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1Nucleic Acid Analysis Techniques
What would you like to know about nucleic acid structure??
1. How much??
2. What size??
3. What structure??
4. What sequence??
1. HOW MUCH ?spectroscopy
common and simple technique to determine the amount / mass of aspectroscopy determine the amount / mass of a DNA or RNA sample
nucleotide bases(absorbs incident radiation - UV light)
absorbance α
2Beer’s Law
A = b cε(different for RNA,
DNA, oligonucleotides)
“rule of thumb” to estimate DNA, RNA, andDNA oligo concentrations
nucleic acid absorbance (A260) concentration (ug/ml)
DNARNA 1.0
1.0RNADNA oligo 1.0
1.0
(Question?? Why the different [C] for the same A260 reading?)
Estimation of DNA and RNA purity:
A260A280
ratioDNARNA
Question?? Why do these values decrease when proteinQuestion?? Why do these values decrease when proteincontaminates a nucleic acid sample?
(Hint: Think about their respective absorption spectra, theirpeak maximums, and their overlap)
3Cot analysis
estimate the [C] of a givenDNA/RNA sequence in alarger population of DNA/RNA molecules
reassociation or base pairingof complementary DNA/RNA strands
Cot analysis of a bacterial genome
bacterial genome
+4-4 -3 00
+1-2 -1 +3+2
What does this curve tell you?
[C] of complementarynumber of gene copies?DNA strands is
number of gene copies?
characteristic of bacterial genomes
4Cot analysis of a eukaryotic genome
shear DNA to short fragments
denature DNA to S S DNA
Cot analysis of i ti ki ti
eukaryotic genome1.0
short fragments to S.S. DNA reassociation kinetics
Co
C 0.5(characteristic of higher
eukaryotic genomes)
+4-4 -3 00
+1-2 -1 +3+2 +4-4 -3 0 +1-2 -1 +3+2log Cot
How many populations of DNA sequences?
Each population is what % of the total genome?
5Detection of specific DNA/gene sequence in a genome?
A single-stranded DNA (or RNA) possessing the sequence
radioactively-labeled(can base pair and detect
sequence In genomic DNA)
incubate with fragmented DNAdenatured to single strandsallow DNA strands reanneal
if the hybridization probebecomes D.S. indicates thepresence of the sequence
reassociationpresence of the sequencein the genomic DNA
2. WHAT SIZE ??sedimentation rate of an DNA/RNA through a sucroseor glycerol gradient is based upon the DNA’s/RNA’s size
10%
centrifuge tube40%
6aftercentrifugation
SDNA/RNA
sedimentationbased on size
S
(values are not linearindication of size)
fractionate the
spectroscopic analysisof the sucrose gradient
A260
fractionate the gradient into individual fxs
of the sucrose gradientfractions
sedimentation
can determine the value of DNA/RNA/protein on asucrose gradient and then estimate the molecular weightusing the following equation:
S
using the following equation:
M =R T ( )S
D ( )1 – v ρ
M = mol. wt.R = gas constantT = temperature (C)D diff i ffi i tD ( )1 v ρ
fV = partial specific volume
D = diffusion coefficient
ρ = density of solvent
f =
(determined by the shape of the molecule)
7protein molecule A protein molecule B
mol wt A mol wt Bmol. wt A mol. wt B
S(A) S(B)however (due to molecule shapes)
frictional coefficient (f) takes shape into account and helps thefrictional coefficient (f) takes shape into account and helps the equation make a more accurate assessment of molecular weight
size determination by
polyacrylamide / agarose gels(different resolution of size ranges)
electricfield
DNA / RNA moleculesare charged
S
S(the larger the molecule
the more retardationin gel matrix)
moleculesS
(again – sedimentation coefficientnot linearly proportional to mol. wt.)migration α y p p )g
potential problem ?? assumption
85’ 3’linear
S.S. strand
5’ 3’foldedD.S. hairpin
folded nucleic acidsaffect migration position
(frequent problemfor RNA)
solution ??
How ??use denaturants such as
denaturing gelspossess these denaturants
3. WHAT STRUCTURE ??
Resolution of S.S. versus D.S. nucleic acids
migration positionin the gradient isbased upon densityof the nucleic acid
9start finish
1.68gm/ml1.55
1.70 gm/ml
gm/ml1.68
gm/ml1.74
gm/ml1.85
fill tube withCsCl solutionof constantdensity
Question ?? What happens uponprolonged centrifugation?
(rho)
for D.S. DNA density ( ) αρ
( l l t %GC t t f DNA i it d it )(you can calculate %GC content of a DNA via its density)
analysis of sheared eukaryotic genomic DNA
A260
ρ
short, highlyrepetitive DNAsequences ofhigher or lower
GC content
10Nuclease Mapping to Determine Structure
utilization of DNases / RNases to probe the structure of folded (or protein-bound and
protected) RNA or DNA molecules
first a brief discussion of nucleases
Nucleases three general categories
1.2.
3.
Properties that define nuclease specificity:
1. recognition of 2’ OH or 2’ HRNA DNARNA DNA
3
2.
3.
5’-specificor
3’-specific
internalcleavage
4. 5’ or 3’ monophosphate products
115.
a. non-specificb. Pu or Pyr-specificb. Pu or Pyr specificc. base-specific (A,C,G,T,U)d. sequence/site-specific
four nucleases we will use:four nucleases we will use:
DNA S.S.D.S.
endo 5’ non-specific5.4.3.2.1.NUCLEASE
DNA/RNA
DNA/RNA
S.S.D.S.
S.S.
endoexo
endoexo 5’
3’ non-specific
non-specific
RNA D.S. endo 5’ non-specific
Now use nucleases to probe and determine RNA structure
2 / 3 structure
single strand-specific determines S.S. regions
double strand-specific determines D.S. regionsg
can used defined cleavage sites to determine the folded structure of an RNA molecule
1235
RNA
(ki d ATP)
35 PO4*
* 3
(kinase enzyme and ATP)
(not labeled andnever measured)
5 PO4 3 35
measure cleavage site from the radiolabeled end to determineth it ( t) l d
measure distance (cleavage nt) by electrophoresis on adenaturing polyacrylamide gelthe site (nt) cleaved denaturing polyacrylamide gel
IMPORTANT TO REMEMBER
(NOT a single molecule)(NOT a single molecule)
THEREFORE under limitingdigestion conditions
generate a population of labeled fragments of different length5
*5 PO4* 3
35 PO4*5 PO4* 3
4
generated fragments (length)
(S1= S.S. V1= D.S)
13An RNA of the following sequence is labeled at the 3’ and then subjectto S1 and V1 nuclease digestion for structural determination
5’ – N G U A G G A U C A G U A G G G A U C C A G U C A G C – 3’
A G U C S1 V1nt-specific nucleases
NOTE: Not everyynt in a given S.S.or D.S. region willnecessarily bedigested with S1 or V1 nucleases.
Therefore, the S1 and V1 digestionpatterns just givepatterns just givean indication ofsingle and doublestranded regions.
From this data, determine this RNA’s folded structure.2
14determination of detailed
2 / 3 structurepowerful approach for
structure determination
(crystal)
(records scatter pattern)
( high energyelectrons )
( can be difficultto crystallizet i l l )certain molecules )
crystal-specific for structure of molecule diffracting electrons
mathematically solve thestructure of the molecule
crystal structures of molecules are described in terms of
(5A / 3A / 1.5A )
ability to resolve atoms in the molecule that are separated by the indicated number of angstroms
(Note: crystal structure may not represent solution structure of the molecule)
154. WHAT SEQUENCE ??general mapping of DNA sequence/genomes to establish “landmarks”
cleave DNA at a specific sequence
Arber / Nathans / Smith Nobel Prize 1977 started the genetic i i l tiengineering revolution
Type I Type II Type III
used for restriction Type II is
1. DNase
mapping and cloning sequence-specific
Properties of Type II restriction endonucleases
2. Double-stranded 3. endonuclease 4. 5’ and 3’ PO45. sequence-specific
enzyme recognition site(binding site)
this property is what distinguishesType II from Type I and Type III enzymes
16restriction endonucleasenomenclature
name indicates
Eschericia coliEschericia coliProteus vulgarisStreptomyces albusBacillus amyloliquefaciens H
Type II restriction site(enzyme recognition site)
exhibits 2-fold axis of symmetry
[ arrows indicate sequence identityabout 2-fold axis ]
Eco R1
palindrome
5’ - G A A T T C – 3’*3’ – C T T A A G – 5’
*
* th l ti it* = methylation sites
when methylatedcleavage is inhibited
5’ - G OH
3’ – C T T A APO4
+ PO4A A T T C – 3’
OH G – 5’
protects endogenousDNA from self-cleavage
(can be useful for cloning)
17
Pvu I 5’ – C A G C T G – 3’3’ G T C G A C 5’3’ – G T C G A C – 5’
restriction enzymes digestat specific sequences/sites
“mapping” or defining restriction sites on a DNA sequence/fragment/genome
restriction mapping of a 10 kb DNA Pvu I fragment
Pvu IPvu I Pvu IPvu I
cut with the following restriction endonucleases
fragments (kb) # restriction sites
Pst I
Eco RI
Bam HI
fragment sizedetermined byelectrophoresis
now to determine these cleavage sites and theirpositions with respect to the other cleavage sites
18Pst I
or
(7+3)
or
Eco RI (6+4)
or
( )
Bam HI (8+2)
or
How to determine the site of each cleavageHow to determine the site of each cleavageand their relationship to the other sites??
uncuT
P E B
P+E
P+B
B+E
T
10
7 86
65
32
434
235
2
YOUR JOB construct a restriction map from this gel data
19A more difficult restriction map to construct:
unc
P+
P+
B+
P+B+
2217
16
uT
PE B E BE E
22
57
10
6
1613
67
5 55 64 4
55 5
Hint:
Hint:
Hint:
Primary Sequence DeterminationTwo Methods:
Maxim-Gilbert sequencing
S iSaenger sequencing
Gilbert and Saenger Nobel Prize in chemistry (1980)
20Maxim-Gilbert Sequencing
basic approach: single strand of DNA
5’ 3’
radioactively end label “tag” end for reference site
*polynucleotide kinase (PO4)*terminal deoxynucleotidyl transferase (pCp)*
aliquot into 4 individual reactionsREMEMBERmany RNA molecules
in each rxn and limiting
G CTAFor each rxn, carry out base-specific nt modification and cleavage
grxns conditions causeone modification per
RNA molecule
this generates end-labeled fragmentsterminating in that specific base
C G A T C G A T G A T C G A5’ 3’C G A T C G A T G A T C G A
****
*****
21run all nucleotide-specific cleavage rxns on a polyacrylamide sequencing gel (denaturing gel)
CON
What is the control?NTROL
A G C T
critical to know forreading the sequence
( Maxim-Gilbert chemical sequencing is complicated and time consuming )
Saenger sequencing
easierfastersequencing
dideoxynucleotide sequencing
ObasePbaseP
OC
C C
CO
C
C C
C
H
C
HOH H
225’ 3’
5’3’
DNA strand to sequence
(complementaryto the 3’ end ofthe DNA strand) (“tagged” end
to “measure”)
(synthesis of a complementaryDNA strand using DNA polymerase)
to measure )
important point: DNA polymerase requires a on the 3’ terminal nucleotide for synthesis of the complementary strand
when a dideoxynucleotide is incorporatedinto the growing chain
A G T CdXTPs
ddXTPsall 4 all 4 all 4 all 4
4 Synthesis Reactions (all contain DNA strand + primer + polymerase)
5’ 3’G TA C
ddXTPs(dideoxy)
5’ 3
*5’
*5’
*5’
*5’
synthesis is terminated when a dideoxynucleotide is incorporated into the chain
electrophoresis on a polyacrylamide gel resolvesprimer extension fragments based on size andallows the determination of DNA sequence
23RNA SequencingA. Nuclease/Chemical Modification Sequencing1. radiolabel the end (5’ or 3’) of the RNA transcript1. radiolabel the end (5 or 3 ) of the RNA transcript
2. digest under limiting conditions the labeled RNAusing
OR
3. resolve the generated end-labeled RNA fragments bygel electrophoresis and determine the RNA sequence
2. chemically modify/cleave under limiting conditions the labeled RNA using
OR
B. Primer-Extension Dideoxy-SequencingUsing “Re erse Transcription”
gel-electrophoresis and determine the RNA sequence
5’ 3’5’3’
Using “Reverse Transcription”
(complementaryto the 3’ end of
the RNA transcript) (“t d” dthe RNA transcript) (“tagged” endto “measure”)synthesis of a complementary
DNA strand using
synthesizes a complementary DNA(cDNA) strand from an RNA template