That was awesome, thanks for such an informative post. The archive of
your post, located at
https://mail.gna.org/public/relax-users/2006-10/msg00020.html
(Message-id: <481156b20610060807v768437bcw2d58f620652c1a0a@xxxxxxxxxxxxxx>),
should be a great resource for later on when asymmetric CSA tensors
are added to relax. I have a small number of questions/points below.
Here's my little tutorial on CSA nomenclature. It's one of the most
annoying things I have found in NMR and I've made it a significant part of
my thesis to boot! So, I'm going to start from scratch and assume almost no
previous understanding (it's a large message board, so why not?). First,
CSA means two things: either 'chemical SHIFT anisotropy' or 'chemical
SHIELDING anisotropy'. The differences come from, I believe, the discipline
of NMR you are most familiar with: solution state people always display
chemical shifts whereas solid state people tend towards the shielding side.
These two terms are synonymous, and yet opposite each other. Shift tensors
are always described as delta(ii) and shielding tensors are sigma(ii) where:
delta(ii) = sigma(reference) - sigma(ii)
The three principle values are then organized as del(11) > del(22) > del(33)
and sig(11) < sig(22) < sig(33), ie del/sig(11) has the largest chemical
shift value (most downfield) while the del/sig(33) value is the most
shielded value (most upfield). To make things even more ridiculous, some
people reverse the sigma organization (sig(11) > sig(22) > sig(33)) where
del11 is now equivalent to sig33, etc! So, beware! As I said before, I
prefer the delta organization as I am in solution state and there isn't the
added sigma confusion.
Now, the CSA tensor that defines the position of a peak in your NMR spectrum
has both the isotropic and anisotropic component:
delta(ij) = delta(iso) + delta(ani,ij)
In relaxation and RCSA studies, the isotropic part is completely irrelevant
and therefore we are only interested in the anisotropic part which is
traceless (SUM[del(ani,ii) from 1to3] = 0) and assumed to be symmetric
(del(ani,ij) = delta(ani,ji)), although this isn't necessarily true, but for
molecular symmetry reasons almost always is true. From here on I'm just
refering to the anisotropic part of the tensor and delta(11) is always
positive and delta(33) is always negative.
For a number of mathematical reasons, the CSA tensor is often reorganized
once again in terms of x,y,z such |deltz(zz)| > |delta(yy)| > |delta(xx)|,
similar to diffusion tensor and alignment tensor definitions. The asymmetry
of the tensor is defined as:
eta = (delta(xx) - delta(yy))/delta(zz)
and varies from 0 (fully symmetric) to 1 (fully asymmetric). This is the
parameter that is ubiquitously assymed to be zero in relaxation in order to
keep the CSA term simple. In the situation where eta ~= 1, del(11) ~=
del(33) or del(zz) ~= del(yy) and del(22) = del(xx) ~= 0.
Should this say "del(22) - del(iso) = del(xx) ~= 0"? And doesn't eta
~= 1 mean that the tensor is approximately axially symmetric about
axis defined by del(xx)? In the Diso, Da, and Dr notation I have used
for the diffusion tensor in relax, a Dr value of one is equivalent to
axial symmetry. A simple change of the axis names (Dx, Dy, and Dz)
will, for the same diffusion tensor, produce a Dr value of zero. It's
all in how you define the axis names. The diffusion tensor Dr
definition in relax is slightly different to the CSA eta definition
though, so eta = 1 may not be axially symmetric.
Now the confusing
part (if this hasn't been confusing enough!): there are many values proposed
in the literature for the principle values of various nuclei and in some
cases |del(11)| > |del(33)| and in others |del(33)| > |del(11)| ! Given the
the CSA is a tensor, the orientation of all 3 components are orthogonal, so
this seemingly harmless difference represents a 90 degree shift in the
dominant principle component value in terms of x,y, and z. So, the question
is, how does this relate to relaxation studies?
In much of the literature (which assumes eta = 0), the ANISOTROPY of the
tensor is given as:
DELTAdelta = delta(zz) - (delta(xx)+delta(yy))/2
or, with a little trivial math:
DELTAdelta = (3/2) delta(zz)
What you find in most relaxation equations for R1 and R2 is that the CSA
constant is:
(2/5) (omega(i) * DELTAdelta)^2 / 3 or equivalently (3/10) (omega(i) *
delta(zz))^2
(For those of you wondering where 2/5 comes from, I simply prefer to pull it
out of the extended model free spectral density function.)
Let's keep things simple for now (ie, isotropic diffusiong). If you want to
include asymmetry, one should really use the following as the CSA constant:
(3/10) omega(i)^2 delta(zz)^2 (1+eta^2/3)
where delta(zz) * sqrt((1+(eta^2)/3)) is the asymmetric anisotropy, or CSAa.
The CSAa is equivalently:
sqrt((3/2) (del(xx)^2 + del(yy)^2 + del(zz)^2))
So, for eta = 0, DELTAdelta = CSAa, but when eta != 0, CSAa > DELTAdelta.
Now, how many biomolecules diffuse isotropically? Can I see a show of
hands? Ok, I don't see a lot of hands, although ubiquitin is sitting in the
back waving his/her arms wildly :). Here's the kicker, when a molecule
tumbles anisotropically the asymmetry term of the CSA can NOT be pulled out
of the spectral density function! The CSA constant for these need to be:
(3/10) omega(i)^2 delta(zz)^2
and the extended model free coefficients contain the asymmetry as well as
euler angles relating the CSA tensor with the diffusion tensor (but let's
not go there for right now).
Awww, that's a pity. Just as you were getting into the really
interesting stuff too!
> Another thing that confuses me is that the dxx eigenvalue is
> approximately zero ppm? Is that possible? Also, doesn't traceless
> refer to a tensor in which the sum of the eigenvalues is zero? That
> would only occur if the isotropic chemical shift is zero ppm. Sorry
> about my ignorance, but what does 'traceless anisotropic part' and
> 'anti-symmetric part' of the CSA tensor mean? Does this involve the
> diagonalisation of the matrix and then the subtraction of the
> isotropic shift times the identity matrix from the diagonalised CSA
> matrix?
I was trying to be a little scary, for some reason. The CSA tensor CAN,
have an anti-symmetric part (delta(ij) != delta(ji)) which leads to terms in
the correlation functions, f_lm(tau), and their Fourier transforms,
g_lm(omega), whith l = 1. !! I'm not very good at describing, let alone
understanding, this but hopefully you know what I mean. It's an interesting
bit of trivia, but for now let's pretend I never said anything ;)
Said what? I didn't hear a thing!
> One last question, if the CSA tensor for RNA/DNA base 13C and 15N
> spins is not rhombic, does this mean that the angle between the unique
> axis of the CSA tensor and the XH dipolar interaction vector is ~90
> degrees?
More often than not, yes, but there's really two principle components to the
tensor that are important. I don't quite understand the rhombic term, but
I'm assuming it means assymetry (?)
Yep, rhombicity and asymmetry are synonymous.
> As you can probably see, I am a bit lost. This is probably because of
> my ignorance of the variety of symbols used to express the same thing
> and the terminology used in each instance.
That's completely understandable! The whole topic is filled with duplicate
and triplicate nomenclature! If I had my way, I would chose yet another way
to describe the CSA tensor that I didn't even discuss that keeps everything
in terms of the delta(11) , (22), and (33) that I prefer. But that would
take a lot of derivations from scratch on my part, so I'm leaving that alone
... for now ;) One day, I hope to confuse the literature even more!!! Mwa
ha ha ha!! I think that's a prerequisite for anyone studying CSAs. I hope
you've enjoyed this mini tutorial of my knowledge on CSA tensors.
It sounds like you have evil plans ;) Thanks again for the tutorial,
it was a very useful summary.
Edward
Re: analysis of limited data sets