# Interpreting difference Patterson Maps in Lab today!Calculate an isomorphous dif

## Interpreting difference Patterson Maps in Lab today!Calculate an isomorphous dif

Von Doviak, Scott, Contributing Writer has reference to this Academic Journal, PHwiki organized this Journal Interpreting difference Patterson Maps in Lab today!Calculate an isomorphous difference Patterson Map (native-heavy atom). We collected 8 derivative data sets in lab3 PCMBS4 EuCl3 1 GdCl3 Did a heavy atom bind How manyWhat are the positions of the heavy atom sitesLets review how heavy atom positions can be calculated from difference Patterson peaks.Patterson synthesisP(uvw)=S hkl cos2p(hu+kv+lw -) hkl Patterson synthesisP(uvw)=S Ihkl cos2p(hu+kv+lw -0) hkl Patterson ReviewA Patterson synthesis is like a Fourier synthesis except as long as what two variables Fourier synthesisr(xyz)=S Fhkl cos2p(hx+ky+lz -ahkl) hkl Difference Patterson synthesisP(uvw)=S FPH(hkl)-FP (hkl)2 cos2p(hu+kv+lw -0) Difference PattersonWe use isomorphous differences as coefficients, so the features in the map correspond to heavy atoms.hkl Contributions of protein atoms cancel out. Much simplified.

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Patterson map= electron density map convoluted with its inverted image.P(uvw)=r(xyz) r (-x-y-z)Peak positions correspond to interatomic vectors in the unit cell. Imagine a vector connecting every pair of heavy atoms.Translate each vector so one atom of the pair is at the origin.Peak is located at position of other atom in the pair.Both as long as ward in addition to reverse vectors included (centrosymmetric).If n atoms in unit cell, then n2 peaks in Patterson. Pt derivative DNA polymerase b x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -zxzyP21212Contains 2 in addition to 21 axes x, y, z (reference) (2) -x, -y, z (2-fold around z)(3) -x + 1/2, y + 1/2, -z (21 around y)(4) x + 1/2, -y + 1/2, -z (21 around x)xzy

One symop per asymmetric unit x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z(4) x + 1/2, -y + 1/2, -zxzyFPH-FP gives Pt contribution x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z(4) x + 1/2, -y + 1/2, -zxzyThese x,y,z coordinates represent the solution we would like to obtain from interpreting the difference Patterson mapDifference Patterson contains peaks between every pair of Pt. x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z(4) x + 1/2, -y + 1/2, -zxzy16 vectors, every pairwise difference between symops. x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z

Each vector translated to origin16 vectors, every pairwise difference between symops. x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z(4) x + 1/2, -y + 1/2, -zxzy x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -zAssign difference vector to peak16 vectors, every pairwise difference between symops.uwy x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -zAssign difference vector to peak16 vectors, every pairwise difference between symops.uwy x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z(4) x + 1/2, -y + 1/2, -zxzy

Assign difference vector to peak16 vectors, every pairwise difference between symops. x, y, z (4) x + 1/2, -y + 1/2, -zuwy x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z(4) x + 1/2, -y + 1/2, -zxzy (4) x + 1/2, -y + 1/2, -z-(1)(x , y , z ½, -2y+1/2,-2zu=1/2 Harker sectionsPlanes perpendicular to a rotation or screw axis.It occurs when one of the coordinates of the difference vector is a constant.They offer clues to assigning a Patterson peak to the correct difference vector equation.In what planes do we expect difference Patterson peaks in P21212Name ops u v wAndrew 1-2 2x 2y 0David 1-3 2x-1/2 -1/2 2zJessica 1-4 -1/2 2y-1/2 2z Hanyoung 2-3 -1/2 -2y-1/2 2zJeff 2-4 -2x-1/2 -1/2 2zJohn 3-4 -2x 2y 0Scott 2-1 -2x -2y 0Steve 3-1 -2x+1/2 1/2 -2z 4-1 1/2 -2y+1/2 -2z 3-2 1/2 2y+1/2 -2z 4-2 2x+1/2 1/2 -2z 4-3 2x -2y 0x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z

In what planes do we expect difference Patterson peaks in P21212Name ops u v wAndrew 1-2 2x 2y 0David 1-3 2x-1/2 -1/2 2zJessica 1-4 -1/2 2y-1/2 2z Hanyoung 2-3 -1/2 -2y-1/2 2zJeff 2-4 -2x-1/2 -1/2 2zJohn 3-4 -2x 2y 0Scott 2-1 -2x -2y 0Steve 3-1 -2x+1/2 1/2 -2z 4-1 1/2 -2y+1/2 -2z 3-2 1/2 2y+1/2 -2z 4-2 2x+1/2 1/2 -2z 4-3 2x -2y 0x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -zIn what planes do we expect difference Patterson peaks in P21212a)w=0b)w=1/2c) v=1/2d) u=1/2x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -zHarker sections in P21212x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z X, Y, Z-X, -Y, Zu=2x, v=2y, w=01. X, Y, Z3. ½-X,½+Y,-Zu=2x-½,v=-½,w=2z1. X, Y, Z4. ½+X,½-Y,-Zu=-½,v=2y-½,w=2z

Which of these difference vectors is likely to correspond to the difference Patterson peak shown here X, Y, Z-X, -Y, Zu=2x, v=2y, w=01. X, Y, Z3. ½-X,½+Y,-Zu=2x-½,v=-½,w=2z1. X, Y, Z4. ½+X,½-Y,-Zu=-½,v=2y-½,w=2za)b)c)d) They are all equally likely.Which of these difference vectors is likely to correspond to the difference Patterson peak shown here X, Y, Z-X, -Y, Zu=2x, v=2y, w=01. X, Y, Z3. ½-X,½+Y,-Zu=2x-½,v=-½,w=2z1. X, Y, Z4. ½+X,½-Y,-Zu=-½,v=2y-½,w=2za)b)c)d) They are all equally likely.Harker section w=0W=0 X, Y, Z-X, -Y, Zu=2x, v=2y, w=00.168=2×0.084=x0.266=2y0.133=y

What is the value of zW=0 X, Y, Z-X, -Y, Zu=2x, v=2y, w=00.168=2×0.084=x0.266=2y0.133=ya)Zero b) x/y c) not specified by this Harker section.What is the value of zW=0 X, Y, Z-X, -Y, Zu=2x, v=2y, w=00.168=2×0.084=x0.266=2y0.133=ya)Zero b) x/y c) not specified by this Harker section.How can we determine the z coordinateHarker Section v=1/2V=1/21. X, Y, Z3. ½-X,½+Y,-Zu=2x-½,v=-½,w=2z0.333=2x-1/20.833=2×0.416=x0.150=2z0.075=z

What are the coordinates x,y,z as long as the heavy atomV=1/20.333=2x-1/20.833=2×0.416=x0.150=2z0.075=z0.168=2×0.084=x0.266=2y0.133=yW=0x=0.084, y=0.133, z=0.075x=0.416, y=0.133, z=0.075None of the aboveWhat are the coordinates x,y,z as long as the heavy atomV=1/20.333=2x-1/20.833=2×0.416=x0.150=2z0.075=z0.168=2×0.084=x0.266=2y0.133=yW=0x=0.084, y=0.133, z=0.075x=0.416, y=0.133, z=0.075None of the aboveResolving ambiguity in x,y,zFrom w=0 Harker section x1=0.084, y1=0.133From v=1/2 Harker section, x2=0.416, z2=0.075Why doesnt x agree between solutions They differ by an origin shift. Choose the proper shift to bring them into agreement.What are the rules as long as origin shifts Cheshire symmetry operators relate the different choices of origin. You can apply any of the Cheshire symmetry operators to convert from one origin choice to another.

Cheshire symmetry X, Y, Z -X, -Y, Z -X, Y, -Z X, -Y, -Z -X, -Y, -Z X, Y, -Z X, -Y, Z -X, Y, Z1/2+X, Y, Z1/2-X, -Y, Z1/2-X, Y, -Z1/2+X, -Y, -Z1/2-X, -Y, -Z1/2+X, Y, -Z1/2+X, -Y, Z1/2-X, Y, Z X,1/2+Y, Z -X,1/2-Y, Z -X,1/2+Y, -Z X,1/2-Y, -Z -X,1/2-Y, -Z X,1/2+Y, -Z X,1/2-Y, Z -X,1/2+Y, Z X, Y,1/2+Z -X, -Y,1/2+Z -X, Y,1/2-Z X, -Y,1/2-Z -X, -Y,1/2-Z X, Y,1/2-Z X, -Y,1/2+Z -X, Y,1/2+Z1/2+X,1/2+Y, Z1/2-x,1/2-Y, Z 1/2-X,1/2+Y, -Z1/2+X,1/2-Y, -Z1/2-X,1/2-Y, -Z1/2+X,1/2+Y, -Z1/2+X,1/2-Y, Z1/2-X,1/2+Y, Z1/2+X, Y,1/2+Z1/2-X, -Y,1/2+Z1/2-X, Y,1/2-Z1/2+X, -Y,1/2-Z1/2-X, -Y,1/2-Z1/2+X, Y,1/2-Z1/2+X, -Y,1/2+Z1/2-X, Y,1/2+Z X,1/2+Y,1/2+Z -X,1/2-Y,1/2+Z -X,1/2+Y,1/2-Z X,1/2-Y,1/2-Z -X,1/2-Y,1/2-Z X,1/2+Y,1/2-Z X,1/2-Y,1/2+Z -X,1/2+Y,1/2+Z1/2+X,1/2+Y,1/2+Z1/2-X,1/2-Y,1/2+Z1/2-X,1/2+Y,1/2-Z1/2+X,1/2-Y,1/2-Z1/2-X,1/2-Y,1/2-Z1/2+X,1/2+Y,1/2-Z1/2+X,1/2-Y,1/2+Z1/2-X,1/2+Y,1/2+ZFrom w=0 Harker section xorig1=0.084, yorig1=0.133From v=1/2 Harker section, xorig2=0.416, zorig2=0.075Apply Cheshire symmetry operator 10To xorig1 in addition to yorig1 Xorig1=0.084½-xorig1=0.5-0.084½-xorig1=0.416 =xorig2yorig1=0.133-yorig1=-0.133=yorig2Hence,Xorig2=0.416, yorig2=-0.133, zorig2=0.075Advanced case,Proteinase K in space group P43212Where are Harker sections

baThe 4 choices of origin are equally valid but once a choice is made, you must remain consistent.Choice 1Choice 2Choice 3Choice 4baX=0.80 Y=0.30X=0.30 Y=0.30X=0.30 Y=0.80X=0.80 Y=0.80X=0.20 Y=0.70X=0.70 Y=0.70X=0.70 Y=0.20X=0.20 Y=0.20Dear Sung Chul,I apologize as long as the lack of clarity. Both x,y,z in addition to -x,y,z can satisfy the vector equations in step 6 because both the Harker sections used thus far contain mirror symmetry that is consistent with either sign of “x” . In other words, the peaks in these Harker sections are related by Patterson symmetry that makes them consistent with either sign of “x”However, there are other difference vector equations in this space group ( as long as example symop 1 – symop 5) that are not reflected by these mirror planes. These can be used to discriminate which sign of “x” is correct. Only one of these choices of the sign of “x” will be able to predict correctly Patterson peaks specified by this difference vector. My thought is that one sign of “x” is consistent with space group P43212, in addition to the other sign of “x” is consistent with space group P41212.MikeOn 01/08/2015 06:21 PM, wrote:>> Dear Michael R. Sawaya,>> >> I have learned a lot basic concept in addition to practical methods using presentation materials of the lectures as long as CHEM M230D Course. THey are prepared well so that anyone can underst in addition to the contents easily even without the explanation by a lecturer.>> >> I have one question about content of the presentation ppt file as long as the lectures as long as CHEM M230D Course. In the ppt file or on the web page as long as Difference Patterson Maps in addition to Determination of Heavy Atom Sites, I dont know why both x,y,z in addition to -x,y,z can satisfy the difference vector equations in the step 6. Could you explain the reason>> >> Best regards,>> SungChul Ha>>

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