S²C² Workshop – Cryo-EM Imaging Process workshop, 2022

EM Images:

what you want vs.

what you get and why

David DeRosier

Brandeis University

EM images

Are

projections

In 1968, Klug and I realized that EM images could be treated as projections of the 3D structure.

But EM images are NOT perfect projections of the desired 3D structure. Here are the problems:

Underfocus affects amplitudes and phases.

Electrons damage the specimen.

Insufficient depth of field alters amplitudes and phases.

Image distortion by the lenses affects phases.

DeRosier and Klug, Nature, 130-134, 1968.

By collecting a set of views, we were able to make the first 3D reconstruction of a structure from electron micrographs.

Beam induced motion limits resolution.

Digitization and masking of the image affects and limits amplitudes and phases.

Interpolation reduces high resolution amplitudes.

Beam tilt alters phases.

Lack of plane parallel illumination alters phases.

Multiple scattering alters amplitudes and phases.

Coherence of the electron beam limits resolution.

Not all scattered electrons are imaged.

The Fourier transform (FT) of an EM image is used to detect many of these problems.

The FT of a sample describes the outcome of all diffraction experiments^* at any wavelength and in any direction of the incident radiation.

The Ewald sphere describes the portion of the FT seen in any one experiment:

* in the linear approximation

To examine what amplitude and phase (if recorded) are measured by a diffracted ray, draw an arrow from the center of the sphere in the direction of the diffracted ray. The point in the FT where the arrow intersects the Ewald sphere is the element of the FT sampled.

By sweeping the direction of illuminating radiation through all angles, one can measure the FT out to a distance of 2/λ. Hence the highest possible Fourier (not Rayleigh) resolution is λ/2.

3D density=ρ(x,y,z)

Fourier transform from Pyimagesearch.com

image space

x,y

FT (reciprocal) space

X,Y

pixel size = Δx

image dimension = NΔx

reciprocal pixel size ΔX =1/(N Δx)

reciprocal dimension = 1/Δx

The radius of the ring (=1/2Δx) defines the Nyquist limit, which gives the highest feasible resolution: (1/(2*0.75) = 1/1.5 A^-1 or 1.5 A resolution).

vs.

FT

IFT

If Δx is 0.75 A and N is 128, then ΔX = 1/(128*0.75) = 0.0104 A^-1.

Digital FTs

The FT is a mathematical operation, which I will not describe.

To get a FT, you feed your digitized image into a FT program.

The question I address is what you do with the FT.

It has information about your sample and about the imaging conditions.

The “short” exposure

The “longer” exposure

Optical FT

tells you about sample.

It’s a crystal.

You can get unit cell dimensions.

You can also get amplitudes and phases.

tells you about imaging conditions.

• Erickson H. P. and Klug A 1971 Measurement and compensation of defocusing and aberrations by Fourier processing
• of electron micrographs Phil. Trans. R. Soc. Lond. B261105–118

Amount of defocus can be determined from the positions of the Thon rings

Amount & direction of astigmatism can be obtained from the ellipticity of Thon rings.

Stronger here

Weaker here

darker

lighter

Amount & direction of instability (e.g., stage vibration) can be determined from the asymmetric intensity of Thon rings

Bammes BE, Jakana J, Schmid MF, Chiu W J Struct Biol 169, 331, 2009

Maximum tolerable electron dose is revealed by the fall off of spot intensity with dose

Resolution is limited by the wavelength, λ, of the radiation. At best, resolution is ~λ/2.

The wavelength of a moving particle, like an electron, is given by λ=h/p (h is Planck’s constant and p is the relativistic momentum). For a 300 keV electron, the wavelength is ~0.02 A.)

If the green ball (28 gm) is dropped from the height of one meter, its wavelength is ~5*10^-23 A. We should see the quarks (10^-11 A ?) in a proton (10^-5 A) with a green ball microscope.

In grad school, I asked my professor: Why can’t we make a microscope using for example green balls?

Resolution and radiation damage

My professor said my question was frivolous and dilatory, but was it?

What is wrong with λ=h/p?

The interaction of a particle with a sample involves two cross-sections: the elastic and the inelastic cross-sections or probabilities.

In an elastic interaction, the electron is scattered but does not lose energy. These electrons remain coherent and contribute to the image of one’s structure.

In an inelastic collision, the electron transfers energy to the structure and loses its coherence. These electrons damage the specimen and contribute to the background/noise.

For every elastically scattered electron, which leaves the sample unchanged, there are ~5 inelastically-scattered electrons that transfer energy to the sample and damage it. Controlling damage is essential to high resolution cryo-microscopy.

Guess the odds that the green ball will pass through an EM grid without banging into it.

Nothing, but wavelength is not the only consideration.

The probability that an electron is scattered is a function of sample thickness.

Mean free path is the distance in which fraction of unscattered electrons is = e^-1 = 0.37

Thinner is better.

=> thickness

fraction of unscattered electrons

Electron Crystallography of Biological Molecules, Glaeser, Downing DeRosier, Chiu, Frank

Questions or comments about Fourier Transforms of images or electron scattering?

Scattering contrast is a form of amplitude contrast

Heavy metals are strong scatterers of electrons and produce contrast by scattering electrons outside the imaging system.

The thicker the heavy metal layer the fewer the number of electrons at that position in the image. Shadowing with platinum, osmium staining of sections, and negative staining with uranyl salts are ways of using heavy metals to produce contrast.

In negative stain, the structure is seen as a hole in the layer of stain. (NOTE: The contrast is opposite to that in frozen hydrated images.)

structure of interest

This is what you see.

Negative stain is particulate, which prevents detail finer than ~15 A from being seen.

The beautiful and important internal features are not seen.

Negative stain

Pro

Fast

Easy

Acts as a fixative

Generates high contrast

Any TEM will do

Gives faithful info on molecular detail (>15A)

Is radiation resistant

Con

Cannot see particle itself

Cannot see secondary structure

Distorts/shrinks/flattens structure

Has opposite contrast to ice images

Some reviewers wrongly do not accept negative stain results

Questions or comments about negative stain?

To understand phase contrast, we can think of electrons as waves in the same way light can be thought of as waves.

​

Waves can be decomposed into parts, which can be altered and then added back.

​

We shall see that generating phase contrast involves such a process.

​

We begin by seeing how a wave front of electrons is altered by our specimen.

Phase contrast

Image formation of particles in ice.

Lens

Transparent

object

Scattered beam

Unscattered beam

Scattered beam

Image plane

Image formation by a perfect lens

The transparent object scatters or diffracts pairs of beams at symmetric angles.

The optical path lengths for the scattered and unscattered beams are the same so that the original wave front is regenerated.

Diffraction plane

Lens

Thus, the imaged wavefront has the same bumps that arose when the plane wavefront passed through the sample. They are bumps in wavefront shape and not in wavefront amplitude. Thus an in-focus image has no contrast.

How can phase contrast be generated in the electron microscope?

An electron wave passing though a particle is advanced compared to the unscattered wave.

The deformed part is the sum of the unscattered wave + the scattered wave:

​

cos(wt-d) ≈ cos(wt) + d*sin(wt) when d is small vs λ

How to think about phase and phase contrast.

The deformed or advanced part of the wavefront can be described by cos(wt – d) compared to cos(wt) for the undeformed part or rest of the wavefront.

(w is frequency, t is time and d is the advance)

sum of blue & red

time ->

Lens

How underfocusing generates phase contrast

At the image plane, the scattered and unscattered beams arrive at the same time as expected.

In underfocus, the image plane lies before the in-focus plane.

At the underfocus plane, however, the scattered beam arrives ahead of the unscattered beam. (It has to arrive earlier because it has farther to go to get to the focal plane on time, that is, in phase.)

In focus i.e., image plane

Underfocus plane

Because the scattered wave arrives ahead of the unscattered wave, it generates destructive interference. The structure appears dark. Here is why:

the sum of the red and blue waves gives a change in phase

The larger the scattering angle, the larger the path difference of the scattered wave relative to the unscattered wave. As a result the contrast varies with scattering angle and explains the Thon rings.

scattered waves

unscattered wave

focal plane

plane of underfocus image

increasing scattering angle

How to interpret the Thon rings

Thon rings appear in the diffraction pattern of the image.

Thus under focusing produces phase contrast but not in a simple way.

JT Finch, J Gen Virol. 24, 359, 1974

Howley PM, Lowy DR. Papillomaviruses. In: Knipe DM, Howley PM. (Eds.) Fields’ Virology, Philadelphia: Lippincott Williams and Wilkins; 2007. p 2300-2354.

Michele Lunardi, Amauri Alcindo Alfieri, Rodrigo Alejandro Arellano Otonel and Alice Fernandes Alfieri (November 20th 2013). Bovine Papillomaviruses Victor Romanowski, IntechOpen, DOI: 10.5772/56195.

negative stain

ice

Scattering contrast + a little underfocus phase contrast

Underfocus phase contrast

ctf(α) = -A(α)sin(2π/λ[α²Δz/2 – α⁴C_s/4])

For a phase object using defocus contrast:

The form of the Thon rings is given by the contrast transfer function (CTF):

How defocus affects the Thon rings in the FT

Where α is the scattering angle,

A(α) is the envelope function,

​

λ is the wavelength,

​

Δz is the defocus,

​

Cs is the spherical aberration coefficient.

By comparing the 1D curve of the CTF² with the observed Thon rings, we can determine the amount of defocus.

​

We may also have to determine the spherical aberration coefficient, Cs.

Questions or comments defocus phase contrast?

Phase plate contrast

Biophysics (Nagoya-shi). 2006 Jun 6;2:35-43. eCollection 2006.

Applicability of thin film phase plates in biological electron microscopy.

Danev R¹, Nagayama K¹.

Back focal (diffraction) plane

sample plane

The unscattered beam passes through a hole while the scattered beam passes through a carbon film, which advances it by ~90^o relative to the unscattered beam.

The Volt Phase Plate (VPP) has a continuous layer of carbon. A negative voltage is generated at the center and delays the unscattered beam by ~90^o, which has the same effect as advancing the scattered beam in the Zernike phase plate.

Even with the phase plate, there is defocus or spherical aberration, which must be corrected.

Where α is the scattering angle,

A is the envelope function,

λ is the wavelength,

Δz is the defocus, and

Cs is the spherical aberration coefficient.

For a phase object using defocus contrast:

The form of the Thon rings is given by

the contrast transfer function (CTF):

CTF for defocus vs. phase plate

ctf(α) = -A(α)sin(2π/λ[α²Δz/2 – α⁴c_s/4])

Images are okay out to the first zero in the CTF but after that, we need to correct for the reversal in contrast.

​

The needed corrections will be covered in Fred Sigworth’s lecture.

Corrections for defocus

Questions or comments about phase plates?

Image defects

How accurate does the determination of defocus need to be?

A typical defocus might be about 1 micron (= 10,000 A), you need to refine that measure to about +/- 100 A or 1%.

Coherence of the electron gun limits resolution

With a LaB₆ gun at 100 keV, it was hard to get below ~10 A resolution.

Beam induced motion and/or drift cause loss of resolution preventing resolutions below 4A.

J Struct Biol. 2012;177:630-7. Beam-induced motion of vitrified specimen on holey carbon film.Brilot AF, Chen JZ, Cheng A, Pan J, Harrison SC, Potter CS, Carragher B, Henderson R, Grigorieff N.

The direct electron detectors made it possible to break a single exposure into a movie of many frames and correct for the motion.

Each averaged frame corresponds to 0.25 s.

​

Dose/frame = 5 e^-/Ų

10-Frame Averages

Beam induced motion causes blurring

Image corrected for beam induced motion

With these new cameras, resolution dropped from 4A to better than 2 A in some cases.

Distortions can affect phases

Capitani et al. Ultramicroscopy 106, 66, 2006

pincushion

barrel

spiral

Elliptical:

Magnification varies with angle

1% variation in one Krios TEM

For a 150 A particle being solved to 1.5 A, variations in the magnification must be corrected to within ~ 0.25%

Digitizing the image

Digitizing an image limits the resolution of the data.

Digitizing the image can also cause aliasing in which high resolution Fourier terms are added into low resolution terms.

unsampled image

FT

finely sampled image

FT of the sampled image is an array of FTs

Nyquist

Limit fixes

resolution

+

+

coarsely sampled image

badly aliased example

There is a slight problem with looking at the FT of a non-square image. While the x,y spacings in the image are the same (0.75 A), the X,Y spacings in the transform are not the same size.

Fixing the dimensions of the FT

The reciprocal pixel sizes are different if the X and Y directions!

Howley PM, Lowy DR. Papillomaviruses. In: Knipe DM, Howley PM. (Eds.) Fields’ Virology, Philadelphia: Lippincott Williams and Wilkins; 2007. p 2300-2354.

Michele Lunardi, Amauri Alcindo Alfieri, Rodrigo Alejandro Arellano Otonel and Alice Fernandes Alfieri (November 20th 2013). Bovine Papillomaviruses Victor Romanowski, IntechOpen, DOI: 10.5772/56195.

When windowing a particle for processing, you might be tempted to use a tight mask to reduce the contribution from the background. That might be a mistake because you could be eliminating high resolution information.

​

Here is why:

In an under focussed image, the highest diffract beams lie outside the apparent image made by the lower resolution diffracted beams.

The mask to cut out a particle needs to be bigger than the apparent image of the particle.

to make the reciprocal pixels have the same size, embed the image into a square array:

The process is called padding, but there is a problem.

Note the big jumps in density at the edges

If you have an oblong particle and you need

To get the input you want, you float the image by subtracting average density at the box’s edge.

the particle density in the array seen in 1D

This is a 1D version.

density

0

This is what you want to input into the FT program.

Floating may include removing a gradient of density.

Padded image

Gradient-removed, floated, apodized, padded image

Apodizing the image edges may also be necessary.

The effect of interpolation

To turn the image upright, you interpolate the densities into an upright box.

But if you have a pixel size of 1 A, and you use bilinear interpolation, the amplitudes at a resolution of 3A will be reduced by a factor of 2 and the 2 A amplitudes are reduced to 0.

There are better ways to interpolate data, but I leave that to others to discuss.

The effect of incident beam tilt on diffracted beams.

The left, L, and right, R, diffracted waves travel different path lengths when the beam is tilted.

L

R

As a result, the higher resolution waves do not interfere as they should, and their measured phases are incorrect.

If the incident beam is convergent or divergent, objects away from the center of the field suffer from beam tilt.

Another way of thinking about the effect of beam tilt.

Corrections for curvature of the Ewald sphere are built into Relion and FREALIGN

Lack of depth of field due to curvature of the Ewald sphere causes amplitude and phase errors.

origin

Waves that are diffracted or scattered more than once have inacccurate amplitudes and phases.

If the probability of a scattering event is p, the probability of the beam being scattered twice is ~p².

The amplitude and phase of the resulting beam is a mixture of two Fourier terms.

We have covered the reasons why images are not perfect projections.

The following talk discusses what can be done to fix the problems.

The End

Questions?

My power point and a treatise I wrote on the FT are available.

The amplitude of the plane wave just before it hits the specimen can be described as

​

A(-δ,y) = 1

After passing through a transparent specimen, the amplitude of the plane wave becomes

​

A(0,y) = e^iρ(y) where ρ(y) describes the variation in path length along y

If ρ is small (the thin phase object approximation), we can write:

​

A(0,y) ≈ 1 + i*ρ(y)

Let us assume for simplicity that ρ(y) = a*sin(y), where a<<1 (i.e., ρ is small).

​

A(0,y) ≈ 1 + i*a*sin(y)

At the perfect defocus plane for this periodicity, we would find destructive interference:

​

A(defocus,y) ≈ 1 – a*sin(y). At y = 0, we would see the maximum contrast.

Suppose instead that ρ(y) = a*sin(y -c), then we would find

A(defocus,y) ≈ 1 – a*sin(y-c). At y = c, we would see the maximum contrast.

The sine wave is shifted by c!

What you see. What you get

Spots You know the specimen is crystalline

What you see. What you get

Spot positions Unit cell size and shape

Spots You know the specimen is crystalline

What you see. What you get

Spot positions Unit cell size and shape

Spot size Size of coherent domains

Intensity relative to background Signal/noise ratio

Distance to farthest spot Resolution

Amplitude and phases of spots Structure of molecules

Spots You know the specimen is crystalline

What you see. What you get

Spot positions Unit cell size and shape

Spot size Size of coherent domains

Spots You know the specimen is crystalline

What you see. What you get

Spot positions Unit cell size and shape

Spot size Size of coherent domains

Intensity relative to background Signal/noise ratio

Spots You know the specimen is crystalline

Movie

Recorded with direct electron detector DE-12 (Direct Electron)

​

Frame rate = 40 fps

Dose/frame = 0.5 e^-/Ų

Duration = 1.5 s

No. of frames = 60

Total dose = 30 e^-/Ų

Accumulating Average

Average shows emerging detail as noise decreases.

Particles are blurred later in the movie due to beam-induced movement.

What you see. What you get

Spot positions Unit cell size and shape

Spot size Size of coherent domains

Intensity relative to background Signal/noise ratio

Distance to farthest spot Resolution

Spots You know the specimen is crystalline

What you see. What you get

Spot positions Unit cell size and shape

Spot size Size of coherent domains

Intensity relative to background Signal/noise ratio

Distance to farthest spot Resolution

Amplitude and phases of spots Structure of molecules

Positions of Thon rings Amount of defocus

Ellipticity of Thon rings Amount of astigmatism

Asymmetric intensity of Thon rings Amount of instability

Direction of asymmetry Direction of instability

Spots You know the specimen is crystalline

How underfocusing generates phase contrast

At the focal plane, the scattered and unscattered beams arrive at the same time as expected.

Lens

Transparent

object

Scattered beam

Unscattered beam

In underfocus, the image plane lies before the in-focus plane.

What you see. What you get

Spot positions Unit cell size and shape

Spot size Size of coherent domains

Intensity relative to background Signal/noise ratio

Distance to farthest spot Resolution

Amplitude and phases of spots Structure of molecules

Positions of Thon rings Amount of defocus

Spots You know the specimen is crystalline

What you see. What you get

Spot positions Unit cell size and shape

Spot size Size of coherent domains

Intensity relative to background Signal/noise ratio

Distance to farthest spot Resolution

Amplitude and phases of spots Structure of molecules

Positions of Thon rings Amount of defocus

Ellipticity of Thon rings Amount & direction of astigmatism

Spots You know the specimen is crystalline

What you see. What you get

Spot positions Unit cell size and shape

Spot size Size of coherent domains

Intensity relative to background Signal/noise ratio

Distance to farthest spot Resolution

Amplitude and phases of spots Structure of molecules

Positions of Thon rings Amount of defocus

Ellipticity of Thon rings Amount & direction of astigmatism

Asymmetric intensity of Thon rings Amount & direction of instability

Spots You know the specimen is crystalline

What you see. What you get

Spot positions Unit cell size and shape

Spot size Size of coherent domains

Intensity relative to background Signal/noise ratio

Distance to farthest spot Resolution

Amplitude and phases of spots Structure of molecules

Positions of Thon rings Amount of defocus

Ellipticity of Thon rings Amount & direction of astigmatism

Asymmetric intensity of Thon rings Amount & direction of instability

Spots You know the specimen is crystalline

Fall off of spot intensity with dose Maximum electron dose