Helical symmetry

Symmetrizing a particle

Helical operator syntax

Symmetrizing a helical particle can be done with the usual tool dsym:

 vh = dsym(v,<operator>);

where operator represents any symmetry operator. The syntax for helical operators is:

 h [<phi>,<dz>,<repetitions[optional]>,<rotational symmetry[optional]>] 

Here:

• phi is the angular increment in degrees
• dz is the axial rise in pixels
• the number of repetitions (defaults to the maximum definible for a given volume).
  • Can be passed as 0 to force Dynamo to create a default value.
• additional Cn rotation to account for n-start helices.

Example:

 vh = dsym(v,'h[15.6,4.7]');
 vh = dsym(v,'h[15.6,4.7,5]');       % uses 5 repetitions
 vh = dsym(v,'h[15.6,4.7,5,8]');    % uses 5 repetitions and adds a C8 symmetrization
 vh = dsym(v,'h[15.6,4.7,0,8]');    % computes internally the maximum of repetitions and creates a C8 symmetrization

Using symmetrization in alignment projects

The simplest way is to input one operator with the syntax described #Helical operator syntax above in the Numerical settings area of the dcm GUI.

This allows to impose a given symmetrization. In order to estimate the symmetry, you will need to create a plugin.

Number of repetitions

Symmetrization of a volume is computed by applying the transformation (phi,dz) in both directions to the original volume a number of times (called repetitions) and then averaging the result. For instance, 17 repetitions imply 8 subsequent applications of (phi,dz) to the original volume and 8 application of -(phi,dz). Each time (phi,dz) is applied, the transformed volume will have a "bottom" of dz layers of pixels that are undefined: (they correspond to the theoretically part of the infinite helix below our volume). The averaging procedure will discard contributions of such pixels. As a result, the central area of the symmetrized volume will be representing contributions of more transformed volumes, attaining thus a higher SNR.

If the user does not specify a number of repetitions, Dynamo will take the maximal number of repetitions for the given volume. If the volume is of size N pixels and the operator carries an axial rise of dz pixels, the maximal number of translations along the z axis is floor(N/dz)