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