crystal

zaw

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Date:22-6-2005\nDear Sir,\nHow do we know that crystal is single and perfect? Why do we refine the\ndata after x-ray diffraction method? I hope your reply.\n\nsincerely yours,\n\nZAW\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>[itex]Date:22-6-2005[/itex]
Dear Sir,
How do we know that crystal is single and perfect? Why do we refine the
data after x-ray diffraction method? I hope your reply.

sincerely yours,

ZAW

Uncle Al

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>zaw wrote:\n>\n> Date:22-6-2005\n> Dear Sir,\n> How do we know that crystal is single and perfect? Why do we refine the\n> data after x-ray diffraction method? I hope your reply.\n\nYou don\'t want a perfect crystal for a diffraction structure. You\nwant slight mosaicity to prevent destructive intereference. When the\nprimary beam makes the proper glancing angle "theta" with the planes\nof a stack to produce a reflection, the reflected ray also also makes\nthis angle with the plane and is therefore reflected back parallel to\nthe primary beam. Each reflection changes the phase of the wave by\n(pi)/2, so the twice-reflected ray is out of phase by "pi" with the\nprimary wave. They subtract amplitudes, and it gets worse the deeper\ninto the crystal you go. Primary extinction is proportional to\n\ntanh(sq)/sq\n\nwhere "s" is the number of planes in perfect registration and "q" is\nthe reflection efficiency of the plane. Primary extinction eats you\nfor reflections having large structure factors (large q) and large\nspacings d(hkl), and is appreciable for paths greater than 10^(-4) cm.\n\nA single crystal is often obvious from its diffraction pattern -\nthough twinning can be tricky. Quartz has at least seven different\nways to twin.\n\nLibrary, Google, your local crystallographer. Show some initiative.\n\n\n--\nUncle Al\nhttp://www.mazepath.com/uncleal/\n(Toxic URL! Unsafe for children and most mammals)\nhttp://www.mazepath.com/uncleal/qz.pdf\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>zaw wrote:
>
> [itex]Date:22-6-2005[/itex]
> Dear Sir,
> How do we know that crystal is single and perfect? Why do we refine the
> data after x-ray diffraction method? I hope your reply.

You don't want a perfect crystal for a diffraction structure. You
want slight mosaicity to prevent destructive intereference. When the
primary beam makes the proper glancing angle "[itex]\theta[/itex]" with the planes
of a stack to produce a reflection, the reflected ray also also makes
this angle with the plane and is therefore reflected back parallel to
the primary beam. Each reflection changes the phase of the wave by
[itex](\pi)/2,[/itex] so the twice-reflected ray is out of phase by "\pi" with the
primary wave. They subtract amplitudes, and it gets worse the deeper
into the crystal you go. Primary extinction is proportional to

[tex]tanh(sq)/sq[/tex]

where "s" is the number of planes in perfect registration and "q" is
the reflection efficiency of the plane. Primary extinction eats you
for reflections having large structure factors (large q) and large
spacings d(hkl), and is appreciable for paths greater than [itex]10^(-4) cm[/itex].

A single crystal is often obvious from its diffraction pattern -
though twinning can be tricky. Quartz has at least seven different
ways to twin.

Library, Google, your local crystallographer. Show some initiative.

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf

Arnold Neumaier

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>zaw wrote:\n> Date:22-6-2005\n> Dear Sir,\n> How do we know that crystal is single and perfect? Why do we refine the\n> data after x-ray diffraction method? I hope your reply.\n\nNo crystal is single and perfect. This is just a convenient idealization.\n\nRaw measurement data must be refined to extract the information of interest.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>zaw wrote:
> [itex]Date:22-6-2005[/itex]
> Dear Sir,
> How do we know that crystal is single and perfect? Why do we refine the
> data after x-ray diffraction method? I hope your reply.

No crystal is single and perfect. This is just a convenient idealization.

Raw measurement data must be refined to extract the information of interest.

Arnold Neumaier

Torsten Becker

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>zaw wrote:\n\n> Why do we refine the data after x-ray diffraction method?\n\nDuring data refinement you use additional knowledge about your system\nto increase the quality of the final structure.\nIf you have for example a crystal that scatters at 1.5 \\AA resolution,\nwhich is already pretty good for large molecules like proteins,\nnaively speaking you can\'t say much about atom positions below 1.5\n\\AA (which is pretty bad if you want to understand how an enzyme\nworks).\n\nThe solution is \'data refinement\'. If you know for example that two\natoms are bonded covalently, it takes an enormous amount of energy to\nstretch such a bond. It\'s most likely that these atoms will be close\nto their minimum bond length (which are known quite often).\nThat way, by starting with a blueprint of your molecule, you can come\nup with an energetically reasonable structure containing more\ninformation than is inherent in the raw electron density.\n(As a rule of thumb, with the above resolution one can trace back\nstructural changes in relative positions of atoms down to 0.3 \\AA.\nBut i\'m not sure about that number. You have to ask a crystallographer\nfor more details)\n\nHope it helps,\nTorsten\n\n\n--\n*******************************************************\nTorsten Becker Structural Biology/Bioinformatics\nUniversity of Bayreuth\n+49-921-55-3547 (office) Universitaetsstr. 30, BGI\n+49-921-7273406 (home) D-95447 Bayreuth; Germany\n\ntorsten.becker@uni-bayreuth.de\n********************************************************\n \n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>zaw wrote:

> Why do we refine the data after x-ray diffraction method?

During data refinement you use additional knowledge about your system
to increase the quality of the final structure.
If you have for example a crystal that scatters at 1.[itex]5 \AA[/itex] resolution,
which is already pretty good for large molecules like proteins,
naively speaking you can't say much about atom positions below 1.5
\AA (which is pretty bad if you want to understand how an enzyme
works).

The solution is 'data refinement'. If you know for example that two
atoms are bonded covalently, it takes an enormous amount of energy to
stretch such a bond. It's most likely that these atoms will be close
to their minimum bond length (which are known quite often).
That way, by starting with a blueprint of your molecule, you can come
up with an energetically reasonable structure containing more
information than is inherent in the raw electron density.
(As a rule of thumb, with the above resolution one can trace back
structural changes in relative positions of atoms down [itex]to .3 \AA.[/itex]
But i'm not sure about that number. You have to ask a crystallographer
for more details)

Hope it helps,
Torsten

--
[itex]*******************************************************[/itex]
Torsten Becker Structural Biology/Bioinformatics
University of Bayreuth
[itex]+49-921-55-3547[/itex] (office) Universitaetsstr. 30, BGI
[itex]+49-921-7273406[/itex] (home) [itex]D-95447[/itex] Bayreuth; Germany

torsten.becker@uni-bayreuth.de
[itex]********************************************************[/itex]

zaw

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Date:\n25-6-2005\nDear Mr Torsten Becker,\nI just received your reply answer today. Thanks you for your reply.\nOne thing I confused is that 1.5\\AA and 0.3\\AA. Could you explain it\nto me? I will be looking forward for your reply.\n\nsincerely yours,\n\nZAW WIN\nResearch Student,\nNagoya Institute of Technology,\nNagoya, Japan\n466-0063\n\n\nTorsten Becker wrote:\n> zaw wrote:\n>\n> > Why do we refine the data after x-ray diffraction method?\n>\n> During data refinement you use additional knowledge about your system\n> to increase the quality of the final structure.\n> If you have for example a crystal that scatters at 1.5 \\AA resolution,\n> which is already pretty good for large molecules like proteins,\n> naively speaking you can\'t say much about atom positions below 1.5\n> \\AA (which is pretty bad if you want to understand how an enzyme\n> works).\n>\n> The solution is \'data refinement\'. If you know for example that two\n> atoms are bonded covalently, it takes an enormous amount of energy to\n> stretch such a bond. It\'s most likely that these atoms will be close\n> to their minimum bond length (which are known quite often).\n> That way, by starting with a blueprint of your molecule, you can come\n> up with an energetically reasonable structure containing more\n> information than is inherent in the raw electron density.\n> (As a rule of thumb, with the above resolution one can trace back\n> structural changes in relative positions of atoms down to 0.3 \\AA.\n> But i\'m not sure about that number. You have to ask a crystallographer\n> for more details)\n>\n> Hope it helps,\n> Torsten\n>\n>\n> --\n> *******************************************************\n> Torsten Becker Structural Biology/Bioinformatics\n> University of Bayreuth\n> +49-921-55-3547 (office) Universitaetsstr. 30, BGI\n> +49-921-7273406 (home) D-95447 Bayreuth; Germany\n>\n> torsten.becker@uni-bayreuth.de\n> ********************************************************\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Date:
[itex]25-6-2005[/itex]
Dear Mr Torsten Becker,
I just received your reply answer today. Thanks you for your reply.
One thing I confused is that 1.[itex]5\AA[/itex] and .[itex]3\AA[/itex]. Could you explain it
to me? I will be looking forward for your reply.

sincerely yours,

ZAW WIN
Research Student,
Nagoya Institute of Technology,
Nagoya, Japan
[itex]466-0063[/itex]

Torsten Becker wrote:
> zaw wrote:
>
> > Why do we refine the data after x-ray diffraction method?
>
> During data refinement you use additional knowledge about your system
> to increase the quality of the final structure.
> If you have for example a crystal that scatters at 1.[itex]5 \AA[/itex] resolution,
> which is already pretty good for large molecules like proteins,
> naively speaking you can't say much about atom positions below 1.5
> \AA (which is pretty bad if you want to understand how an enzyme
> works).
>
> The solution is 'data refinement'. If you know for example that two
> atoms are bonded covalently, it takes an enormous amount of energy to
> stretch such a bond. It's most likely that these atoms will be close
> to their minimum bond length (which are known quite often).
> That way, by starting with a blueprint of your molecule, you can come
> up with an energetically reasonable structure containing more
> information than is inherent in the raw electron density.
> (As a rule of thumb, with the above resolution one can trace back
> structural changes in relative positions of atoms down [itex]to .3 \AA.[/itex]
> But i'm not sure about that number. You have to ask a crystallographer
> for more details)
>
> Hope it helps,
> Torsten
>
>
> --
> [itex]*******************************************************[/itex]
> Torsten Becker Structural Biology/Bioinformatics
> University of Bayreuth
> [itex]+49-921-55-3547[/itex] (office) Universitaetsstr. 30, BGI
> [itex]+49-921-7273406[/itex] (home) [itex]D-95447[/itex] Bayreuth; Germany
>
> torsten.becker@uni-bayreuth.de
> [itex]********************************************************[/itex]

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