This class covers the fundamental principles underlying cryo-electron microscopy (cryo-EM) starting with the basic anatomy of electron microscopes, an introduction to Fourier transforms, and the principles of image formation. Building upon that foundation, the class then covers the sample preparation issues, data collection strategies, and basic image processing workflows for all 3 basic modalities of modern cryo-EM: tomography, single particle analysis, and 2-D crystallography. Philosophy: The course emphasizes concepts rather than mathematical details, taught through numerous drawings and example images. It is meant for anyone interested in the burgeoning fields of cryo-EM and 3-D EM, including cell biologists or molecular biologists without extensive training in mathematics or imaging physics and practicing electron microscopists who want to broaden their understanding of the field. The class is perfect as a primer for anyone who is about to be trained as a cryo-electron microscopist, or for anyone who needs an introduction to the field to be able to understand the literature or the talks and conversations they will hear at cryo-EM meetings. Pre-requisites: The recommended prerequisites are college-freshman-level math, physics, and biochemistry. Pace: There are 14.5 hours of lecture videos total separated into 40 individual “modules” lasting on average 20 minutes each. Each module has at the end a list of “concept check” questions you can use to test your knowledge of what was presented. As the modules are grouped into seven major subjects, one reasonable plan would be to go through one major subject each day. That would mean watching a couple hours of lecture and spending another hour or so thinking through the concept check questions each day for a week. Another reasonable plan would be to go through one module each day for a little over a month, or even three modules a week (Monday, Wednesday, and Friday) for a 3-month term. It is likely that as you then move on to actually begin using a cryo-EM or otherwise engage in the field, you will want to repeat certain modules.
CTF Correction
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来自 Caltech 的课程
冷冻电子显微镜（cryo-EM）入门
198 个评分
从本节课中
Image Formation
与讲师见面
Grant J. JensenProfessor of Biophysics and Biology; Investigator, Howard Hughes Medical Institute
Biology and Biological Engineering
Given that the contrast transfer function is like a very complicated
band pass filter, it's clear that EM images do not exactly
reproduce the density distributions present in the sample.
Instead, the images are blurred somewhat by missing and
damped and sometimes inverted Fourier components.
A term that we use to describe this blurring
effect is the Point [SOUND] spread [SOUND]
function [SOUND] or I'll abbreviate it as PSF.
What the point spread functions means,
is that if you had a single point in the sample.
And you imaged it, in an electron microscope.
What would the picture of that single point look like?
You'd like it, to look exactly like the original point,
that would be a perfect microscope.
But in fact, because of the problems inherent in any kind,
any real microscope, that point gets blurred a little bit and
it can sometimes even be surrounded by some halos, some fringes.
If we were to draw a cross section through that, and
plot it instead of looking like a delta function.
If we took an image of a delta function and
instead of looking like the delta function.
Instead it would look like a point with some spread, and
it might even have some fringes around it.
And, so that's, the shape of that is what we mean by the point spread function.
Just to show you what an actual point spread function looks like
in an electron microscope image.
I'm showing here,
a single slice from a Three-dimensional cryotomagram of a bacterial cell.
So here's the bacterial cell.
And as we'll talk about later,
to produce these cryotomagrams we often add to the sample gold beads.
This one, this gold bead, this gold bead.
These, these dark spots are the images of
gold beads that we've added to the sample that serve as fiducial markers.
And when you look carefully at this gold bead, you see a dark spot in the middle.
But it's surrounded by a white halo, and actually white streaks on either side.
This illustrates the point spread function,
in fact it was a single spherical gold ball with sharp edges.
But here in the image, it appears as a dark spot that's a little wider than
it should be with a white halo surrounding it and white streaks beyond that.
And this is the point spread function of this particular cryotomagram.
And as you can see, each of the gold beads which were each spheres
is surrounded by the same kinds of halos and streaks around it.
Now the relationship between the point spread function and
the contrast transfer function, is that they are fourier transforms of each other.
So the point spread function is equal to the fourier
transform of the contrast transfer function.
So in the example of a delta function,
if the object that we're trying to image is a delta function.
The question is, what will it look like in our image?
[SOUND] And we can make a little chart that compares
the CTF of the microscope that produces this image,
to the point spread function that will result.
And so in the first case, let's suppose that
the contrast transfer function is perfect,
meaning that all spatial frequencies are delivered.
All the spatial frequencies present in the object are delivered into the image.
Well, then the image that we'll get will look just like a delta function.
In other words, if the contrast transfer function is perfect,
delivering all the spatial frequencies at their full strength.
Then the image will look just like the original object.
And so if we start with the delta function,
the image will be a delta function.
Now in fact, the fourier transform of a delta function
is a straight line and vice versa.
The fourier transform of a straight line is a delta function.
So we see that the relationship between the contrast transfer function and
the point spread function, is that they are the fourier transform of each other.
What if our contrast transfer function wasn't perfect?
What if instead, it was gradually attenuated at high spatial frequencies.
Then we would be missing some of the high frequency components
that define the sharp up and down nature of this delta function.
What we would get.
Is a somewhat blurred peak representing the delta function.
And in fact if this contrast transfer function is a broad gaussian peak.
Then the point spread function will be a slim gaussian peak
because the 4A transform of a gaussian function is another gaussian function.
Similarly, if the contrast transfer function is heavily attenuated,
meaning that only the very lowest spatial frequencies are transferred.
And all of the higher spatial frequencies are lost,
then the image of the delta function will look very broad.
Because there's only low-frequency components available to represent it.
And again, if this contrast transfer function is a gaussian,
then its point spread function will also look like a gaussian.
But a, sharply attenuated gaussian in, as a contrast transfer
function leads to a very broad, wide point spread function.
Finally, if our contrast transfer function is very complicated,
like in the case of an electron microscope.
The Point spread function will also be complicated.
It'll be a smearing with ripples around the peak,
meaning that every dense object in the sample will appear
as a somewhat broadened density peak in the image surrounded by some ripples.
The first minimum will look like a white halo and
then there may be additional maxima and minima beyond that that are visible.
And that's of course exactly what you see over here in this image.
The dark gold bead which is somewhat
representative of a sharp delta function kind of an object.
Appears in the image as a broadened peak with a white halo around it and
some ripples beyond that.
The other idea about the Point spread function, is that there's a single
point spread function in general across the entire image.
So every sharp peak in the object like this gold bead and that one and
that one, as well as this one and this one and all the.
All of the densities, in fact,
in the image will be blurred by the same point spread function.
And we call this convulsion.
[NOISE] We would say that this image is equal
to the object convolved with the point spread function,
and this is the symbol that we used to mean convolution.
What this means is that every density in the original sample appears
blurred by the same pattern and the same point spread function.
So convolution applies the same blurring effect to every density present.
So if the image is equal to the object convolved with a point spread function,
then the Fourier transform of that
image will be equal to the Fourier transform
of the object convolved with the point spread function.
I've just taken the Fourier transform of both sides of the equation.
Using whats the convolution theorem, this side of the equation
can be rewritten as the Fourier transform of the object times
the Fourier transform of the point spread function.
The convulsion theorem is that the four eight transform of two functions that
are convolved is equal to their separate four eight transforms multiplied.
Well now we can rewrite this is the Fourier transform the object times
the contrast transfer function because remember we just established that.
The relationship between the contrast transfer function and
the point spread function is that they are each others Fourier transform.
So the Fourier transform of the point spread function is the CTF.
So carrying forward this side of the equation, Fourier transform of the image.
If we now take the inverse Fourier transform of both sides of the equation,
the inverse foray transform of the Fourier transform of the image
just brings us back to the image.
The image is then the inverse Fourier
transform of the Fourier transform
of the object times the CTF.
Now this equation makes sense when we think back again about image formation
in the microscope starting with a sample.
We've talked about how radiation is scattered by the sample,
and the scattering process physically
reveals the Fourier transform of the density of the sample
because each scattering angle represents a different Fourier component of the sample.
And then the lens of the microscope gathers
this scattered radiation through.
The back focal plane of the electron microscope and
brings it together to form an image.
Now we've talked already how the physical process of scattering
reveals the Fourier transform of the sample because on the back
focal plane of the objective lens is present a wave function
that represents the Fourier transform of the object.
In a perfect microscope, all of these Fourier components would be transferred
into the image forming a perfect image, but as we've discussed, in real
electron microscopes that's not the case because of the contrast transfer function.
The contrast transfer function selects certain of these spatial
frequencies to deliver into the image and others are attenuated or even blocked.
And so, we might depict that by erasing some of these
rays representing different spatial frequencies that might have
made it to the image plane, but because of the contrast transfer function, don't.
Now finally, we've understood that as the wave
progresses from the back focal plane of the lens to the image plane,
another Fourier transform occurs, bringing us back into real space.
Now let's look back at this equation.
What we've said is that the image,
this image, is equal to an inverse Fourier transform.
That's this step.
Of what?
Of the wave function that exists here, and that is the Fourier
transform of the object as we've described times the CTF.
And so you see this equation just tells us that the image that is formed
is the inverse transform not of just the object but
the object times the contrast transfer function.
Now building from this equation, let's take another step,
and let's try to solve this equation for the object because
what we'd really like to know is the structure of the object.
So rearranging this equation, we can write that the Fourier transform of
the object is therefore the Fourier transform
of the image divided by the CTF.
And if we take the inverse Fourier transform of both sides of that equation,
we arrive at an, a very important result, which is that the object
is equal to the inverse Fourier transform of the Fourier transform
of the image divided by the CTF.
And this gives us an idea of how we can improve
the images that we get from the electron microscope.
We record an image, but
unfortunately the image is degraded by the point spread function.
The fact that the contrast transfer function wasn't perfect
means that all the features are a little bit blurred.
This equation teaches us that if we take the Fourier transform of our image and
we divide that in reciprocal space by the contrast transfer function,
and then we take the result and do an inverse Fourier transform,
we will recover a more true version of the structure of the object.
And this process we call CTF correction.
Now lets look more carefully at how this process might work.
Suppose that we're trying to take an image of an object, and just for fun,
let's suppose it looks something like a tree.
And we put that in our microscope, and we try to record an image.
And we've already talked that.
On the back focal plane of the electron microscope will
be a wave function that represents the Fourier transform of that object.
So we'll say Fourier transform of the object.
However, because of the contrast transfer function,
it may be that we will lose all of the high spatial frequencies.
Here.
Those will be lost.
I'll draw them blue as if they were zeroed.
As well as there may be bonds within this region
where the contrast transfer function diminishes or loses those values, and
it may even lose all the values right in the middle.
And so in fact what we have on the back focal plane of the microscope
is the Fourier transform of the object times this CTF.
And then what is recorded as the image is the Fourier transform of that product.
And so here I'll draw as the image something again,
a local, like a tree, I'll draw it as fuzzy.
Because it's a corrupted version of the object.
And so here is our image.
Now, the process of CTF correction tells us
to first calculate in the computer a Fourier transform of that image.
And so here, we'll calculate a Fourier transform of that image.
And we need to divide the Fourier transform of the image by the CTF.
Now what that means.
Let's draw an example CTF.
That varies from negative 1 to positive 1.
So here's the CTF of the microscope that recorded this image.
And what it means to CTF correct the image
is that in the Fourier transform of the image, we consider each pixel.
And remember, each pixel represents one Fourier component of the image,
and it has an amplitude and sign.
So, if we were to make a little chart,
we might include let's suppose that this pixel
in the true Fourier transform of the real object,
let's suppose it should have had an amplitude
with a value of 10, in some arbitrary units.
Now, the CTF at that same spatial frequency.
Now we're considering a spatial frequency that's very close to the origin.
So maybe it's, it's here.
The CTF at that spatial frequency
may have been quite low like negative 0.2 for instance.
And so.
That's what the CTF is.
And because of that the amplitude that we observe in
the Fourier transform of the image [NOISE] will not be 10,
but rather 10 times negative 0.2, or
negative 2 is what we'll actually observe as the amplitude in this position.
What it means to CTF correct, is to take this value, and
divide it by the value of the CTF at that location.
And so in the CTF corrected [NOISE]
version of the Fourier transform of the image.
We will put a value of minus 2, the observed divided by the contrast transfer
function at that spatial frequency, and this of course will give us
an amplitude of 10, which is the original that we were looking for.
So now, divide by the CTF to produce
another version of this transform.
And we'll call this the CTF corrected [NOISE]
version of the Fourier transform of the image.
And in this case [NOISE] the value of this particular pixel,
its amplitude, instead of the minus 2 that was present here,
we'll replace it with a value of 10.
Now let's look at another example pixel.
Let's suppose that this pixel, at a higher spatial frequency,
let's suppose it's over here, and in the actual transform of the object,
let's suppose that it should have had an amplitude of 5.
And let's say that this pixel is at a spatial frequency right here
where the contrast transfer function has a value of 0.3.
The CTF then is 0.3 at that position, so the value that we'll actually
observe in the Fourier transform of the image will be 1.5.
CTF correction, in this case, will take that amplitude of 1.5,
and if we divide it by 0.3, [NOISE] we'll recover
the original amplitude that it should've been of 5.
And so we'll replace the amplitude here with the number 5.
One more example.
Suppose, we'll next consider a pixel here.
Of course we do this process for
every pixel in the Fourier transform of the image.
But if we consider another pixel over here, let's suppose that in the Fourier
transform of the actual object that should have had an amplitude of, let's say 2.
And this pixel, the CTF at that pixel, may have been over here.
Let's suppose that the value of the CTF at that position is also,
is a negative 0.3.
[SOUND] Then the amplitude that we actually
observe here is going to be a negative 0.6.
And so we'll replace that value in the CTF corrected version of the Fourier
transform of the image with a negative 0.6 divided by negative 0.3,
or in other words, the value 2, the original value.
And okay, so we continue this process for all the pixels in this transform.
And then finally the last step of CTF
correction is to calculate the Fourier
transform of this CTF corrected version and
produce what we'll call a CTF corrected image.
And I'm going to draw as more nearly like the original tree
that we were trying to image in the first place.
So this is the process of CTF correction, and it can work very well.
Except you might ask what do we do in those spatial
frequencies at which the CTF is actually 0.
For instance at this 0 of the CTF, and
this spatial frequency and this spatial frequency,
the value of the CTF is actually 0, and we can't divide by 0.
And so, in fact, what's commonly done, there's a number of
ways to handle this problem with different advantages and disadvantages.
As one example of a way to handle this problem,
is to simply exclude the spatial frequencies anywhere near 0 and
just exclude them from the process, and zero them out.
And if you do this, what you'll do is CTF correct
those regions that you can and restore the effective
contrast transfer function back up to the value of one.
And then you'll be missing the values here around CTF zeroes.
And you'll get a final CTF that looks like this,
Etc.
Now there's recovering those regions of the CTF where it's significantly non 0,
and in order to prevent division by numbers close to 0.
Just ignoring, or multiplying by 0, the values near any zeroes of the CTF.
A second way that this has been handled is by,
instead of dividing by the actual value of the CTF.
Simply phase flipping.
Those fourier components that have negative amplitudes in the CTF.
So in other words, you would take these spatial frequencies and
either multiply these amplitudes by negative 1 or you could add 180
degrees to their phases, essentially advancing that sine wave by 180 degrees.
And as a result you get a net CTF that still oscillates.
But you flip those regions of the CTF that are below zero and
you flip those back to be positive.
So at least they're contributing contrast in the same sense
as the other regions of the CTF.
Now there's no way to recover
the information at true zeroes of the CTF because
that information simply was not delivered into the actual image.
That was lost in the microscope and
it cannot be recovered from this particular image.
And we, as we talk about the different modalities of Cryolium, we'll talk
about one way to handle this problem is to take pictures at different defocus values.
So if you can take pictures of the same object with different defocus values,
one image may give you this CTF.
But if you took another image at a different defocus,
you might get a CTF that looked more like that.
You can use the information from this image to discover the amplitudes and
phases of these fourier components that are lost in the first.
Now, in order to divide by the CTF, it has to be determined precisely from the image.
As an example, here is, again, an image we recorded of graphite.
And a, a enlarged version where you can actually see the lines of graphite here.
In order to CTF correct this image,
we would first calculate its fourier transform.
And when viewed as a power spectrum, it looks like this,
which we've looked at before, showing all of the tong rings of the CTF.
Now if we take this power spectrum and
we calculate a radial average as a function of position from the origin.
This is a screen shot of a program called CTF in the software package Bsoft,
which we used to look at the pattern of tong rings in this power spectrum.
And so it's a plot, here's the vertical axis.
And it's a plot along spatial frequency, in one over Angstroms.
And, the blue lines here, are a plot of
the power in this power spectrum is a function of radius from the center.
And you can see it oscillate in each of the tong rings.
Then, the red curve is a calculated CTF.
Based on the voltage of the microscope, which in this case was 300 kilovolts,
based on the known coefficient of spherical aberration.
If you remember, that was an element of the CTF curve.
And here, there's a slider that allows a user to
move the slider up and down changing the defocus
that's used to calculate this theoretical CTF curve.
And we've moved the slider back and
forth until at the position 1.99 microns under focus.
The calculated theoretical CTF matches the oscillations of
the observed power spectrum very well.
And this allows one to calculate the CTF at any
particular spatial frequency, allowing one to do a precise CTF correction.
So in summary, CTF correction proceeds by taking the image
produced by the microscope, calculating its Fourier transform.
Given that Fourier transform, one first fits
a theoretical CTF curve to the patterns that's observed.
And this allows one to calculate the CTF at any particular spatial frequency.
Using that curve, you then divide the amplitude in each pixel of this
Fourier transform of the image by the theoretical CTF value at that position.
Excluding the areas around zeroes to avoid division by zero to
produce a CTF corrected version of the transform.
And then through an inverse Fourier transform, you recover a CTF corrected
image which is closer to the object than the image that you first recorded.
