl of the clear supernatant from the top of your isolated cheek cell
DNA into one tube of a strip of 8 tubes: Na) do not disturb the pellet of cell debris and Chelex
beads at the bottom of the tube; b) insure that no DNA remains in the tip after pipetting!
5.
Student: please write down your place number in the sample sheet in order to know
which (of the 8) reaction is yours!
6.
Assistant: hand in the labelled PCR tubes for PCR reaction. Mark them with your name.
The tubes are then placed in a PCR cycler, using the following profile:
95°C for 5 min
95°C for 20 sec
58°C for 20 sec
30 cycles
72°C for 20 sec
72°C for 20 sec
4°C until tubes are removed from PCR machine
11
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
A) Human genetic diversity → Experiment 1: “Bitter” taste perception_the PTC gene
1.4.3. Setting up sequencing reaction
Each student (and each assistant) will set up 1 sequencing reaction using his/her own PCR
amplified DNA of part of the TAS2R38 (= PTC) gene.
Reagents
• PCR amplified DNA of part of the PTC gene from part 2 (protocol 1.4.2.)
• DNA primer (10 µM stock solution) for sequencing:
TasterSNP_fwd (pos 669 – 688): 5’-CATGAGGACAATGAAGGTCT-3’
The sequencing reaction is done in a final volume of 10 µl. Keep the tubes always on ice!
1.
Assistant: set up a mastermix for the sequencing reaction in an Eppendorf tube
containing all components except the PCR template. Use always new tips for every
component.
If you have less than 8 students → calculate the correct amount for the mastermix!
Mastermix for sequencing (keep on ice):
Reagent
Amount for 1 or 10,
1x
10x
ddH2O
Primer_fwd (10µM)
6 µl
1 µl
60 µl
10 µl
Total volume
7 µl
70 µl
..........
or different number of tubes:
students + 1 (assistant) + 1 “reserve” = ........... x
Assistant: • mix mastermix by tapping the tube
• spin down mastermix for a few seconds
2.
Assistant: dispense 7µl of sequencing mastermix into as many PCR tubes as there are
students in your group (+ one tube for yourself)
3.
Each student: pipet/add 3µl of your PCR product into one of the tubes containing
sequencing mastermix and close the lid. NInsure that no DNA remains in the tip after
pipetting!
4.
Student: label your tube with your place number, give it to your assistant.
→ You will get the sequenced part of your PTC gene back in the next course.
5.
Assistant: hand in the labelled PCR tubes for sequencing.
→ Collect all tubes from your group of students and put them into the zip lock bag that
you got handed out.
→ Write your name, date and all tube numbers on the bag and bring it to the table in
the front.
12
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
A) Human genetic diversity → Experiment 1: “Bitter” taste perception_the PTC gene
1.5. Analysis of PTC genotype and phenotype
1.5.1. Determining your PTC-genotype
Each student will get his/her own sequence from the amplified part of the TAS2R38 gene.
Example of part of such a sequenced PCR product:
Nucleotide color code: A (green), T (red), G (black), C (blue); N = base cannot be determined
first part of sequence is
always badly readable
--> many “unclear” (N) bases
a)
“double peak”: red and blue peaks have a similar height
--> at this (SNP-2) position both bases T (red) and C (blue) are present in this “example” PCR product
In the „wildtype“ taster sequence (Fig. 3) on page 7, ...
1. Mark the positions of the three SNPs (see also Figures 4 + 5 on page 5)
2. At each SNP, write the „mutant“ nucleotide triplet leading to the altered amino
acid above the „wildtype“ taster sequence.
3. Indicate the positions of the two primers that were used for amplification of part
of the TAS2R38 gene. (For primer sequence see protocol 1.4.2. and 1.4.3.)
4. Indicate which of the two primers was then later used for sequencing.
b) Compare your sequence with the one from the „wildtype“ taster allele from Fig. 3.
1. What bases (A, T, G, C) do you have at the polymorphic positions SNP-2 and
SNP-3 ?
Bases at SNP-2: .........................................................
Bases at SNP-3: .........................................................
2. Are you homozygous or heterozygous at SNP-2 and SNP-3?
3. Determine your PTC genotype (PAV/PAV; PAV/AVI; AVI/AVI; other...?)
--> My genotype = ........................................................................................................................................
c)
According to your genotype, what prediction can you make for your PTC phenotype?
£ PAV/PAV = homozygote taster (T/T)
£ PAV/AVI = heterozygote taster (T/t)
£ AVI/AVI = homozygote non-taster (t/t)
£ other ........ = ?
13
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
A) Human genetic diversity → Experiment 1: “Bitter” taste perception_the PTC gene
1.5.2. Testing your PTC-phenotype
Determine your „real“ PTC phenotype
1.
Place a strip of PTC taste paper in the center of your tongue for several seconds.
How would you describe the taste of the PTC paper: strongly bitter, weakly bitter,
or no taste other than paper-like?
2.
Correlate your PTC genotype with your phenotype. Hand in your data by
answering the “Klicker-survey” à we will record class results
Genotype
strongly bitter
--> Supertaster
Phenotype
bitter
--> Taster
not bitter
--> Non-Taster
T/T = (P)AV/(P)AV
T/t = (P)AV/(A)VI
t/t
= (A)VI/(A)VI
Other ?
1.6. Questions
à course-1: provide answers to questions 1-3 / course-2: provide answer to questions 4, 5
1.
Indicate on page 11 what happens during the various steps of the PCR reaction.
2.
Why does the annealing temperature of a PCR reaction depend on the length and
sequence of the primers?
3.
DNA replication and PCR are both DNA amplification events. What is the difference
between the two?
4.
Give an explanation how the other rare PTC-chromosome variants that exist in the
human population (AAV, AAI, PVI, PAI, AVV, PVV) could have arisen.
5.
Why are there some students in which genotype and phenotype does not match?
a) genotypically predicted homozygous AVI (non-taster) that do smell PTC.
b) genotypically predicted homozygous or heterozygous PAV (taster) that do not smell PTC.
14
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 2. Isolation of pB30 plasmid DNA
B) Recombinant DNA Technology
Recombinant DNA technology is a set of molecular techniques for locating, isolating,
altering, and studying DNA segments. The term “recombinant” is used because, frequently,
the goal is to combine DNA from two distinct sources: e.g. a gene from Drosophila, C.
elegans, mouse or human might be inserted into a bacterial or viral vector.
A first step in the molecular analysis of a DNA segment or gene is to isolate it from the
genome and to make many copies of it so that further analyses can be carried out.
One way to amplify a specific piece of DNA is to place the fragment in a bacterial cell and
allow the cell to replicate the DNA. This procedure is termed gene cloning because identical
copies (clones) of the original piece of DNA are produced.
Overview of gene cloning:
a) Isolation of DNA fragment from any organsim
b) Insert DNA fragment into DNA vector (e.g. plasmid) → recombinant DNA molecule
c) Introduce recombinant DNA molecule into bacteria (= transformation)
d) Detection of bacteria that contain a recombinant DNA molecule
e) Further growth of bacteria with recombinant plasmid → amplification of rec. plasmid
f) Isolation of recombinant DNA plasmids from bacteria → Experiment 2
g) Test of recombiant DNA plasmid, e.g.: - inserted fragment is DNA segment of interest?
→ Experiment 3
- orientation of inserted fragment in plasmid?
- etc.
see also: • Griffiths 10ed, international; chapter 11.2, Fig. 11-7, -8, -9
• Griffiths 11ed; chapter 10.2, Fig. 10-7, -8, -9
• lectures L21 and L22 (Prof. Hengartner)
2. Isolation of pB30 plasmid DNA
2.1. Introduction
Many purposes in molecular biology (such as cloning) require the amplification of plasmid
DNA. The first step in amplifying DNA is to introduce the plasmid into bacteria (e.g.
Escherichia coli). Amplification of plasmid DNA is achieved by growing bacterial cells
containing the plasmid in liquid cultures to obtain large numbers of cells. From these cells,
the plasmid DNA can be separated from the bacterial chromosomal DNA and selectively
isolated.
In this experiment, E. coli cells have been transformed with the plasmid pBE30 that contains
the C. elegans ced-9 cDNA, and the Ampicillin resistance gene.
2.1.1. General Principle of Plasmid DNA Isolation
The first step is to culture the E.coli cells and then separate the cells from the culture. This is
done by centrifugation. The cells are then re-suspended in a re-suspension buffer (= buffer
P1) to enable further processing. The re-suspension buffer has a more or less neutral pH (pH
= 8.0). It contains EDTA (Ethylenediaminetetraacetic acid), which chelates divalent metal
ions like Mg2+ and Ca2+, thereby rendering DNases inactive. DNA cannot be degraded and
remains intact.
15
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 2. Isolation of pB30 plasmid DNA
The next step is the disruption of the plasma membrane to release the DNA out of the cells.
This is achieved by using the lysis buffer (= buffer P2) that contains a detergent like SDS
(sodium dodecyl sulphate) that disrupts the plasma membrane. The lysis buffer also contains
sodium hydroxide (NaOH), which makes the lysis buffer strongly basic with a pH of 12.0 –
13.0. Because this is a huge contrast to the normal physiological pH, proteins and DNA
denature, cannot be solute any more and precipitate. In denatured DNA (both chromosomal
and plasmid DNA), the two strands get separated from each other (in plasmid DNA the two
separated single stranded DNA molecules remain interlocked).
In the next step the plasmid DNA has to be re-natured and separated from the bacterial
chromosomal DNA. The lysate is neutralized by addition of acidic potassium acetate (buffer
P3). The high salt concentration leads to co-precipitation of denatured proteins,
chromosomal DNA and cellular debris. In this step, the pH is lowered again reaching
physiological conditions. The plasmids are able to renature first, because they are small and
the two DNA strands find each other with a much higher probability than the ones of the big
chromosome. If the renaturising step is kept short, only the plasmids are able to build double
helices and solute again. The chromosomal DNA and the proteins stay precipitated. They
can be separated from the plasmids by centrifugation. (The white fluffy material on the
bottom of the tube after centrifugation are the precipitated proteins, DNA and other cell
components.) The plasmid DNA is dissoved in the clear supernatant.
Finally, the plasmid DNA has to be taken out of this salty solution. This can be done in two
ways:
a) the classical method: adding isopropanol (an anti-solvent) to the solution. DNA is less
soluble in isopropanol than in water and therefore will precipitate after centrifugation. The
pelleted plasmid DNA is washed with ethanol and then dissolved in TE buffer. The EDTA
in the buffer prevents plasmid DNA from degradation and RNase could be added to remove
any RNA contamination.
b) using spin column: Plasmid DNA is first bound to silica-based membranes on the spin
column, washed with high salt buffers containing ethanol and finally released (eluted) from
the membrane with low salt, neutral pH buffers, like TE buffer.
2.2. Goal of the experiment
In this experiment we will isolate DNA from a bacterial liquid culture. The bacteria contain
the plasmid pB30 with the inserted C. elegans ced-9 cDNA. This isolated plasmid will then be
used further in experiment 3 to establish a restriction map and to draw conclusions about the
inserted DNA fragment.
16
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 2. Isolation of pB30 plasmid DNA
2.3. Protocol: Plasmid Miniprep Kit
Plasmid DNA is isolated from E.coli bacteria by a modified alkaline lysis method, according
to the protocol provided by ZYMO RESEARCH (ZR Plasmid MiniprepTM – Classic).
Each student will isolate plasmid DNA from 1ml of a freshly grown bacterial culture.
Reagents
• Bacterial culture of E. coli. Bacteria carry the pB30 plasmid with inserted C. elegans ced-9 cDNA.
Aliquots of 8-10ml are prepared.
• Buffer P1 (red) = Re-suspension buffer (stored at RT)
’neutral pH
trade secret; contains glucose, Tris-HCl, pH 8.0, EDTA (ethylenediaminetetraacetic acid, chelating
agent)
• Buffer P2 (blue) = Lysis buffer (stored at room temperature)
’basic (alkaline) solution (high pH)
trade secret; contains NaOH (alkali: denatures DNA and proteins), SDS (sodium dodecyl sulphate, a
detergent: solubilizes the phospholipids and protein components of the cell membrane, leading to
lysis of the cell membrane), pH 12.0 – 13.0
• Buffer P3 (yellow) = Neutralizing buffer (stored at 4°C)
’acidic solution (low pH)
trade secret; contains potassium chloride and guanidinium chloride (= chaotropic reagent)
(Converts soluble SDS into PDS, decreases alkalinity of mixture and allows renaturation of DNA)
• Wash buffer-1 (stored at RT)
trade secret; contains guanidinium chloride, isopropanol
’high salt, alcohol
• Wash buffer-2 (stored at RT)
trade secret; contains Tris-HCl and ethanol
’high salt
• Elution buffer = TE buffer (stored at RT)
10mM Tris-HCl, pH 8.5, 0.1mM EDTA
’neutral pH, low salt
• During all centrifugation steps: a) Balance tubes and b) put hinge of the tube towards
top/outside of the centrifuge, so that you know where the pellet will be !!
• Keep buffer P3 (yellow colour) always on ice!
STEP-1: Separation of plasmid DNA from other cell components and genomic DNA
1. Transfer 1ml of a freshly grown bacterial culture to an Eppendorf tube. Label the tube
with your seat number.
2. Spin down the bacterial cells by centrifugation at max. speed for 1 min.
3. Decant the supernatant completely, shake out remaining drops. NDo NOT disturb pellet!
4. Add 200µl of red P1 buffer (= re-suspension buffer). Disperse the pellet completely in the
liquid by vortexing. Make sure there aren’t any little lumps visible.
Ê This step leads to resuspension of the cells so that they are evenly distributed in the buffer.
IMPORTANT for steps 5 and 6: the given (short) reaction times shouldn’t be exceeded
→ one person (student or assistant) should pipet the two steps for all the group samples:
5. Add 200µl of blue P2 buffer (= lysis buffer) and mix gently by inverting the tube 6-8
times. NDo NOT vortex!
Let the tube stand at room temperature for 1min. NDo not exceed 1 min!
Ê This step lyses the cells: the plasma membrane will get disrupted and the contents of the bacterial cells leak out
giving the solution a viscous (purple) and less opaque appearance. The alkali denatures DNA and proteins.
6. Add 400µl of yellow P3 buffer (= neutralizing buffer; on ice) and mix by gently inverting
the tube until the solution turns entirely yellow, about 6-8 times. NDo NOT vortex!
The solution will get cloudy. Let the tube stand for 1 min. NDo not exceed 1 min!
Ê Chromosomal genomic DNA, denatured proteins, lipids and cell debris will stick together.
Small plasmid DNA gets renatured (→ selective part of the procedure)
17
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 2. Isolation of pB30 plasmid DNA
The following steps can be done again individually by the group members:
7. Centrifuge for 2 min at maximum speed.
Ê Chromosomal genomic DNA, denatured proteins, lipids and cell debris will form a fluffy white
precipitate. The plasmid DNA is in the yellow supernatant, that you are going to clean up in
step-2.
STEP-2: Cleaning and concentration of plasmid DNA
8. While the centrifuge is running, insert a spin column in a collection tube without
touching the outlet of the spin column. Label the spin column with your seat number.
9. Pipet the yellow, clear supernatant (approx. 800-900µl) to the spin column. NMake sure
not to disturb or transfer any of the precipitated cell debris and proteins that form a fluffy
pellet at the bottom of the tube!
10. Centrifuge the column-collection tube assembly for 30 sec at full speed.
11. Disassemble column and collection tube and discard the flow-through in the collection
tube. NMake sure the flow-through does not touch the bottom of the column.
Return the spin column to the collection tube.
Ê The plasmid DNA is bound to the white silica-based membrane on the spin column.
12. Add 200µl of Wash Buffer-1 to the spin column and centrifuge for 30sec at full speed.
Ê The plasmid DNA bound to the to silica-based membrane is washed with high salt buffers
containing alcohol. Plasmid DNA is not soluble in high salt-alcohol solutions and stays
bound to the membrane. Proteins, cell debris, etc. are washed away from the membrane.
13. Add 400µl of Wash Buffer-2 to the spin column and centrifuge for 1min at full speed.
14. Transfer the spin column into a clean Eppendorf tube, labelled with your seat number.
15. Add 30µl of Elution Buffer directly in the middle of the membrane at the bottom of the
column. NDo NOT touch the membrane with the tip of your pipet!
It is important to pipet the buffer in the center of the membrane in order to elute enough
DNA!
Ê This step will release the bound plasmid DNA from the membrane by adding low salt (watery),
neutral pH elution buffer.
16. Centrifuge for 30 sec at full speed.
17. Discard spin column, close Eppendorf tube.
18. Store the Eppendorf tube with the plasmid DNA at room temperature.
® Use 2µl to determine the concentration and purity of the isolated plasmid DNA
(see below: 2.4. Quantification of isolated DNA)
® Afterwards, hand in your properly labeled tube (seat number) to your assistant.
You will get it back in the next course (course-5).
18
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 2. Isolation of pB30 plasmid DNA
2.4. Quantification of the isolated DNA
We will determine the concentration and purity of the isolated plasmid DNA by ultraviolet
absorbance spectrophotometry using the NanoDrop spectrophotometer. The NanoDrop is a
cuvette free spectrophotometer, which requires a sample size of only 1 µl, and calculates
automatically the concentration of the sample (see example picture below).
2.4.1. DNA concentration
DNA absorbs UV light with a maximum absorbance of 260nm. All 4 bases contribute to the
absorbance.
An absorbance(A260) of 1.0 corresponds to 50µg of double stranded DNA/ml.
® A260, 1 unit ≈ 50µg/ml (equal to 50ng/µl)
x 50 ng/µl
ratios 260/280 (and 260/230) to
determine the purity of the sample
= DNA concentration in sample
2.4.2. Purity of DNA
In most DNA preparations, one step is the separation of DNA from protein. Carryover
protein during a DNA prep could lead to problems with subsequent operations, such as
cutting with a restriction endonuclease. The most commonly used assay is the A260/A280 ratio.
Ratio 260/280
The ratio of absorbance at 260nm and 280nm is used to assess the purity of DNA and RNA.
As a "rule of thumb": A ratio of ~1.8 is generally accepted as “pure” for DNA; a ratio of ~2.0
is generally accepted as “pure” for RNA (RNA will typically have a higher 260/280 ratio due
to the higher ratio of Uracil compared to that of Thymine). If the ratio is appreciably lower in
either case, it may indicate the presence of protein, phenol or other contaminants that absorb
strongly at or near 280 nm.
A260/A280 = 1.6 – 2.0
® pure DNA sample
A260/A280 < 1.6
® residual proteins
and other contaminants
A260/A280 > 2.0
® residual RNA,
degraded DNA
19
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 2. Isolation of pB30 plasmid DNA
2.4.3. Concentration and purity of “your” DNA sample
Note the concentration and purity of the plasmid DNA you have isolated, using the
NanoDrop spectrophotometer:
Concentration of plasmid DNA in your sample: ………………………………………………………….…………..
Purity of your plasmid DNA sample: A260/A280 ratio = …………………………………………………….……
2.5. Questions
1.
Why does the re-suspension buffer (buffer P1) contain glucose and EDTA?
2.
Why does the lysis buffer (buffer P2) contain sodium hydroxide (NaOH) and the
detergent sodium dodecyl (lauryl) sulfate (SDS)?
3.
How does adding neutralizing buffer (buffer P3) help to separate the plasmid DNA
from the bacterial genomic DNA?
20
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 2. Isolation of pB30 plasmid DNA
4.
The elution buffer contains low salt Tris-buffer (≈pH 8) and EDTA. Why is the elution
buffer made of this two components?
5.
In some protocols, RNase is added after elution of the plasmid DNA. Why? What
advantage could the addition of RNase have?
21
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 3. Restriction analysis of plasmid pB30
3. Restriction analysis of plasmid pB30
3.1. Introduction
3.1.1. Plasmid vectors
In medical terminology, a vector is an organism that carries a pathogen from one host
organism to another. In molecular biology, a vector is a DNA molecule that is used as a
vehicle to carry foreign DNA sequences into a host cell. Plasmids are the simplest bacterial
vectors. Ranging in length from 1’000 to 200’000 bp, they are circular DNA molecules that
exist separately from the main bacterial chromosome.
For a plasmid to be propagated through successive bacterial generations, the plasmid vector
must contain a specific DNA sequence, called the origin of replication (ori), which allows it to
be replicated within the host cell. E. coli DNA polymerase and other proteins required to
initiate DNA synthesis bind to the ori.
Plasmids can be divided into two broad groups, according to how tightly their replication is
regulated. Plasmids under ‘stringent control’ replicate once per cell division, along with the
bacterial chromosome. ‘Relaxed’ plasmids replicate their DNA autonomously throughout
the cell cycle of the bacteria and accumulate in up to hundreds of copies per cell. Relaxed
plasmids are therefore useful for amplification of large amounts of cloned DNA. The
plasmid pBluescript II SK (+) is an example of this latter class of plasmids.
The plasmids routinely used as vectors carry a gene for drug resistance and a gene to
distinguish plasmids with and without DNA inserts:
a) These drug-resistance genes (e. g. antibiotic resistance sequence to ampicillin) provide a
convenient way to select for bacterial cells transformed by plasmids: those cells still alive
after exposure to the drug must carry the plasmid vectors.
b) However, not all the plasmids in these transformed cells will contain DNA inserts. For this
reason, it is desirable to be able to identify bacterial colonies with plasmids containing DNA
inserts. DNA inserts disrupt a gene (lacZ) in the plasmid that encodes an enzyme (βgalactosidase) necessary to cleave a compound added to the agar (X-gal) so that it produces a
blue pigment. Thus, the colonies that contain the plasmids with the DNA insert will be white
rather than blue, since they cannot cleave X-gal because they do not produce β-galactosidase.
3.1.2. Restriction endonucleases
Restriction endonucleases are possibly the most powerful tools in biotechnology. They allow
scientists to precisely cut DNA in a predictable and reproducible manner. They are
components of the restriction-modification systems that bacteria developed to protect
themselves from viral infections. From the three classes of restriction endonucleases, type II
enzymes are ideal as tools for manipulating DNA for several reasons: (i) they contain only
restriction activities, but no modification activity; (ii) they cut in a predictable and consistent
manner, either within or right next to their recognition sequence; (iii) they require only Mg++
as a cofactor, ATP is not needed. The names for restriction endonucleases reflect their origin:
EcoR I:
E
= genus (Escherichia)
co
= species (coli)
R
= strain RY13
I
= first endonuclease identified in E. coli
22
23
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 3. Restriction analysis of plasmid pB30
Type II restriction endonucleases recognize sequences that are generally four to eight
nucleotides in length, and are usually palindromic - that is, they have the same sequence
(read 5' > 3') on both (complementary) strands.
EcoR I acts as a homodimer: Two EcoR I molecules align at the recognition site in opposite
orientation on the DNA and work by accessing the bases on the inside of the DNA molecule.
Specific amino acids within the EcoR I enzyme form hydrogen bonds with the DNA
recognition sequence, GAATTC. Once bound, other residues within the enzyme catalyze a
hydrolysis reaction that uses water to break the phosphodiester linkage on each strand of the
DNA helix within the recognition sequence. The DNA molecule is cut in two, with a
phosphate group at the 5’ end and a hydroxyl group at the 3’ end. By working as a dimer,
the enzyme is able to cut the DNA simultaneously on both strands.
The Figure (Fig. 1) below shows the graphical map of the plasmid pB30. Into this plasmid the
C. elegans ced-9 cDNA containing 5’ and 3’ upstream sequences (UTRs) has been integrated.
Ced-9 is a gene that prevents programmed cell death in the nematode C. elegans [1, 2, 3, 7].
The ced-9 cDNA has been inserted into the plasmid as an EcoRI fragment: The plasmid as
well as the ced-9 fragment have been cut by the EcoRI restriction enzyme and then the ends
have been ligated. The pB30 plasmid with the integrated ced-9 insert at the EcoRI site has a
length of 4254bp.
The DNA sequence of the plasmid is depicted in Figure 3 (next page).
Fig. 1
The pB30 plasmid contains a multiple cloning site (MCS; containing many restriction sites
but only the one for EcoRI is shown), an antibiotic resistance sequence to ampicillin (AmpR)
an E.coli origin of replication (ori) and a lacZ gene.
The multiple cloning site sequence (MCS) is located within the LacZ gene, that encodes the
β-galactosidase enzyme necessary to cleave the compound X-gal. If no insert is present in the
MCS, the enzyme is produced and cleaves X-gal, converting it into a blue dye (--> bacterial
colonies on the agar plate turn blue). If the LacZ gene is disrupted by successful insertion of
a DNA sequence, X-gal can no longer be cleaved and the bacteria exhibit a white color (Fig.
2).
Fig. 2, blue-white selection for bacteria with insert
24
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 3. Restriction analysis of plasmid pB30
Figure 3: DNA sequence of the pB30 plasmid with inserted ced-9 cDNA
1 CACCTGACGC
GTGGACTGCG
61 TGACCGCTAC
ACTGGCGATG
121 TCGCCACGTT
AGCGGTGCAA
181 GATTTAGTGC
CTAAATCACG
241 GTGGGCCATC
CACCCGGTAG
301 ATAGTGGACT
TATCACCTGA
361 ATTTATAAGG
TAAATATTCC
421 AATTTAACGC
TTAAATTGCG
481 GCGCAACTGT
CGCGTTGACA
541 AGGGGGATGT
TCCCCCTACA
601 TTGTAAAACG
AACATTTTGC
661 GCCCCCCCTC
CGGGGGGGAG
721 ACGCTGCACG
TGCGACGTGC
781 TGGCGAGATG
ACCGCTCTAC
841 TAGTGATGCT
ATCACTACGA
901 CGGAGAGTCA
GCCTCTCAGT
961 ATTTGTGGTC
TAAACACCAG
1021 ACCGGGATTG
TGGCCCTAAC
1081 ATTCGAGAAG
TAAGCTCTTC
1141 CAGAATCTCA
GTCTTAGAGT
1201 TCAATGTCCA
AGTTACAGGT
1261 TGCAAAAATG
ACGTTTTTAC
1321 ATCGCTGTTC
TAGCGACAAG
1381 CGACTTCATG
GCTGAAGTAC
1441 AGTGGGACGC
TCACCCTGCG
1501 AGCCATTGGA
TCGGTAACCT
GCCCTGTAGC
CGGGACATCG
ACTTGCCAGC
TGAACGGTCG
CGCCGGCTTT
GCGGCCGAAA
TTTACGGCAC
AAATGCCGTG
GCCCTGATAG
CGGGACTATC
CTTGTTCCAA
GAACAAGGTT
GATTTTGCCG
CTAAAACGGC
GAATTTTAAC
CTTAAAATTG
TGGGAAGGGC
ACCCTTCCCG
GCTGCAAGGC
CGACGTTCCG
ACGGCCAGTG
TGCCGGTCAC
GAGGTCGACG
CTCCAGCTGC
GCGGACAACT
CGCCTGTTGA
AAGGAGTTTC
TTCCTCAAAG
CAGGACTTGC
GTCCTGAACG
ATTGATGGAA
TAACTACCTT
GACTATTTCA
CTGATAAAGT
CCGTGTGGAG
GGCACACCTC
AAGCACGCGG
TTCGTGCGCC
TTTTCACTGT
AAAAGTGACA
ATGTCTTATG
TACAGAATAC
ATGGAATCCG
TACCTTAGGC
ATCAAAACGC
TAGTTTTGCG
ACACTCGGAA
TGTGAGCCTT
CGGAAGCAGA
GCCTTCGTCT
ATCGTTGGAG
TAGCAACCTC
GGCGCATTAA
CCGCGTAATT
GCCCTAGCGC
CGGGATCGCG
CCCCGTCAAG
GGGGCAGTTC
CTCGACCCCA
GAGCTGGGGT
ACGGTTTTTC
TGCCAAAAAG
ACTGGAACAA
TGACCTTGTT
ATTTCGGCCT
TAAAGCCGGA
AAAATATTAA
TTTTATAATT
GATCGGTGCG
CTAGCCACGC
GATTAAGTTG
CTAATTCAAC
AATTGTAATA
TTAACATTAT
GTATCGATAA
CATAGCTATT
CGCTGACGAA
GCGACTGCTT
TGGGGATAAA
ACCCCTATTT
CATCACCGAG
GTAGTGGCTC
AAATCAATGA
TTTAGTTACT
CGCACCGAAT
GCGTGGCTTA
TGCAACCGGA
ACGTTGGCCT
AAAATTTTGA
TTTTAAAACT
ATCAGGATGT
TAGTCCTACA
GACGTTTGAT
CTGCAAACTA
TGGAACTGCA
ACCTTGACGT
GGATCCGCAA
CCTAGGCGTT
AACAAATGAA
TTGTTTACTT
ACAGACGGTG
TGTCTGCCAC
TCGTCGTGTG
AGCAGCACAC
GCGCGGCGGG
CGCGCCGCCC
CCGCTCCTTT
GGCGAGGAAA
CTCTAAATCG
GAGATTTAGC
AAAAACTTGA
TTTTTGAACT
GCCCTTTGAC
CGGGAAACTG
CACTCAACCC
GTGAGTTGGG
ATTGGTTAAA
TAACCAATTT
CGCTTACAAT
GCGAATGTTA
GGCCTCTTCG
CCGGAGAAGC
GGTAACGCCA
CCATTGCGGT
CGACTCACTA
GCTGAGTGAT
GCTTGATATC
CGAACTATAG
TCCGGCGTAT
AGGCCGCATA
AGGCACAGAG
TCCGTGTCTC
TAGGCAGGCT
ATCCGTCCGA
TTGGGAAGAG
AACCCTTCTC
CCGGCAAAAC
GGCCGTTTTG
GCACGAAATG
CGTGCTTTAC
GACCTTCTGT
CTGGAAGACA
GGTTCGGACG
CCAAGCCTGC
AGGTCTAATC
TCCAGATTAG
GGGACAAGTG
CCCTGTTCAC
CAACTGGAAG
GTTGACCTTC
AGAGGACTAC
TCTCCTGATG
GTCGATGATT
CAGCTACTAA
TGGGCGGATG
ACCCGCCTAC
TGTGGTGGTT
ACACCACCAA
CGCTTTCTTC
GCGAAAGAAG
GGGGCTCCCT
CCCCGAGGGA
TTAGGGTGAT
AATCCCACTA
GTTGGAGTCC
CAACCTCAGG
TATCTCGGTC
ATAGAGCCAG
AAATGAGCTG
TTTACTCGAC
TTCCATTCGC
AAGGTAAGCG
CTATTACGCC
GATAATGCGG
GGGTTTTCCC
CCCAAAAGGG
TAGGGCGAAT
ATCCCGCTTA
GAATTCCGGT
CTTAAGGCCA
CGGCGACGAA
GCCGCTGCTT
CCCACCGATT
GGGTGGCTAA
TCGACGCGAA
AGCTGCGCTT
CCAAGGCTTG
GGTTCCGAAC
GGAATGGAAT
CCTTACCTTA
ATGCGAGTTA
TACGCTCAAT
GAGCAGCTGC
CTCGTCGACG
GTTGGAAATG
CAACCTTTAC
TCGTTCGGCG
AGCAAGCCGC
CGAAACCTCT
GCTTTGGAGA
GAACACAATC
CTTGTGTTAG
GAACGAGCAG
CTTGCTCGTC
GGCGCTGGAG
CCGCGACCTC
ATGTTCAGCT
TACAAGTCGA
ACGCGCAGCG
TGCGCGTCGC
CCTTCCTTTC
GGAAGGAAAG
TTAGGGTTCC
AATCCCAAGG
GGTTCACGTA
CCAAGTGCAT
ACGTTCTTTA
TGCAAGAAAT
TATTCTTTTG
ATAAGAAAAC
ATTTAACAAA
TAAATTGTTT
CATTCAGGCT
GTAAGTCCGA
AGCTGGCGAA
TCGACCGCTT
AGTCACGACG
TCAGTGCTGC
TGGGTACCGG
ACCCATGGCC
TTGAGATGAC
AACTCTACTG
CGATGGCGAC
GCTACCGCTG
TTGGAATCAA
AACCTTAGTT
GAATGTCCAT
CTTACAGGTA
ATATCGAGGG
TATAGCTCCC
GGTTTGGAGC
CCAAACCTCG
TGGGAACGAT
ACCCTTGCTA
TCGCAGTGCC
AGCGTCACGG
CACAGACAGA
GTGTCTGTCT
GTTTCGTAGC
CAAAGCATCG
TCGTTTACAC
AGCAAATGTG
GGAGCTGGGA
CCTCGACCCT
AAGCTGAAAA
TTCGACTTTT
TAACAGCTGG
ATTGTCGACC
TGAAGTAACG
ACTTCATTGC
Ori
PvuI
lacZ
EcoRI
ced9
25
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 3. Restriction analysis of plasmid pB30
1561 TATTCAATTT
ATAAGTTAAA
1621 GCTCACTGAT
CGAGTGACTA
1681 GGCTCTCTGT
CCGAGAGACA
1741 CAAACCCTAC
GTTTGGGATG
1801 TGAAACCCTA
ACTTTGGGAT
1861 TACCCCTGTA
ATGGGGACAT
1921 TACGCGTACG
ATGCGCATGC
1981 AAAACGGTTC
TTTTGCCAAG
2041 GAGCGGCCGC
CTCGCCGGCG
2101 AGCTTGGCGT
TCGAACCGCA
2161 CCACACAACA
GGTGTGTTGT
2221 TAACTCACAT
ATTGAGTGTA
2281 CAGCTGCATT
GTCGACGTAA
2341 GCGAGTGACT
CGCTCACTGA
2401 AGGCGGTAAT
TCCGCCATTA
2461 AAGGCCAGCA
TTCCGGTCGT
2521 TCCGCCCCCC
AGGCGGGGGG
2581 CAGGACTATA
GTCCTGATAT
2641 CGACCCTGCC
GCTGGGACGG
2721 CTCATAGCTC
GAGTATCGAG
2841 GTGTGCACGA
CACACGTGCT
2801 AGTCCAACCC
TCAGGTTGGG
2861 GCAGAGCGAG
CGTCTCGCTC
2941 ACACTAGAAG
TGTGATCTTC
3001 GAGTTGGTAG
CTCAACCATC
3061 GCAAGCAGCA
CGTTCGTCGT
GTGTAAATAA
CACATTTATT
TCTCTCATCC
AGAGAGTAGG
GTCGATTTAC
CAGCTAAATG
TTCCGCGTAA
AAGGCGCATT
ACTTTTCTCG
TGAAAAGAGC
TCTCAATAAT
AGAGTTATTA
CGACTTTGTA
GCTGAAACAT
AAAAAAAAAA
TTTTTTTTTT
CACCGCGGTG
GTGGCGCCAC
AATCATGGTC
TTAGTACCAG
TACGAGCCGG
ATGCTCGGCC
TAATTGCGTT
ATTAACGCAA
AATGAATCGG
TTACTTAGCC
GAGCGACGCG
CTCGCTGCGC
ACGGTTATCC
TGCCAATAGG
AAAGGCCAGG
TTTCCGGTCC
TGACGAGCAT
ACTGCTCGTA
AAGATACCAG
TTCTATGGTC
GCTTACCGGA
CGAATGGCCT
ACGCTGTAGG
TGCGACATCC
ACCCCCCGTT
TGGGGGGCAA
GGTAAGACAC
CCATTCTGTG
GTATGTAGGC
CATACATCCG
GACAGTATTT
CTGTCATAAA
CTCTTGATCC
GAGAACTAGG
GATTACGCGC
CTAATGCGCG
TTAATTTATG
AATTAAATAC
TTTGAACTGG
AAACTTGACC
GATTTTACTG
CTAAAATGAC
TATCAACTTT
ATAGTTGAAA
CCGTGGCCTA
GGCACCGGAT
TCATCTTCAC
AGTAGAAGTG
TTTATTTTTT
AAATAAAAAA
AAAACGGAAT
TTTTGCCTTA
GAGCTCCAGC
CTCGAGGTCG
ATAGCTGTTT
TATCGACAAA
AAGCATAAAG
TTCGTATTTC
GCGCTCACTG
CGCGAGTGAC
CCAACGCGCG
GGTTGCGCGC
AGCCAGCAAG
TCGGTCGTTC
ACAGAATCAG
TGTCTTAGTC
AACCGTAAAA
TTGGCATTTT
CACAAAAATC
GTGTTTTTAG
GCGTTTCCCC
CGCAAAGGGG
TACCTGTCCG
ATGGACAGGC
TATCTCAGTT
ATAGAGTCAA
CAGCCCGACC
GTCGGGCTGG
GACTTATCGC
CTGAATAGCG
GGTGCTACAG
CCACGATGTC
GGTATCTGCG
CCATAGACGC
GGCAAACAAA
CCGTTTGTTT
AGAAAAAAAG
TCTTTTTTTC
TACAACTCCT
ATGTTGAGGA
AAGAAGTGGG
TTCTTCACCC
CAATTTTTTC
GTTAAAAAAG
TCCGTGTTCT
AGGCACAAGA
GCCTCCCGCT
CGGAGGGCGA
TTTAACTGTC
AAATTGACAG
TCAAATTGTT
AGTTTAACAA
TCCTGCAGCC
AGGACGTCGG
TTTTGTTCCC
AAAACAAGGG
CCTGTGTGAA
GGACACACTT
TGTAAAGCCT
ACATTTCGGA
CCCGCTTTCC
GGGCGAAAGG
GGGAGAGGCG
CCCTCTCCGC
GGCTGCGGCG
CCGACGCCGC
GGGATAACGC
CCCTATTGCG
AGGCCGCGTT
TCCGGCGCAA
GACGCTCAAG
CTGCGAGTTC
CTGGAAGCTC
GACCTTCGAG
CCTTTCTCCC
GGAAAGAGGG
CGGTGTAGGT
GCCACATCCA
GCTGCGCCTT
CGACGCGGAA
CACTGGCAGC
GTGACCGTCG
AGTTCTTGAA
TCAAGAACTT
CTCTGCTGAA
GAGACGACTT
CCACCGCTGG
GGTGGCGACC
GATCTCAAGA
CTAGAGTTCT
TACATTTGAA
ATGTAAACTT
AAAGCTAGGC
TTTCGATCCG
CGATTGCCTT
GCTAACGGAA
GTACATTTCG
CATGTAAAGC
TCTCTTCCAC
AGAGAAGGTG
TCTTTTCGTG
AGAAAAGCAC
TTCTCTCTAC
AAGAGAGATG
CGGGGGATCC
GCCCCCTAGG
TTTAGTGAGG
AAATCACTCC
ATTGTTATCC
TAACAATAGG
GGGGTGCCTA
CCCCACGGAT
AGTCGGGAAA
TCAGCCCTTT
GTTTGCGTAT
CAAACGCATA
AGCGGTATCA
TCGCCATAGT
AGGAAAGAAC
TCCTTTCTTG
GCTGGCGTTT
CGACCGCAAA
TCAGAGGTGG
AGTCTCCACC
CCTCGTGCGC
GGAGCACGCG
TTCGGGAAGC
AAGCCCTTCG
CGTTCGCTCC
GCAAGCGAGG
ATCCGGTAAC
TAGGCCATTG
AGCCACTGGT
TCGGTGACCA
GTGGTGGCCT
CACCACCGGA
GCCAGTTACC
CGGTCAATGG
TAGCGGTGGT
ATCGCCACCA
AGATCCTTTG
TCTAGGAAAC
TCTCATTTTK
AGAGTAAAAM
CACAAATTAC
GTGTTTAATG
TTTTTTTGGC
AAAAAAACCG
TCAAAAACCC
AGTTTTTGGG
ATTTCCAAAG
TAAAGGTTTC
TGGCCTCTTC
ACCGGAGAAG
AACAACAAAA
TTGTTGTTTT
ACTAGTTCTA
TGATCAAGAT
GTTAATTTCG
CAATTAAAGC
GCTCACAATT
CGAGTGTTAA
ATGAGTGAGC
TACTCACTCG
CCTGTCGTGC
GGACAGCACG
TGGGCGCTCT
ACCCGCGAGA
GCTCACTCAA
CGAGTGAGTT
ATGTGAGCAA
TACACTCGTT
TTCCATAGGC
AAGGTATCCG
CGAAACCCGA
GCTTTGGGCT
TCTCCTGTTC
AGAGGACAAG
GTGGCGCTTT
CACCGCGAAA
AAGCTGGGCT
TTCGACCCGA
TATCGTCTTG
ATAGCAGAAC
AACAGGATTA
TTGTCCTAAT
AACTACGGCT
TTGATGCCGA
TTCGGAAAAA
AAGCCTTTTT
TTTTTTGTTT
AAAAAACAAA
ATCTTTTCTA
TAGAAAAGAT
ced9
MscI
EcoRI
lacZ
prom
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 3. Restriction analysis of plasmid pB30
3121 CGGGGTCTGA
GCCCCAGACT
3181 CAAAAAGGAT
GTTTTTCCTA
3241 GTATATATGA
CATATATACT
3301 CAGCGATCTG
GTCGCTAGAC
3361 CGATACGGGA
GCTATGCCCT
3421 CACCGGCTCC
GTGGCCGAGG
3501 GTCCTGCAAC
CAGGACGTTG
3561 GTAGTTCGCC
CATCAAGCGG
3621 CACGCTCGTC
GTGCGAGCAG
3681 CATGATCCCC
GTACTAGGGG
3741 GAAGTAAGTT
CTTCATTCAA
3801 CTGTCATGCC
GACAGTACGG
3861 GAGAATAGTG
CTCTTATCAC
3921 CGCCACATAG
GCGGTGTATC
3981 TCTCAAGGAT
AGAGTTCCTA
4041 GATCTTCAGC
CTAGAAGTCG
4101 ATGCCGCAAA
TACGGCGTTT
4161 TTCAATATTA
AAGTTATAAT
4201 GTATTTAGAA
CATAAATCTT
CGCTCAGTGG
GCGAGTCACC
CTTCACCTAG
GAAGTGGATC
GTAAACTTGG
CATTTGAACC
TCTATTTCGT
AGATAAAGCA
GGGCTTACCA
CCCGAATGGT
AGATTTATCA
TCTAAATAGT
TTTATCCGCC
AAATAGGCGG
AGTTAATAGT
TCAATTATCA
GTTTGGTATG
CAAACCATAC
CATGTTGTGC
GTACAACACG
GGCCGCAGTG
CCGGCGTCAC
ATCCGTAAGA
TAGGCATTCT
TATGCGGCGA
ATACGCCGCT
CAGAACTTTA
GTCTTGAAAT
CTTACCGCTG
GAATGGCGAC
ATCTTTTACT
TAGAAAATGA
AAAGGGAATA
TTTCCCTTAT
TTGAAGCATT
AACTTCGTAA
AAATAAACAA
TTTATTTGTT
AACGAAAACT
TTGCTTTTGA
ATCCTTTTAA
TAGGAAAATT
TCTGACAGTT
AGACTGTCAA
TCATCCATAG
AGTAGGTATC
TCTGGCCCCA
AGACCGGGGT
GCAATAAACC
CGTTATTTGG
TCCATCCAGT
AGGTAGGTCA
TTGCGCAACG
AACGCGTTGC
GCTTCATTCA
CGAAGTAAGT
AAAAAAGCGG
TTTTTTCGCC
TTATCACTCA
AATAGTGAGT
TGCTTTTCTG
ACGAAAAGAC
CCGAGTTGCT
GGCTCAACGA
AAAGTGCTCA
TTTCACGAGT
TTGAGATCCA
AACTCTAGGT
TTCACCAGCG
AAGTGGTCGC
AGGGCGACAC
TCCCGCTGTG
TATCAGGGTT
ATAGTCCCAA
ATAGGGGTTC
TATCCCCAAG
CACGTTAAGG
GTGCAATTCC
ATTAAAAATG
TAATTTTTAC
ACCAATGCTT
TGGTTACGAA
TTGCCTGACT
AACGGACTGA
GTGCTGCAAT
CACGACGTTA
AGCCAGCCGG
TCGGTCGGCC
CTATTAATTG
GATAATTAAC
TTGTTGCCAT
AACAACGGTA
GCTCCGGTTC
CGAGGCCAAG
TTAGCTCCTT
AATCGAGGAA
TGGTTATGGC
ACCAATACCG
TGACTGGTGA
ACTGACCACT
CTTGCCCGGC
GAACGGGCCG
TCATTGGAAA
AGTAACCTTT
GTTCGATGTA
CAAGCTACAT
TTTCTGGGTG
AAAGACCCAC
GGAAATGTTG
CCTTTACAAC
ATTGTCTCAT
TAACAGAGTA
CGCGCACATT
GCGCGTGTAA
GATTTTGGTC
CTAAAACCAG
AAGTTTTAAA
TTCAAAATTT
AATCAGTGAG
TTAGTCACTC
CCCCGTCGTG
GGGGCAGCAC
GATACCGCGA
CTATGGCGCT
AAGGGCCGAG
TTCCCGGCTC
TTGCCGGGAA
AACGGCCCTT
TGCTACAGGC
ACGATGTCCG
CCAACGATCA
GGTTGCTAGT
CGGTCCTCCG
GCCAGGAGGC
AGCACTGCAT
TCGTGACGTA
GTACTCAACC
CATGAGTTGG
GTCAATACGG
CAGTTATGCC
ACGTTCTTCG
TGCAAGAAGC
ACCCACTCGT
TGGGTGAGCA
AGCAAAAACA
TCGTTTTTGT
AATACTCATA
TTATGAGTAT
GAGCGGATAC
CTCGCCTATG
TCCCCGAAAA
AGGGGCTTTT
ATGAGATTAT
TACTCTAATA
TCAATCTAAA
AGTTAGATTT
GCACCTATCT
CGTGGATAGA
TAGATAACTA AmpR
ATCTATTGAT
GACCCACGCT
CTGGGTGCGA
CGCAGAAGTG
GCGTCTTCAC
GCTAGAGTAA
CGATCTCATT
ATCGTGGTGT
TAGCACCACA
AGGCGAGTTA
TCCGCTCAAT
ATCGTTGTCA PvuI
TAGCAACAGT
AATTCTCTTA
TTAAGAGAAT
AAGTCATTCT
TTCAGTAAGA
GATAATACCG
CTATTATGGC
GGGCGAAAAC
CCCGCTTTTG
GCACCCAACT
CGTGGGTTGA
GGAAGGCAAA
CCTTCCGTTT
CTCTTCCTTT
GAGAAGGAAA
ATATTTGAAT
TATAAACTTA
GTGC
CACG
26
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 3. Restriction analysis of plasmid pB30
3.1.3. Agarose gel electrophoresis
DNA fragments of different sizes, such as PCR fragments or fragments resulting from digestion
with restriction endonucleases, can be separated by agarose gel electrophoresis. This method
takes advantage of the fact that DNA is negatively charged. When placed in an electric field,
DNA molecules are attracted by the positive pole (anode), and repelled by the negative pole
(cathode). During electrophoresis, DNA fragments migrate at different rates through the agarose
gel. The gel matrix acts as a molecular sieve through which smaller molecules can move more
easily (rapidly) than larger ones. Thus, agarose gel electrophoresis will sort DNA fragments
according to their size. An agarose matrix can efficiently separate DNA fragments ranging in size
from 100 bp to more than 50 kb.
DNA fragments in different size ranges can be separated by adjusting the agarose concentration:
A low concentration (down to 0.3 %) produces a loose gel that separates larger fragments,
whereas a high concentration (up to 2%) produces a stiff gel that resolves small fragments.
To load the DNA onto the gel and monitor the migration of the unseen DNA bands in the gel
matrix, a dense sugar solution or glycerol and one or more visible dyes are added to the samples.
The dense sugar solution weights the DNA sample, helping it to sink when loaded into the well.
The negatively charged visible dyes migrate towards the cathode, for easy visual tracking of
DNA migration: e.g. the markers xylene cyanol and bromophenol blue migrate at a rate equivalent
to a DNA fragment of about 4kb and 400bp, respectively (in a 1% gel and TAE running buffer,
see picture below).
Visualization of the DNA in the agarose gel is achieved by a fluorescent dye, ethidium bromide
(EtBr), which usually is added directly to the buffer with which the agarose gel is made of (When
handling the gel, be sure to always wear protective gloves as the EtBr in the gel is classified as
carcinogenic!) The planar EtBr molecule can intercalate between the stacked nucleotides of the
DNA helix. Because EtBr that is bound to DNA fluoresces much more strongly than free
ethidium bromide, the DNA bands in the gel become readily apparent as fluorescent bands in a
dark background. The fragment pattern is viewed and recorded directly under UV light: the
DNA/EtBr complex strongly absorbs UV light at 300 nm, and emits visible light in the orange
range at 590 nm, which can be captured by a camera or on a photographic film (see picture
below). This technique is extremely sensitive; as little as 5 ng of DNA can be detected! It is
important to understand that a band of DNA seen in a gel is not a single DNA molecule, but
rather a collection of millions of identical DNA molecules, all of the same length.
To estimate the length of the DNA fragments, DNA size markers (ladder) are loaded in a well of
the gel, and the experimental fragments are compared with the marker fragments of known size.
Day light: Only the added dye is visible
on the gel.
UV light: The intercalated EtBr becomes
visible, indicating the DNA bands
Xylene cyanol
(ca. 4kb)
migration
wells
Bromphenole blue
(ca. 400bp)
Photo of gel under UV light:
The DNA bands are in white
DNA marker (ladder)
3000bp
1000bp
500bp
3.2. Goal of the experiment
In this experiment, we will establish in which orientation the EcoRI fragment containing the
ced-9 gene is inserted in pB30. The plasmid is digested with different combinations of
restriction enzymes, and the size of the resulting restriction fragments is determined by
agarose gel electrophoresis.
27
28
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 3. Restriction analysis of plasmid pB30
3.3. Protocol
The plasmid that has been isolated in experiment 2 (protocol 2.3.) will now be digested with
different combinations of restriction enzymes, and the size of the resulting restriction
fragments is determined by agarose gel electrophoresis.
3.3.1. Restriction digest
Each student in a group is responsible for 1 restriction digest using his/her isolated plasmid
DNA from the previous course.
The whole set of digests of the group is later used to establish the orientation of the ced-9
insert. In addition, each group will set up one control tube in which no restriction enzymes
are added, in order to see what the original undigested plasmid DNA looks like.
Reagents
• Isolated pB30 plasmid DNA with inserted C. elegans ced-9 cDNA. Tubes labeled with place numbers.
• Restriction buffer: NEB-Buffer 3 (10x), pH 7.9
1X Buffer: 100mM NaCl, 50mM Tris-HCl, 10mM MgCl2, 1mM DTT + BSA (Bovine Serum Albumin:
prevents adhesion of the enzyme to reaction tubes and pipette surfaces. BSA also stabilizes some
proteins during incubation).
• Restriction enzymes: EcoRI, MscI, PvuI (temperature sensitive → keep always on ice!)
• ddH2O
Per assistant one complete set of reactions (= 5 different digests + 1 undigested control) has
to be done. Each student sets up one restriction digest. If there are more than 5 students per
group, some digests are done twice (see below). One of the students (in addition to a
restriction digest) sets up also an undigested control reaction without restriction enzyme (if
needed, this can also be done by the assistant).
The complete set of digests (per assistant) comprises the following reactions:
Restriction enzyme
Single digests
Double digests
Control tube
Number of tubes in a group of 8 students
1. EcoRI
2x
2. MscI
1x
3. PvuI
2x
4. EcoRI + MscI
1x
5. PvuI + MscI
2x
6. without enzyme
1x
1. Calculate the amount of plasmid DNA and ddH2O you need for the digest (and for the
undigested control tube: if you are the one to set this up).
’ Note the amounts and the restriction enzyme you use in the table (next page).
a) For each reaction you will need 500ng plasmid DNA. According to the concentration
of “your” plasmid DNA (see course-4, experiment 2.4.3.), calculate the amount (µl)
you have to add to your reaction tube:
Concentration of “your” plasmid DNA sample = ………………ng/µl
→ amount you have to add of your plasmid DNA to the digest reaction =
…………µl
b) Finally calculate the amount of ddH2O you have to add to your reaction tube. The
final/total amount for the digest is 20µl.
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 3. Restriction analysis of plasmid pB30
2. Set up the restriction digest:
• Spin down all reagents before adding them, to make sure that all the liquid is at the
bottom of the tubes.
• label an Eppendorf tube with your place number and/or reaction number (1 – 6)
• add the following components in the order shown below (fill in the blank according to
the concentration of “your” plasmid DNA sample):
control
Amount for
Single digest Double digest
reagents
ddH2O
2 µl
2 µl
2 µl
Buffer, 10x (including BSA)
pB30 plasmid DNA, 500ng
-- no --
1 µl
20 µl
20 µl
2 µl (1µl each)
Restriction enzyme(s): ……………….……..
¬ Total amount per tube
20 µl
3. Mix gently by tapping the tube with a finger.
4. Spin down in a centrifuge for a few seconds.
5. Incubate the sample at 37 °C for ca. 45 – 60 minutes in the thermomixer/thermoblock.
3.3.2. Separating DNA fragments by agarose gel electrophoresis
After the restriction digest is completed, the different DNA fragments will be separated by
agarose gel electrophoresis. This method sorts DNA fragments according to their size.
To estimate the length of the DNA fragments, a DNA size marker is loaded in a well of the
gel, and the experimental fragments are compared with the marker fragments of known size.
1. Add 4µl of the 6x loading buffer (= Tis buffer, glycerol, blue dye (bromphenol blue, xylene cyanol)) to each
Eppendorf tube (containing the digest or the control reaction) and mix well by tapping
the tube gently.
3.
uncut plasmid DNA
DNA ladder
2. Load 10µl into the assigned well of a 1% agarose gel (see below).
NExpel any air from the tip before loading! NBe careful not to push the tip of the pipette through
the bottom of the sample well! NWhen handling the gel always wear protective gloves (see 3.1.3.)
Well 1:
Well 2:
Wells 3 – 10:
different digests
load 5µl of 2log DNA Marker (ladder)
load 10µl of control reaction (uncut)
load 10µl each of digests
!! Note directly in the “gel picture” (left) in which
well you loaded your sample
’ make sure you have a complete overview of the
order of the loaded samples from your group !!
If all samples are loaded, let the gel run for ca. 60min. at 100-120V.
NRemember the gel that belongs to your group!
’ Assistant: write your name on a tape and put it on or next to the electrophoresis
chamber in which “your” gel is running. If you remove the gel from the chamber
to take a photograph, take the tape with your name with you!
29
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 3. Restriction analysis of plasmid pB30
3.4. Analysis of agarose gel
3.4.1. Fragment size expectations of restriction digest
1. Below you find two plasmid maps with two different possible orientations of the ced-9
cDNA insert (“clockwise” or “counterclockwise” orientation).
The numbers given in the plasmid maps correspond to basepair (bp) positions for the
cutting sites of the three restriction enzymes you used in this experiments (EcoRI, MscI,
PvuI). The numbers 1 and 4254 indicate the start and the end, resp. of the plasmid.
For each of these sites indicate the corresponding restriction enzyme (use Fig. 3).
“Clockwise” ced-9 orientation
“Counterclockwise” ced-9 orientation
2. Determine the length of the DNA fragments you expect after the digestion of the pB30
plasmid with the different restriction enzymes for both possible orientations of the ced-9
insert.
Fill in the bands of the expected sizes in the sketches of “hypothetical agarose gels” below:
Expected fragments in “clockwise” ced-9 orientation
Expected fragments in “counterclockwise” ced-9 orientation
3. Which restriction digest(s) are most useful to discriminate between the two possible
orientations?
30
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 3. Restriction analysis of plasmid pB30
3.4.2. Actual fragment sizes and orientation of ced-9 insert
After the DNA fragments are separated, each assistant and his group will take a photograph
of the gel ’ analyze the different fragment sizes obtained by the complete set of digests and discuss
the following points:
1. How do I see that a DNA fragment has inserted into the pB30 plasmid vector?
2. Is the inserted DNA fragment the one of interest (namely ced-9)?
3. In what orientation has the ced-9 fragment been inserted in the vector (clockwise or
counterclockwise)?
3.5. Questions
1.
When undigested plasmid DNA is loaded on a gel, a complex banding pattern can be
visible: Sometimes up to 5 bands can be distinguished on the gel after electrophoresis.
Indicate, which of the following forms of plasmid DNA could correspond to each of
the five putative bands. Quickly explain how these forms arise:
five possible forms of plasmid DNA: plasmid multimer, doublestranded open-circular,
singlestranded open-circular, supercoiled, linear
(See also lecture L3).
31
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
B) Recombinant DNA technology _ Experiment 3. Restriction analysis of plasmid pB30
2.
On an Ethidiumbromide (EtBr) gel, small fragments are not as bright as long
fragments. Sometimes they are even hardly detectable. Why? Give a short explanation.
3.
Why can fragments of similar or very similar size represent problems in mapping
experiments (by gel electrophoresis)?
4.
There is a site for the restriction enzyme BamHI (GGATCC) in the MCS as well as at
position 1342 and at position 1029 (see Fig. 3). Would this enzyme also be useful to
clone the ced-9 cDNA into the plasmid pB30?
5.
Before inserting a DNA fragment into a plasmid vector, why is it important to have a
map displaying the cutting sites of all possible restriction enzymes?
6.
Why is it important to analyse in which orientation a fragment has been inserted into a
(plasmid) vector?
32
BIO 111 – Classical and Molecular Genetics
Practical course, part-2: Molecular Genetics, Prof. Michael Hengartner
4. References
4. References
Exp. 1:
“Bitter” taste reception
Bufe, B. et al (2005). The Molecular Basis of Individual Differences in Phenylthiocarbamide
and Propylthiouracil Bitterness Perception. Current Biology, Vol. 15, 322–327
Kim U. et al (2003). Positional Cloning of the Human Quantitative Trait Locus Underlying
Taste Sensitivity to Phenylthiocarbamide. SCIENCE VOL 299
Meyerhofer et al. (2009). The Molecular Receptive Ranges of Human TAS2R Bitter Taste
Receptors. Chem Senses. 2010 Feb;35(2):157-70
Risso, E.S. et al. (2016). Global diversity in the TAS2R38 bitter taste receptor: revisiting a
classic evolutionary PROPosal. Scientific Reports 6
Santa-Cruz Calvo, S. and Egan, J.M. (2015). The endocrinology of taste receptors. Nature
Reviews Endocrinology
Exp. 2 and 3: Recombinant DNA technology
1.
M. O. Hengartner, H. R. Horvitz (1994). The ins and outs of programmed cell death
during C. elegans development. Philos. Trans. R. Soc. Lond. B Biol. Sci, 345(1313): 243246. Review
2.
M. O. Hengartner, R. E. Ellis, H. R. Horvitz, Nature 356, 494-499 (1992).
3.
M. O. Hengartner, H. R. Horvitz, Cell 76, 665-676 (1994).
4.
Kim, U., Jorgenson, E., Coon, H., Leppert, M., Risch, N., and Drayna, D. (2003).
Positional Cloning of the Human Quantitative Trait Locus Underlying Taste Sensitivity
to Phenylthiocarbamide. Science 299:1221-1225.
5.
M.C. Campell, A. Ranciaro, A. Froment, J. Hirbo, S. Omar, J.-M. Bodo, Th. Nyambo, G.
Lema, D. Zinshteyn, D. Drayna, P. A. S. Breslin, S. A. Tishkoff, Mol. Biol. Evol. 29(4),
1141–1153 (2012)
6.
Wooding, S., Bufe, B., Grassi, C., Howard, M. T., Stone, A. C., Vazquez, M., Dunn, D.
M., Meyerhof, W., Weiss, R. B., and M. J. Bamshad. (2006). Independent evolution of
bitter-taste sensitivity in humans and chimpanzees. Nature, 440, 930-934.
7.
Jovanovic M, Hengartner M O (2006). Developmental apoptosis in C. elegans: a
complex CEDnario. Nat Rev Mol Cell 7(2): 97-108
Books:
D. A. Micklos, G. A. Freyer, DNA Science-A First Course, Second Edition, Cold Spring
Harbor Laboratory Press (2003).
33