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running CTFFIND4 on tilted data

6 replies [Last post]
shabih
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Joined: 25 Jan 2016

Dear All,

Is there any special precaution (e.g. switching on a particular flag) which we have to take while running CTFFIND4 on tilted data. We had preferential orientation problem so we tilted the stage to 20 and 30 degrees.

Any comments or suggestions regarding processing are welcome too!

Cheers,
Shabih

himesb
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Joined: 26 Jul 2017
ctf estimation on tilted data

Hi Shabih,

The accuracy of your defocus estimation will be reduced on the tilted data, however, you should be able to recover this by refining your ctf in cisTEM later in processing.

For example, if you have a 1 Ang pixel sampling on a k2, then at 30 degrees tilt you will have a defocus gradient that results in a change of ~ 43 nm midway from the tilt axis and ~ 86 nm at the edge of the image. This gradient will blur the Thon rings and limit the accuracy of the defocus fit to something around 8-10 Angstroms. (If my memory serves me right...the exact numbers aren't so important though.)

What is important is to make sure that the default search range in your CTF refinement is set large enough to account for tilting. The current default values in cisTEM search a range of 500 Ang, which in this example case would not be sufficient to correct particles far from the tilt-axis.

You can check your needed range by using the formula
pixelSize * max_image_size_in_pixels/2 * tan(tilt_angle)

shabih
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Joined: 25 Jan 2016
Hi, The formula you provided:

Hi,

The formula you provided: pixelSize * max_image_size_in_pixels/2 * tan(tilt_angle) gives a -13642 for a pixel size of 1.04 Å, image dimension 4096 X 4096 and tilt angle of 30 degrees. Just wondering whether the search range of 13642 Å is too big?

And thanks for posting the tilt data processing on twitter, that's very helpful.

Regards,
Shabih

himesb
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Joined: 26 Jul 2017
2x your calculation

Hi shabih

Tan(30)~0.5 so in you case the max defocus would be ~1024 Å

How did you calculate your numbers? Calculator, python, abicus? Lol

Np, thanks for the question. I appreciate it when people aren't intimidate by a public forum!

It looks like you 1) used a calculator that expected radians (so use 30*π/180.0) and 2) did not divide the image dimension by 2

shabih
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Joined: 25 Jan 2016
My bad. Used the good old

My bad. Used the good old calculator.

himesb
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Joined: 26 Jul 2017
2x your calculation

Duplicate, see above

shabih
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Joined: 25 Jan 2016
Thanks a lot. Very helpful.

Thanks a lot. Very helpful.