14. Practical: Electron Tomography#
14.1. Introduction#
Unlike single-particle analysis, electron tomography (cryo-ET) reconstructs a 3D volume from images of a single specimen tilted to many angles by physically rotating the sample stage. We work through the problem in 2D: the unknown object is a 2D image, and we acquire 1D projections at various tilt angles, then reconstruct the 2D image from those projections.
Axis convention: The electron beam travels along \(z\). The tilt axis is \(y\) (pointing through the imaging plane). Tilting the sample about \(y\) rotates the \(xz\)-plane, so each tilt angle \(\alpha\) produces a 1D projection along \(x\) of the rotated image.
A typical cryo-ET experiment collects tilt images from \(-60°\) to \(+60°\) in steps of \(1°\)–\(3°\), limited by grid-bar occlusion and increased sample thickness at high tilt.
Interactive elements
Click Live Code in the top toolbar to activate the kernel, then expand each Show code toggle and click ▶ to run the cell.
14.1.1. Helper functions#
14.2. The 2D test image#
We work with a 2D phantom that mimics a biological cross-section: a hollow ring (microtubule), two line segments (membrane sheets), and a small dot (protein complex).
14.3. Tilting the sample and acquiring projections#
To acquire a tilt series, the specimen is physically rotated about the \(y\)-axis. At each tilt angle \(\alpha\), the transmitted electrons form a 1D projection (the column sum of the rotated image). The interactive below shows the specimen at a chosen tilt angle together with the resulting 1D projection profile.
14.4. The sinogram#
Collecting projections over the full tilt range and stacking them row by row produces the sinogram — so named because a point object traces a sinusoidal curve across it as the tilt angle varies.
14.5. Task 1 – Backprojection#
The simplest reconstruction method is backprojection: for each projection at angle \(\alpha\), repeat the 1D line into a 2D slab, rotate it back by \(-\alpha\), and sum over all angles.
This is correct in principle but overweights low-spatial-frequency components (every point in the image receives contributions from every projection), producing a blurry, star-shaped reconstruction artifact.
Task 1
Take a 1D projection
projection.Repeat it \(N\) times along a new axis using
repeat_lineto get a 2D slab of shape \((N, N)\).Rotate the slab back by \(-\alpha\) using
nd_rotate.Sum over all angles. Divide by the number of angles.
14.6. Task 2 – Filtered backprojection#
The blurring artifact of plain backprojection is caused by the over-sampling of low frequencies: every projection adds a constant along its direction, contributing to the low-frequency region of Fourier space more than to high frequencies. The fix is to pre-filter each projection with a ramp filter \(|\nu|\) before backprojecting. This exactly counteracts the \(1/|\nu|\) weighting introduced by backprojection.
Task 2
Take a 1D projection.
Compute its Fourier transform using
np.fft.rfft.Multiply by the ramp filter \(|\nu|\) (use
np.fft.rfftfreqfor the frequency axis).Inverse-Fourier-transform back.
Use the filtered projections for backprojection.
14.7. The Fourier Slice Theorem#
The projection-slice theorem (also called the Fourier slice theorem) states that the 1D Fourier transform of a projection at angle \(\alpha\) equals a central slice through the 2D Fourier transform of the object at the same angle:
Each projection therefore fills one line (passing through the origin) in 2D Fourier space. To reconstruct the full 3D (or 2D) object, we need enough projections to cover Fourier space densely.
The interactive below shows, for a chosen tilt angle:
The rotated specimen and its 1D projection
The 2D Fourier transform of the specimen with the corresponding central slice highlighted
A verification that the 1D FT of the projection matches the highlighted central slice
14.8. Missing wedge#
In practice the tilt range is limited to roughly \(\pm 60°\) because:
Grid bars obstruct the beam at high tilt
Increased effective sample thickness at high tilt degrades image quality
This means the central region of Fourier space is never filled — the missing wedge. Features oriented along the tilt axis (perpendicular to the tilt direction) are most affected: they appear elongated in the reconstruction.
The interactive below shows the Fourier coverage at different tilt ranges. The missing wedge is the dark triangular gap.
14.9. Task 3 – The Crowther criterion#
To fully reconstruct an object of diameter \(D\) at resolution \(r\), how many tilt angles \(m\) do we need? The answer follows from the Fourier slice theorem and the sampling theorem.
Each projection fills one central line in 2D Fourier space. At resolution \(r\), we need the outermost Fourier shell (radius \(1/r\)) to be densely sampled. Two adjacent central lines at the edge of the Fourier disk must be separated by at most \(1/D\) (the Nyquist frequency across the object). The angular separation between adjacent projections at radius \(1/r\) is therefore:
Since the total range is \(\pi\) radians (projections at \(\alpha\) and \(\alpha+\pi\) are identical), the number of tilts needed is:
This is the Crowther criterion. It tells us that to achieve resolution \(r\) from an object of diameter \(D\), we need at least \(\pi D/r\) projections — independent of whether we are doing tomography or SPA (in SPA, particles in random orientations naturally fill the full angular range).
14.10. Summary#
Concept |
Key idea |
|---|---|
Tilt series |
Physical rotation of specimen; each tilt gives one 1D projection |
Sinogram |
Stack of 1D projections; point objects trace sinusoids |
Radon transform |
Mathematical projection operator \(p_\alpha(t) = \int A(t\cos\alpha - s\sin\alpha, t\sin\alpha + s\cos\alpha)\,ds\) |
Backprojection |
Smear each projection back along its direction and sum |
Ramp filter |
$\hat h(\nu) = |
Fourier slice theorem |
\(\mathcal{F}_1\{p_\alpha\}(\nu) = \mathcal{F}_2\{A\}(\nu\cos\alpha, \nu\sin\alpha)\) |
Missing wedge |
Triangular gap in Fourier coverage from limited tilt range |
Crowther criterion |
\(m \geq \pi D / r\) tilts for resolution \(r\) from object of diameter \(D\) |