1

A brief introduction to SAXS

M.H.J. Koch

2

SAXS

02

20

02

22

0

12

2cos12 I

r

rI

r

)θ(r)θ(Ie =

+=

classical electron radius = 2.82 10-15 m

For SAXS this factor =1

sin2θ = 2θ cos 2θ = 1

I0=Iinexp(-µt)

r>>t

∆λ/λ~0

For a single electron:very small!!

3

Electrons in an electromagnetic fieldare accelerated and therefore emit radiation: they scatter. The spatial distribution of the scattered intensity depends on the geometry of theexperiment. For unpolarized incident radiation the spatial distributionon the equator is:

2Ө

(1) I)ω(ω

ωr1

2)θ2(cos1

r)I(2 02220

4

2

220

−⋅⋅

+⋅=θ

where 2θ is the angle between the incident and the scattered beam.

corresponds to the

natural frequency (ν0=ω0/2π) of the oscillator and ω to the frequency of the incident radiation.

Obs. m

k=0ω

4

) ωω (ω

220

2

−

in the previous equation is the one describing the frequencydependence. For the AMPLITUDE (E with I= E·E*)

The natural frequency of the oscillator (ω0) corresponds to the binding strength of electrons in atoms and lies somewhere in the UV to X-ray region. If the incident radiation is visible light (λ ≈ 500 nm),ω << ω0 and the factor reduces to:

20

2

ωω

The amplitude of the scattered radiation at r is proportional to ω2 andin phase with the incident radiation. This is

Rayleigh scattering.

The scattered intensity is proportional to ω4 hence the blue sky.

The most interesting factor

5

X-RaysIf the incident radiation is X-rays (λ ≈ 0.1 nm), ω0 << ωand the factor

) ωω (ω

220

2

−

The scattering amplitude is independent of the frequency and its phase is shifted by 180 degrees relative to the incident radiation. This is

Thomson scattering

= -1

As it is independent of the frequency of the incident radiation, the worldof X-rays is colorless with shades of gray (i.e. contrast) only and Eq.1 above simplifies to:

radiuselectron classical10817.2mce

r 152

2

0 m−==

02

220 I

12

)2(cos1r)I(2

r⋅

+⋅=

θθ

6

Energy out /energy in

0220e I

r1

r)(0I =For unpolarized X-rays at 2Ө = 0:

I0

1m2

r

Ie(2θ) r2dΩ

Detector

sample

22

0

e brI

) (2θIdΩdσ

=⋅=

b: scattering length =

r0 = 2.810-15 m for the electron

Energy scattered/unit solid angle/unit time

Energy incident/unit area/unit time

The differential scattering cross-section

has the dimension of an area and represents

For one electron: the amplitude of scattering |Ie(0)|1/2= 2.810-15|I0|1/2 and as the scattering amplitude ≡ f =

amplitude scattered by an objectamplitude scattered by an electron in identical conditions

fe =1

7

Particle size (D) and wavelength of the radiation

When λ >> D all N electrons in the particle are accelerated in phasethe scattering amplitude is N times that of one electron.

When λ < D the electrons in the particle are no longer moving in phaseand one has to take the phase shift of the waves into account.

ObserverObserver

λ >> D Light scattering λ < D X-ray scattering

8

Waves and Interference

Interferences lead to fringe patterns. This is

illustrated here with water waves.

When “solving” a structure the problem is to

go from the fringe pattern – in the case of

X-ray diffraction from the intensities of the

fringes – to the distribution of sources i.e.

of scatterers.

Similar effects are observed with optical

transforms obtained by shining coherent visible

(laser) light through small apertures (see e.g.

Cantor and Schimmel , Biophysical Chemistry,

Part II, Ch. 13).

9

Interference and coherent scattering

r

2Θ

r.S

r.S0

λ)( 0SS

s−

=

λ2sinθ

=s

S/λ

S0/λ

The total amplitude from two centers (one at the origin and one at r) is thus:

)rs i exp(2πff)sr i exp(2πfF(s) 2eei

2

1ie ⋅+==∑

=

0SrSr ⋅−⋅Path difference =

In Thomson (coherent scattering) the scattered wave is 180° out of phasewith the incident wave.

Phase difference = )SrS(r 0⋅−⋅λ

π2

10

The sum of amplitudes for N electrons:

F(s) is the Fourier transform of the distribution of electrons

) i exp(2πf)F(N

1ie irss ⋅=∑

=

The average of the exponential factor over all orientations of r relative to s

for randomly oriented particles (e.g. in solution) is:

sr

sr

π

π

2)2(sin

) i π2(exp =>⋅< irs

As this is a real number there is no phase problem but one has lost most of the structural information.

11

Scattering factor

For λ = 0.15 nm f’ = 0.3641f” = 1.2855

drsrπ2

)srπsin(2rρ(r)π4f(s)F(s) 2

0∫∞

=≡

For an atom with a continuous radial electron density ρ(r):

1sin

lim0

=

→xx

x

and since

f(0) = Z

( )( ) ( )( )sRsRsR

sRsf sphere πππ

π2cos22sin

2

3)( 3 −=

For modeling purposes one oftenuses larger spherical subunits (beads,dummy residues, etc) for which:

12

Anomalous scattering

The scattering factor fe must be modified to take anomalous scattering into account

fe = 1 + fe’ + ife”(f” is always π/2 ahead of the phase of the real part)

Note that for most practical purposes, f’ and f” are independent ofs = 2sinθ/λ

KronigKramers'dωω'ω

)'(ω"fω'π

2)(ω'f

weightatomicA number, sAvogadro' :N

t,coefficien absorption tric)(photoeleclinear :µ )ω(µcNr4πωA

0)ω,("f

22

A

A0

−−

=

=

∫

Near an absorption edge, the dissipative effects due to the rearrangement of the electrons can no longer be neglected.

13

Selenium f‘ and f‘‘

SeK

)ω(µcNr4πωA

0)ω,("fA0

=

'dωω'ω

)'(ω"fω'π

2)(ω'f

22∫ −=

Note the sign and absolute value of the corrections.

14

Scattering from N spherical atoms:

F(s) is the Fourier transform of the distribution of the spherical atoms.Crystallographers call this the structure factor. Note that in SAXS the structure factor refers to structure of the solution. The intensity is, of course:

) i (s)exp(2πf)F(N

1ii irss ⋅=∑

=

ij

ijj rsπ2

)rsπ2(sin)-( i π2(exp =>⋅< rrs i

))-( i (s)exp(2π(s)ff)I(1i

j1j

ji∑∑= =

⋅=N N

rrss i

For random orientation

ij

ijN

1i

N

1jji sr2π

)srsin(2π)(s(s)ffI(s) ∑∑

= =

=

and

Debye (1915)

Crystal structure -atomic coordinates

Solution –Distance distribution only!

This is a real number!

15

Short distances >> low frequencies dominate athigh anglesLarge distances >> high frequencies contribute only at low angles.

In isotropic systems, each distance d = rij contributes a sinx/x –like term to the intensity. A scattering pattern is a continuous function of s.

SAXS:ij

ijN

1i

N

1jji sr2π

)srsin(2π)(s(s)ffI(s) ∑∑

= =

=

16

The wider a function in real space the narrower its transform in reciprocal space

1) The Fourier transform of the Dirac delta function

is the 1(x) function (i.e. the function which has a constant value of 1 over the interval [-∞,∞].2) Obviously, the Fourier transform of 1(x) is δ(x).

∫∞

∞−=∞= 1δ(x)dx δ(0)

3) The Fourier transform of a Gaussian is also a Gaussian

)/exp()())(exp( 222ak

akFaxFT π

π−==−

Note the relationship between the widths. If the Gaussian has a widthσR=(1/2a)½, its transform has a width σF=(a/2π2)½ and σRσF=1/2π.

The δ-function is an infinitely narrow Gaussian.

17

Inte

nsi

ty

10.0

1.0

0.10.02 0.04 0.06Q (Å-1)

0.1 s ID2 (ESRF)

Kinetics of the Ca2+-dependent swelling transition of Tomato Bushy Stunt Virus

Perez et al.

Larger objects scatter at lower angles!

18

In an ideal solutionThe solute particles are randomly oriented and their positionsare uncorrelated in space and time. Consequently their scattering inisotropic and incoherent. The total scattering intensity is the sum ofthe coherent scattering intensity of all molecules. It is a function of the scattering angle or modulus of the scattering vector only: I(s).

Usually one plots log(I(s)) vs s, because the intensity falls off rapidly due to the interferences.

i.e. for atoms one can neglect the s-dependence of fi

If one uses a continuous density distribution ρ(r) this becomes

ij

ijN

1i

N

1jji1 sr2π

)srsin(2πff(s)i ∑∑

= =

=

2112

1222111 2

)2sin()()()( rrrr dd

sr

srsi

V π

πρρ∫ ∫=

19

Interactions of X-rays with matter

Sample

fluorescence

Incident beamTransmitted beam

Coherent scattering

I0

I=I0 exp(-µt)

Incoherent scattering

Structural information at the atomic/molecular level is in: coherent scattering

and to a limited extent in absorption/fluorescence near edges (EXAFS, XANES)

At lower resolution transmission/phase contrast imaging is also useful

20

Choice of wavelength

Absorption!!Incoherent!!

Optimal thickness of the sample: topt =1/µ ≈ 1 mm for H2O @ 1.5Å

For oxygen

21

Background

solution

buffer

with air gap

vacuum

Background arises from the incoherent(Compton) scattering from the sample and from coherent and incoherent scattering due to air gaps, windows, solvent ….

To obtain the coherent scattering of the solute normalize the intensities ofsolution and buffer to transmitted beam and subtract :

I(s)= [I(s)/I0T]solution – [I(s)/I0T]buffer

Divide by c to normalize for concentration

Lysozyme : 14 289 Da MW5 mg/ml solution in Acetatebuffer pH 4.5

22

CONTRAST: < ρparticle(r) > - ρbuffer

Homogeneous solvent

Particle:

Solvent:

Only fluctuations in electron density contribute to the scattering:

rrsrs )diexp(2π)(ρ)(FV

pp ⋅= ∫

rrss )diexp(2πρ)(FV

bb ⋅= ∫ Ideally, a Dirac δ(0) and constantin practice not

Iobs(s) = Ip(s) –Ib(s)

23

Isolution(s) Isolvent (s) Iparticle(s)

♦ To obtain scattering from the particles, solvent scattering must be subtracted to yield the excess scattering

ρp(r) = ρsolution(r) - ρb

where ρb is the scattering density of the solvent

Solvent scattering and contrast

24

Scattering arises from fluctuations

In a perfectly homogeneous body the contribution of each smallscattering volume element to the scattering amplitude would always becancelled out by that of another one which is out of phase. This is why perfect crystals do not scatter visible light.Solutions of macromolecules or colloidal suspensions strongly scattervisible light because of fluctuations in concentration. The fluctuations are also what links scattering to diffusion (e.g. DLS) andthermodynamics (compressibility).

25

EXCESS SCATTERING DENSITY

(Stuhrmann and Kirste, 1965)

)(ρρ-)(ρ)(ρρ) ρ( cbscx rrrr +=

shape

internalstructure

equivalent volume of solvent

)(ρ)(ρ )ρ-ρ( )ρ( scbx rrr += )(ρ)(ρ ρ sc rr +=

contrast

(s)I(s)Iρ(s)IρI(s) scsc2 ++=The corresponding intensity is:

For a homogeneous particle: Is(s) =Ics(s) =0

ρc (r) has a value of 1 inside the particle and 0 outside and thusrepresents the shape. ρs(r), the internal structure, represents the fluctuations around the average electron density of the particle ρx.

26

A communication problem

(s)I(s)Iρ(s)IρI(s) scsc2 ++=

For a homogeneous particle (i.e. a shape): Is(s) =Ics(s) =0

Is(0) =Ics(0)=0 always!

These two terms are the mostimportant in SAXS

This is essentially whatcrystallographers talk about

27

Scattered intensity

)F()F()(F)F()(i *1 sssss −⋅=⋅=

)()()]([

)()]([* ssr

sr

FFFT

FFT

=−=−

=

ρ

ρρ(r) is the excess electron density

Hence )]()([)]([)]([)( *1 rrrrs −=−⋅= ρρρρ FTFTFTi

γ(r): Autocorrelation function of the excess electron density and:

γ (r)

∫ ⋅==

rV

rdViFTsi )2exp()())(()(1 rsrr πγγ

drrs

rsrpsi ∫

∞

=0

1 2)2sin(

)(4)(π

ππ )()( 2

rrrp γ=

After spherical averaging:

with( ) ( )rγ γ= r

28

Autocorrelation: Shift-Multiply-Integrate

∫ +==−=

u

uuurrrrrrV

dV)()()()()()()( * ρρρρρργ o

-r

u

Dmax

2Dmax

r

ρ(u)

γ(r) p(r)

r

× r2Dmax

0

29

2 2 20( ) ( ) ( )p r r r r r Vγ γ ρ= =

rij ji

r

p(r)

Dmax

For a homogeneous body ρ(r) = ρ, γ(0) = ρ2V

γ(r) : probability of finding a point at r from a given point (i).Number of volume elements i ∝ V;Number of volume elements j ∝ 4πr2.Number of pairs (i,j) separated by the distance r ∝ 4πr2Vγ0(r)=(4π/ρ2) p(r).

0( ) ( ) (0)r rγ γ γ= Characteristic function

Distance distribution function

30

I(s) and p(r)dr

sr

srrpsI

D

π

ππ

2)2sin(

)(4)(max

0∫=

drsr

srsIrrp ∫

∞

=0

2

2)2sin(

)(2)(π

π

For a homogeneous particle p(r) represents the histogram of distancesbetween pairs of points within the particle.

31

At low angles

( ) ( )L−+−=

1202

62

12

)2sin( 42srsr

sr

sr ππ

π

π

ij

ijN

1i

N

1jji1 sr2π

)srsin(2πff(s)i ∑∑

= =

=

At very low angles one can use the approximation: exp(x) = 1+ x + x2/2 +… which yields the Guinier formula

)sRπ34

(1Vρ(s)i 22g

2221 K+−=

massMolecular ~Vρ)0( )sRπ34

exp( I(0))( 2222g

2 =−= IwithsI

Since2

∑

i

if

Rg is the mean squared distance to the centre of scattering massweighted by the excess electron density.

32

Valid for a sphere for

0 < 2ππππRgs<1.2

The plot of ln[I(s)] vs s2

[ ] [ ]24

3π

≅ − 2 2

gln I(s) ln I(0) R s

yields two parameters :

y-intercept: I(0)Rg: from the slope

Guinier plot for ideal solutions

33

Forward scattering I(0) and molar mass (M)

2

´

´

´)()()0(excessi

ir

V V

r fdVdVrrI

r r

== ∑∫ ∫ ρρ

2

02

0 )()()0(

−=−= ρρν ρ

AN

MmmI

VN

NMc

A

=

[ ]20 )()0( ρρν ρ −=

AN

cMVI

concentration (w/v)

Molar mass

Avogadro’s number

Number of electrons contrastPartial specific volume

∫=max

0

)(4)0(D

drrpI πor

number ofmolecules

Tip: Use a sample of similar contrast e.g.a known protein ( BSA) as standard to calibrate the measurements:MMsample= MMstand.I(0)/c)sample/(I(0)/c)stand.

34

Radius of gyration

2121

21

2

21212

)()(2

)()(

rrrr

rrrrrr

dd

ddRg

∫ ∫∫ ∫ −

=ρρ

ρρ

If one places the origin of the coordinates at r0,,the centre of scatteringmass of the particles, this yields the expression for the radius of gyrationwhich is obtained from a Guinier plot, Or even better, calculated fromthe whole experimental scattering pattern:

∫

∫=

V

Vg

d

dr

Rrr

rr

)(

)( 2

2

ρ

ρ

∫

∫=

max

max

0

0

2

2

)(2

)(

D

D

g

drrp

drrpr

R

Rg is the second moment (standard deviation) of the electron density distribution.

35

Guinier’s formula and the contrast

1sRπ34

for sRπ34

exp Vρ )sRπ34

(1VρI(s) 2222222222222 ≤

−=+−= K

where R is the radius of gyration and I(0) the forward scattering is: 22Vρ

If the contrast changes so do I(0) and R:

2

2c

2

ρ

β

ρ

αRR −+=

second moment ofinternal structure

displacement of centre of scattering mass with contrast

α = 0α = 0α = 0α = 0 α > 0α > 0α > 0α > 0 β=0β=0β=0β=0 β>0β>0β>0β>0αααα < 0

radius of gyration at infinite contrast

36

Contrast: X-rays

Substance

X-

rays

Proteins 2.5

Nucleic acids 6.7

Fatty acids -1.1

Carbohydrates 4.5

Average contrast (X 1010 cm-2) of biological macromolecular assemblies in water.

One can change the contrast e.g. by adding salts like CsBr but this has disadvantages like increasing absorption,fluorescence and changing ionic strength.

The alternative is to use anomalous scattering (see e.g Stuhrmann HB.Acta Crystallogr A. 2008, 64:181-91)but beware of radiation damage.

TIP: If you need to change the contrastUSE NEUTRONS!

37

Typical values of radii of gyrationMw Rg(nm)

Ribonuclease 12700 1.48Lysozyme 14800 1.45B-lactoglobulin 36700 2.17Bovine serum albumin 68000 2.95

Myosin 493000 46.8Brome mosaic virus 4.6 106 13.4Tobacco mosaic virus 3.9 107 92.4

For a sphere of radius R: 22

53

RRg =

38

For an infinite rod

)(1)()(),0,0( zyxf δδ=∞ )()(1)(1)0,,( ZYXF δ=∞∞

or a long fiber with its axis along z, the transform is limited to theX,Y plane (i.e. the cross-section)

z

xy

Z

X

YFT

Remember δ(x)!

In such a case the radius of gyration of the cross-section and themass/unit length can be derived using a representation analogousto the Guinier plot with a plot of sI(s) vs s2 to obtain

)2exp()( 222sR

L

mssI cπ−= For a circular cross-section

Rc=R/√2

39

For a flat object like a membrane

)()(1)(1)0,,( zyxf δ=∞∞ )(1)()(),0,0( ZYXF δδ=∞

With a small thickness (T) along z, the transform is limited to theZ-axis (i.e. the thickness)

z

xy

Z

X

YFT

Remember δ(x)!

In such a case the radius of gyration of the thickness and themass/unit area are obtained from a plot of s2I(s) vs s2 to obtain

)4exp()( 2222sR

A

msIs Tπ−=

12/TRT =with

40

Shape scattering

At larger s-values the scattering ofa particle with ρs ≠ 0 oscillates around a straight line given byPOROD’s law:

s4I(s)= Bs4 +A

Subtract a constant equal to theslope of the Porod plot toobtain an approximation to the

SHAPE scattering

41

Mixtures

)0(/)0( )()(N

1i

22

1

IRInRsInsI iii

N

i

ii ∑∑==

==

ni: number concentration of the ith species, with forward scatteringIi(0) and radius of gyration Ri.

These formula illustrate that a small quantity of high molecularmass or hight contrast particles will have a large influence onthe scattering curve.

For modeling it is thus indispensable to make sure that the solutions do not contain aggregates!

42

Proteins at low resolutionHydration shell

Crowding max. conc. 300-500mg/ml

IONS:Kosmotropes e.g. Na+

Chaotropes e.g. K+

OSMOLYTESe.g. free amino acidspolyhydroxy alcoholsmethylated ammoniumand sulfonium compoundsurea.

solvent

Interactions/ stability/activitymodulated by

FOLDING

Coupled equilibriaNon-contact interactions

43

In the case of an unfolded protein : models developed for polymers

Gaussian chain : linear association of N monomers of length l with no persistence length (no rigidity due to short range interactions between monomers) and no excluded volume (i.e. no long-range interactions).

Debye formula : where

I(s) depends on a single parameter, Rg .

Valid over a restricted s-range in the case of interacting monomers

Limit at large s :

2

( ) 2( 1 )

(0)xI s

x eI x

−= − + ( )2

2 gx R sπ=

22

2 2

1 1 (2 )lim [ ( )]

2g

s

g

R ss I s

R

π

π→∞

−=

I(s) varies like s-2 instead of s -4 for a globular particle (Porod law).

Scattering by an extended chain

44

arrows : angular range

used for Rg determination

Pérez et al., J. Mol. Biol.(2001)

308, 721-743

Debye law for extended chains: NCS unfolding

2

( ) 2( 1 )

(0)xI s

x eI x

−= − +

( )2

2 gx R sπ=

Neocarzinostatine : small (113 residue long) all-β protein.

45

Is sensitive to the degree of compactness of a protein.

Globular particle : bell-shaped curve

Gaussian chain : plateau at large s-values but a plateau does not imply a Gaussian chain!

Kratky plot: s2I(s) vs s

Pérez et al., J. Mol. Biol.(2001), 308, 721-743

In thermal unfolding of NCS the Kratky plot has a plateau althoughunfolded NCS is not a Gaussian chain when unfolded.A thick persistence chain is a bettermodel in this case.

46

Protein folding: cytochrome c

Akiyama, S. et al. (2002) Proc. Natl. Acad. Sci. USA, 99, 1329-34

47

Protein folding: cytochrome c

Akiyama, S. et al. (2002) Proc. Natl. Acad. Sci. USA 99, 1329-34

?InitialEnsemble ?

expect >900 Å2

for randomchain !

48S. Akiyama et al. (2002), PNAS, (2002), 99, 1329-1334.

160 µs after mixing

44 ms after mixing

cytochrome c folding

49

Labeling

D L1 L2D X X XL1 X X XL2 X X X

D L1 L2D X 0 0 L1 0 0 0L2 0 0 0

D L1 L2D X X 0L1 X X 0L2 0 0 0

D L1 L2D X 0 XL1 0 0 0 L2 X 0 X

+ +-

L1 D L2

L1-D or L2-D

L1-D-L2

D InterferenceFT

Distance betweenL1 and L2

Matthew-Fenn R.S. et al. (2008)Science 232, 446-9

50

35

30252015

10bp

Note the difference in position and width(variance) of the distribution of end to end distances as a function of the number ofbase pairs between labels. In absence of applied force DNA is at least1000 softer than in single molecule stretching experiments. Stretching is cooperative over more than two turns of thedouble helix. DNA is not an elastic rod.

Interference

FT

51

What SAXS/SANS have to offer

SAXSSANS

Electron microscopy

NMR

Modelling

Light scattering

Molecular Dynamics

ultracentrifugation

viscosity

Electric or flowdichroism

chromatography

Perturbation methodsLabeling methods