PLB161A Laboratory XI a
Genome Mapping
Restriction Digests and Agarose Gel Electrophoresis of Genomic DNA.
A. Restriction Digests.
Introduction
Restriction enzymes are a class of DNA endonucleases, which occur in nature in
various microorganisms. They have evolved as a defense mechanism in bacteria to
degrade foreign DNA such as that introduced by a bacteriophage. Such exogenous
DNA is rapidly cleaved into smaller fragments by the restriction enzyme and the
bacterium is thus protected from viral infection. The bacterial DNA is protected from
cleavage by methylation of sequences that would otherwise be cut. Type II restriction
endonucleases are the type used most often in molecular biology. The proteins bind
specifically to double stranded DNA and then cleave it, and after this has occurred, the
enzyme dissociates from the DNA molecule. Some enzymes cleave exactly at the axis
of symmetry, generating fragments of DNA that carry blunt ends. Other enzymes
cleave each strand at similar locations, on opposite sides of the axis of symmetry,
creating fragments of DNA that carry protruding single-stranded termini also referred to
as “sticky ends”. The vast majority of type II restriction endonucleases recognize
specific sequences that are four, five, six or eight nucleotides in length and display two
fold symmetry:
Eco RI (from Escherichia coli) cleaves GAATTC
Pst I (from Providencia stuartii) cleaves CTGCAG
Taq I (from Thermus aquaticus) cleaves TCGA
Some restriction enzymes (rare-cutters) cut the DNA very infrequently generating
a small number of very large fragments (several thousand to a million bp). These
enzymes are generally used for physical mapping. For genetic studies using RFLPs
restriction enzymes that cut DNA more frequently are generally used. These restriction
enzymes generate a large number of small fragments. Restriction enzymes with 4-base
recognition sites will yield pieces of 256 bases long on average and those with 6-base
recognition sites will yield pieces of 4,096 bases long on average. These numbers are
affected by the differential abundance of nucleotides in different organisms. Since
hundreds of different restriction enzymes have been characterized and are available
commercially, DNA can be cut into many different small fragments.
Methylation effect on restriction enzymes
Plant DNA has a high content of 5-methylcytosine. Approximately 80% of the CG,
and CNG sequences are methylated in plants. Some restriction enzymes will not cut
5m
CG sequences (i.e. HhaI, HpaII, SalI, XhoI) and others will not cut 5mCNG (i.e.
EcoRII, PstI, PvuII). I some cases there are isoschizomers (restriction enzymes from
different sources but with the same recognition site) that differ in their sensitivity to
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methylation. BstNI is an isoschizomer of EcoRII but BstNI is not blocked by 5mCNG
methylation.
Sometimes, the 5mCG site is not within the restriction site, but can be formed with
the adjacent bp in the DNA sequence. For example, Apa I (GGGCCC) does not have a
CG site, but if the next bp is a G or the previous base pair is a C, there will be a CG pair
that if methylated (5mCG) will not be cut by Apa I. This is called overlapping
methylation.
Methylation and overlapping methylation affect the average size fragments
produced by a restriction enzyme. In the case of internal methylation, if we assume
random distribution of the 4 base pairs (=0.25 probability of occurrence for each one),
the expected average size of a six-cutter restriction enzyme sensitive to 5mCG
methylation (e.g. SnaBI TACGAT) the probability of finding an un-methylated C is 0.25 *
0.20=0.05 (because 80% of the CG are methylated). Therefore the expected size will be
1/[(0.25)6*0.2]= 20,480-bp. Therefore, the average fragment size will be five times larger
larger than in a restriction enzyme without sensitivity to methylation (1/(0.25)6= 4,096bp.
In the case of overlapping methylation, you need to include in the calculation
the probability that the adjacent bases will generate a 5mCG or 5mCNG. For example if
your enzyme recognition sequence is GATC, and the enzyme is sensitive only to 5mCG
methylation, the GATC site will have adjacent C or G in 7/16 of the combinations
indicated in red below. In 80% of the 7/16 the CG will be methylated and the enzyme
will not cut (7/16 * 0.80= 0.35).
aGATCa
aGATCt
aGATCc
aGATCg
tGATCa
tGATCt
tGATCc
tGATCg
cGATCa
cGATCt
cGATCc
cGATCg
gGATCa
gGATCt
gGATCc
gGATCg
The probability of a GATC site cut, based on the adjacent bases is 9/16 +
7/16*0.2 = 0.65. Therefore the expected size will be 1/[(0.25)4*0.65]= 394-bp. If the
enzyme is sensitive to both 5mCG and 5mCNG overlapping methylation the denominator
needs to be multiplied again by 0.65: 1/[(0.25)4*0.652]= 605-bp
Once a plant DNA segment is cloned and introduced into a bacteria, for example
in a BAC clone, the methylation pattern of the plant disappears and is replaced by the
methylation of the bacteria. Therefore, when digesting a BAC clone, you need to
consider the effect of Dam (GmATC) and Dcm (CmC(A/T)GG). For example if a BAC
clone is cut with a restriction enzyme sensitive to overlapping Dam methylation as Xba I
(TCTAGA), the sites followed by TC (TCTA GmATC) will not be cut in the bacteria but
will be cut in the plant genomic DNA producing differences in the observed restriction
fragments.
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Technical information:
•
Homogenization: One of the primary problems encountered with digestions
of high molecular weight DNA is that the DNA is sometimes not evenly
distributed in solution, leading to localized regions of high DNA concentrations.
These “clumps” of DNA are relatively inaccessible to restriction enzymes, and
result in only the “partial” digestion of the DNA. If clumps of DNA are seen, it is
necessary to homogenize the solution before the digestion. Mixing high
molecular weight genomic or nuclear DNA requires gentle handling in order to
minimize mechanical shearing of the DNA during extraction. Use wide-bore pipet
tips prepared by removing the tip of a regular pipet tip with a sterile razor blade.
Solutions of high molecular weight DNA are quite viscous and are therefore
difficult to pipet.
•
Time and number RE units: By definition one unit of restriction
endonuclease will completely digest 1 µg of substrate DNA (lambda for example)
in a 50 µl reaction in one hour. Restriction digestion of high molecular weight
genomic DNA requires longer digestion times (2 to 3 hours) and 3 units of
restriction enzyme per µg of genomic DNA. If after digesting the DNA for three
hours the restriction is not complete; a second aliquot of enzyme can be added to
continue to digest the DNA.
•
DNA amount and concentration: Wheat has a very large genome with huge
amounts of repetitive DNA. Consequently, to obtain an adequate detection of low
copy number RFLP sequences it is necessary to load more DNA in each well
than in RFLP studies of species with small genomes like rice or tomato. For
wheat RFLPs, 10-12 µg of DNA are loaded in each well. In our class we will load
only 5 µg because we will use a high copy number probe. The maximum volume
that can be loaded in one well is generally lower than 80 µl. To load this amount
of DNA in this volume the DNA concentration of the samples should be higher
than 150 ng/µl (10µg /150 ng/µl= 67µl). Although DNA solutions of high molecular
weight DNA are viscous, due to the length of the molecules, the concentration of
genomic DNA preparations can be quite low. If the concentration is lower than
150-ng/µl the restriction digest can be performed in a larger volume and then
concentrated in a vacuum concentrator before loading.
Procedures:
Restriction Digests of Wheat Genomic DNA
1.
Determine the total µl of each DNA sample to be digested with an enzyme in a
single tube as follows:
Total µl DNA = (Total µg DNA) / (DNA concentration, µg/µl).
We will digest 5 µg of DNA with restriction enzyme Taq I (65 C). Example: if a
DNA concentrations is 200-ng/µl add 25 µl of DNA.
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Some enzymes as TaqI work better with BSA (Bovine Serum Albumin). Use 1/100 of
the total volume
2.
Determine the units (U) and volume of enzyme necessary to digest each DNA
sample. In general, it is best to use 3.0 U/µg DNA to prevent partial digestions.
Total U Enzyme = (Total µg DNA) x 3.0
Total µl Enzyme = (Total U enzyme) / (enzyme concentration, U/µl)
For 5 µg of DNA use 15 U. TaqI is at 20 U/µl so use 0.75 µl of enzyme per
sample. Note: Do NOT add enzyme in excess of 10% of the reaction volume.
Many enzymes exhibit “star” activity and cut the DNA at nonspecific sites in the
presence of glycerol > 5%. The restriction endonucleases are supplied in 50%
glycerol.
3.
Based on the DNA and enzyme volumes, determine the total reaction volume. In
this example total volume is 50 µl.
4.
Determine the volume of ddH20 per tube as follows:
µl ddH20 = 50µl – 5 µl buffer - µl DNA - µl enzyme -µl BSA
For example if DNA concentration is 200 ng/µl, and the restriction enzyme is
10U/µl add 50-5-25-0.75-0.5= 18.75 µl of ddH20.
5.
Use permanent black marker to label 20 sterile PCR tubes. Set up a reaction for
each of the 14 nulli-tetrasomic lines. Add the proper amount of water first to each
tube (you can use the same tip) and then the DNA samples (using new sterile
tips for each sample). If all DNA samples were adjusted to the same
concentration, the water can be included in the digestion mix with the buffer and
the enzyme.
6.
Calculate a bulk digestion mix containing the buffer, and enzyme needed for the
total number of different DNA samples to be digested by the same enzyme.
Prepare bulk mix on ice, adding enzyme last; mix well. To allow for pipetting
errors prepare extra bulk mix (usually one extra every 20 samples).
5 µl buffer * 15=
75 µl
0.75 µl enzyme * 15= 11.25 µl
0.5 µl BSA * 15= 7.5 µl
7.
Add 6.25 µl of mix to each tube. An extremely important, yet often overlooked,
element of a successful restriction digest is mixing. The reaction must be
thoroughly mixed to achieve complete digestion. Gently pipet the reaction mixture
up and down (avoid bubbles). Close tube tops and pool reagents by pulsing in a
microfuge or by sharply tapping the tube bottom on lab bench.
8.
Place reaction tubes in a water bath or PCR machine with the optimum
temperature for the enzyme.
Note: The recommended incubation temperature for most restriction
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endonucleases is 37 °C. restriction endonucleases isolated from thermophilic
bacteria require higher incubation temperatures ranging from 50 °C to 65 °C.
Always check catalog for optimum incubation temperature.
9.
Following incubation, add 5 µl of loading dye and freeze reactions at –20 °C until
ready to continue.
10 X Loading Dye:
[FINAL]
50 ml
100 ml
15%
7.5 g
15.0 g
Bromophenol blue
0.25%
0.125 g
0.25 g
Xylene cyanol
0.25%
0.125 g
0.25 g
10 X Tae
10 ml
20 ml
To 50 ml
To 100 ml
Ficoll type 400
50 X TAE buffer
ddH20
B. AGAROSE GEL ELECTROPHORESIS
Introduction
Electrophoresis through agarose or polyacrilamide gels is the standard method to
separate, identify and purify DNA fragments. Although agarose gels have a lower
resolving power than polyacrilamide gels, they have a greater range of separation and
are much simpler and easier to handle than the polyacrilamide ones, and have been
more widely used in DNA electrophoresis. Agarose, a purified form of agar isolated
from seaweed, is a linear polymer. Various grades and forms of agarose are available
commercially. Agarose gels are cast by melting the agarose in the presence of the
desired buffer until a clear, transparent solution is achieved. The melted solution is then
poured into a mold and allowed to harden. Upon hardening, the agarose forms a
matrix, the density of which is determined by the concentration of the agarose. The gel
is then submerged in “running buffer” and the DNA samples are loaded into wells that
were cast into the gel using a plastic comb at the time of preparation. An electrical field
is then applied across the length of the gel and the negatively charged DNA molecules
migrate toward the anode of the electrophoretic system. The rate at which linear DNA
molecules migrate within the gel depends on their length: larger molecules migrate
more slowly than smaller ones because it is more difficult for them to pass through the
agarose pores. Thus, the DNA molecules become separated according to their size.
At high gel concentrations the resolution of smaller fragments is favored, while at
low agarose concentrations that of larger fragments is favored. DNA is visualized
by soaking the gel in a solution of ethidium bromide; this fluorescent dye intercalates
between the two strands of DNA and makes them visible under ultraviolet light.
When plant or animal DNA is digested with restriction endonucleases and the
DNA fragments are separated by gel electrophoresis, the result is a smear of millions of
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bands. In the DNA hybridization laboratory we will learn how to visualize specific
sequences within this smear. It has been shown empirically that individuals are not
identical with respect to the distance (in bases of DNA) between restriction
endonucleases cleavage sites. We will use these differences in sizes as molecular
markers for genetic studies.
Optimum agarose concentrations for different DNA sizes.
% Gel
Optimum Resolution for Linear DNA (kb)
0.5
0.7
1.0
1.2
1.5
30 to 1.0
12 to 0.8
10 to 0.5
7 to 0.4
3 to 0.2
Procedure
1. Prepare a 1% agarose gel cast in 1X TAE using a gel rig and a 2-mm comb with
15-wells. For 100-ml:
i) Weigh 1g of SeaKem LE agarose, transfer to a 250-ml flask
ii) Add 1XTAE to 100 ml
iii) Melt agarose in microwave oven
iv) Pour molten agarose (cooled to 60oC) into rig with comb; let gel cool 20 minutes
to harden. Be sure that the comb is at least 1mm above the surface of the rig.
vi) Cover gel with 1X TAE buffer
2.
Heat the λ marker at 60oC for 3-5 minutes, cool on ice, and briefly
microcentrifuge to collect sample in bottom of tube. If the restriction digests have
been stored at 4oC, heat the digests at 55 - 60oC for 2 -3 minutes to disrupt any
base pairing between protruding cohesive termini before loading the gel.
3.
Load the molecular weight standard and the samples slowly.
4.
Run the gel slowly overnight at constant voltage (35 Volts). This maximizes
resolution of restriction fragments, prevents smearing of bands, and reduces
distortion or “smiling” of the DNA bands.
5.
After electrophoresis is completed, stain the gel for 10 minutes in an ethidium
bromide staining solution (0.25 µg/ml) if the ethidium bromide was not added to the
gel. Rinse the gel briefly in ddH2O and photograph the gel using the still-image
video system. Be sure to place a fluorescent ruler along the side of the gel before
recording the image.
6.
Using permanent waterproof ink (i.e. India ink) and an 18-gauge needle, mark the
molecular weight markers. Take precautions to prevent nicking the UV filter! Place
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a sheet of UV-transparent Plexiglas under the gel before marking the lanes.
Restriction Enzymes Exercises
1) What is the optimal digestion temperature?
Taq I:
Bsa I:
Bst EII:
2) Which of the following enzymes need BSA?
ClaI:
XbaI:
SpeI:
3) Which is the best “New England BioLabs” buffer?
For NdeI:
Double digestion EcoRV –PstI:
EcoRI – HindIII:
4) What is the average size fragment generated by RE assuming random distribution of
base pairs?
DdeI:
AscI:
AccI:
AvaII:
5) Calculate the expected average size fragment generated by Dra I assuming a)
random distribution of base pairs and b) using the known base pair composition of the
wheat genome: A=27.2%, T=27.1%, C=22.7%, and G=23%.
6) What of the following enzymes will have problems to digest plant DNA because of
methylation and at which base methylation will occur?
SalI:
PvuII:
TaqI:
7) Calculate the expected average size fragment generated by BglII (AGATCT)
considering that this enzyme is sensitive to overlapping methylation at the beginning
and the end of the recognition site. Assume random distribution of base pairs and that
80% of the CG pairs are methylated.
8) Calculate the expected average size fragment generated by BstZI (CGGCCG)
considering that this enzyme is sensitive to internal and overlapping 5mCNG
methylation. Assume random distribution of base pairs and that 80% of the CG and
CNG are methylated.
9) Explain why you can get different fragment sizes for a segment form total plant
genome DNA and from a BAC from the same species and from the same region when
digested with Taq I.
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10) Find the isoschizomers of the following enzymes and indicate the differences in
methylation sensitivity between them
BstNI:
MspI:
11) Calculate the number of copies of a single copy gene present in a restriction digest
of 10 µg of diploid wheat DNA and in 5 µg of rice DNA.
Assume that there is no restriction site within the gene. The size of the wheat genome is
5.6 pg per haploid genome and that of the rice genome 0.5 pg per haploid genome.
12) How do you know if a RE is methylation sensitive by looking at a digestion of
genomic DNA in the gel?
13) Indicate if there are higher possibilities to find polymorphisms with a four- cutter
restriction enzyme or a six-cutter RE in the following two scenarios:
a) All the polymorphisms originated by insertion deletions (no point mutations)
b) All point mutations (no insertions or deletions). How does the size of your probe
affect your answer?
There are 6 additional questions for this Lab report after the Southern chapter.
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Exercise 1. Indicate restriction maps compatibles with the observed
polymorphisms. Use insertion/ deletions
or base pair mutations
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