THE UNIVERSITY
LEEDS 2
Dear Professor Pauling:
Since returning to Leeds I have received your letter from New York including the copy
sent care of Kathleen. I am sorry the time for discussion was so restricted, for there
were many things that could have been said if there had been time. Bob took notes
on all the papers and will no doubt report them to you. I will at this time only mention
the synthetic polypeptide work, as it is nearer to my own line of work. Except that
I will say that the papers presented by Bragg and Perutz were substantially what they
told me at Edinburgh a month ago and already sent to you in my letter last month.
They have now fixed the sign choice for two of the seven layers so there are only
five left before they can prepare a Fourier projection. I will also add that Astbury
and his people are saying that everyone familiar with the field is sure that the 5.1
Å reflection of alpha keratin is a true meridional spot and not a doublet- and that
they doubt if the Lotmar-Picken material is alpha keratin or even a protein. They
express the same doubt over the fibrin for which they sent you pictures. These are
matters that can be settled only by more experiments.
Bamford and Co. restated their criticisms of the March 1 Nature paper with the new
point that they now believe their pictures are best fitted by a cell with co = 43.2
Å. This makes the old 5th layer now the 8th and all layers are given new indices which
are the closest possible to 8/5 times the old indices. They also state that there
are definitely weak spots, four of them, not indexed by this cell and not compatible
with hexagonal packing. There is also a streak which, if part of the alpha patter,
will require doubling co to give 86.4 Å. They give the Perutz spacing as 1.50 and
their value for the old co was 27.0. These are notably larger than Yakel’s values.
Their pictures were made in a vacuum camera with monochromatized rays and silver was
plated on the sample to give the silver pattern for calibration purposes. As I do
not know what precautions Harry took with regard to calibration and film shrinkage
I do not know what to say about these discrepancies. But I do note that harry said
to me in a letter several months ago that there were some spots that didn’t seem to
fit the cell. I discussed their remarks about the infrared results in my letter last
month. These same remarks were made in part by some Englishman, whose name I didn’t
catch, immediately after Bob’s brief rebuttal but when I wanted to amplify them Adrian
turned the meeting over to Edsall for the final summary.
Since then I have considered the new indexing. It corresponds to twenty-nine residues
in eight turns. Upon applying the Cochrane-Vand formula one finds that for the layer
lines observed the Bessel function orders are exactly those found for the old indexing
while for the missing layers the orders are equal to or greater than those for the
old indexing. A twenty-nine residue helix can’t have hexagonal symmetry. The cell
might be orthorhombic with axial ratios almost exactly corresponding to hexagonal
packing but with the C centering not exactly obeyed. The strong spots would then have
spacings fitting a hexagonal cell but there could be some weak ones not permitted
by a hexagonal cell. If at the same time the over all scale is nearer to Harry’s determination
than to theirs, the density would fit and there is no argument against the alpha helix.
Checking over the above calculations suggested another interesting point to me. Yesterday
I calculated the Fourier transform of a spiral of atoms with no true repeat. I assumed
a fixed pitch p per atom and P per turn and assumed a large number of turns. By this
time I was not surprised to find that
Even with no repeat a helix gives discrete layer lines. But in general there is no
constant single layerline spacing. The structure factor equation (per atom) is like
that given by Cochrane with the conditional equation being
P(ζ - k/P ) = m
where m is any integer, k is the order of the Bessel function involved and ζ is the
reciprocal lattice coordinate parallel to the axis. This rearranges to
ζ = m/p + k/P
Thus for Bessel functions of a given order there is a system of layer lines with separation
l/p . But with m constant one gets a system with separation l/P on which the Bessel
function order increases by one between layer lines. If p and P have a l. c. m. c
such that c = ap = bP with a and b integers, then
ζ = ma/c + kb/c = l/c,
which is the case for a true repeat co , l being the layer line index. This is the
case discussed by Cochrane.
If one knows approximate values for p and P one can from the observed ζ values find
the m and k values for each observed layer, making use of the fact pointed out by
Cochrane that the order k must be low. One can then carry out at least squares determination
of p and P . Bamford’s data in Nature are given as spacings rather than as layer line
separations. I have converted them to ζ values using a slightly smaller value of ao
than they give, a change suggested by the agreement between calculated and observed
values on the equator. They give no numerical data at all for two very weak layers.
From an analysis as outlined above I get p = 1.499 and P = 5.42 . . I suggest that
such a calculation should be made for Harry’s data after the calibration has been
checked if that appears necessary.
I believe it was Dorothy Crowfoot who mentioned to me at Oxford in February that rotation
pictures about a certain axis of a certain protein- I think she said Bernal’s ribonuclease-
has this characteristic appearance. She asked at that time what it might mean. From
the above it appears that it might mean there are spirals parallel to the axis with
P much larger than p so that as you go up the layer line series the Bessel function
order increases regularly and one must go farther and farther from the meridian to
find spots.
I am still thinking about all this and may send a note to Nature.
We are expecting Bijvoet here next week and Kratky near the end of the month. The
Corey’s will arrive about the time Bijvoet leaves. They are driving a rented car and
left London yesterday for Oxford and Cambridge.
I suppose you will have heard of the letter Sir Robert Robinson wrote to the Times
Monday. People here are generally baffled or angry or both about the State Department’s
action. Kathleen told me that one of her colleagues said that any day now he would
expect a modern Mayflower bringing back persecuted Americans to England. He overlooked
the fact that modern Mayflowers require passports. If you are planning to take any
action involving legal or other expense I would be glad to send a check to help and
will do anything else I can to be of assistance.
With regard to my appointment- if I understand the implications I think we would rather
stay on the nine month basis. All of our relatives are either in the Eastern United
States or in Europe. We would therefore like to be free to leave about one summer
in three to come over here. These visits would be arranged at the times of the International
Union Congresses or some other equally important scientific meeting but would still
be primarily visiting. Otherwise we would remain in Pasadena most of the summer-so
what we need is to alternate, two years on full time basis and one year on nine months
basis but that of course is too fancy an arrangement. We do not think at the moment
that we need the extra money and would rather feel free to be away, as outlined above,
if you do not think it contrary to the good of the Division.
Best wishes and regards to you and Mrs. Pauling.
Sincerely,
P. S.
I wish also to raise a question regarding Figure 3 of your MS and the statement on
page 9 that Harry has calculated the intensities for the fifth layer. The only reference
is to the published work of Cochrane Crick and Vand which I have not seen (i.e., their
forthcoming Acta paper. I do not know what they give but the full structure factor
for one residue is given by
[lengthy handwritten equation]
where ξ, ζ, α are reciprocal space cylindrical coordinates, rj, xj and βj are the
cylindrical coordinates of the jth atom of the residue relative to some origin in
the helix, k is the order of Bessel function and the equation given above relates
ζ , k, p, P and the arbitrary integer m. Now to get a proper Figure 3 one would split
this structure factor equation into real and imaginary parts, sum over the j’s with
appropriate coordinates for each atom, square and add the two squares and then average
over alpha. As nearly as I can tell Figure 3 was actually computed for just one atom
and accordingly represents only approximately and only near the meridian the places
where the intensity can be large and in particular where it must be zero. Because
of the Bessel function argument I feel that Figure 3 is useful only at small values
of the argument and then only for showing where the intensity must be zero. I also
wonder if Harry went through this full complicated process, taking into account also
the position of the helix in the cell?
