A widely used display of protein shapes is based on the coordinates of the alpha carbons - - C α

Transcription

1 Nice plottig of proteis: I A widely used display of protei shapes is based o the coordiates of the alpha carbos - - C α -s. The coordiates of the C α -s are coected by a cotiuous curve that roughly follows the directio i space of the protei chai. Each amio acid has oly oe alpha carbo ad the distace betwee sequetial C α -s is about.8 agstrom. There is oly oe eceptio to the.8 rule, which is the cis isomer of prolie for which the distace is smaller tha the above. The uiform distributio of the C α -s alog the protei curve makes plots of the protei backboe relatively easy to do. The simplest solutio (that we used already) is to coect the coordiates by a straight lie iterpolatig from oe C α to the et. This procedure creates a zigzag lie, which is ok, but ot great ad ot pleasig to the eye. Starig at proteis shapes ad lookig for iterestig features ca be difficult (eamples for iterestig features are structural domais that are shared betwee the proteis, similar active sites of evolutioary related proteis, etc). There are may automated algorithms to look for the features metioed above. However, the huma eye is i may cases a better detectio device. Therefore, producig better pictures of protei chais is likely to help basic research i this area. Oe thig that our eyes do ot like is discotiuous derivatives. The eye ca detect secod order derivative that is ot cotiuous. The curve of the protei chai, which we plot i three dimesios by coectig liearly the C α -s, is discotiuous i the first derivative. Here we seek aother represetatio that will make the protei curve differetiable at least to a secod order. That is, we are facig with the topic of iterpolatio betwee poits (the C α -s positios) that represet a curve. Iterpolatio The first approach to iterpolatio that we cosider is the use of polyomials. A set of N { i i } i= poits ( y) ca be approimated by a polyomial y ad a variable y = c + c + c + c + + c 0... A alterative represetatio of the th order polyomial is ( 0 ( ( (...)) ) ) y = c + c + c + or yet aother represetatio usig the roots of the polyomial ( ) ( ) y = c... r r

2 (Note that i creatig a protei curve we eed to cosider three fuctios. A fuctio for each of the coordiates, y, ad z. The curve will be parameterized with idepedet cotiuous variable t that is equal to the amio acid ide i at the coordiates of Cα ( i). Hece, we compute the cotiuous curves (), t y(), t z() t. The polyomial is cotiuous ad differetiable at all orders ad therefore suggests itself as a plausible represetatio of the protei curve). I MATLAB we ca obtai the roots of the polyomial by the roots commad. For eample, cosider the polyomial below: y = y + y + 5 The roots commad fids the zeroes of the polyomial >> roots([5 0 ]) as = i i The roots determie the polyomial up to a costat multiplier. The commad poly creates the polyomial from the root: >> poly(roots([5 0 ])) as = The polyomial so created (always) has a coefficiet c =, to recover the origial polyomial we eed to multiply all coefficiets by the origial c, ad by 5 i the specific eample above. How to compute polyomials efficietly? Here is a MATLAB loop that computes y = c0 + c + c + c c The coefficiets are stored i a vector c of legth + le= legth(c); polyomial = c(); ym=; for i=:le

3 ed ym = y * ym; polyomial = polyomial + c(i)*ym; The umber of operatios to compute the value of the polyomial is (le-)* It is possible to compute the polyomial more efficietly by usig the equivalet formula ( 0 ( ( (...)) ) ) y = c + c + c + We have: le= legth(c); polyomial = c(le); for i=le-:-: polyomial = polyomial*y + c(i); ed Oly (le-)* operatios are required this time, so this is clearly a better way of computig values of the polyomial. Yet aother formulatio (Newto represetatio) is give below. Cosider the iterpolatio betwee four poits {(, )} 4 i yi. I the Newto represetatio we write the i= polyomial as: ( ) ( )( ) ( )( )( ) y = c + c + c + c 4 To determie the coefficiets c i we use the four poits y = c y = c+ c y = c+ c + c y = c + c + c + c ( ) ( ) ( )( ) ( ) ( )( ) ( )( )( ) This set of liear equatios (for the c i ) is relatively easy to solve. I a matri form we write: c y ( ) 0 0 c y = ( ) ( )( ) 0 c y ( 4 ) ( 4 )( 4 ) ( 4 )( 4 )( 4 ) c4 y4 It is obvious that c = y, substitutig we obtai the followig liear equatio

4 0 0 0 c y 0 ( ) 0 0 c y y = 0 ( ) ( )( ) 0 c y y 0 ( 4 ) ( 4 )( 4 ) ( 4 )( 4 )( 4 ) c4 y4 y By dividig lie,, ad 4 by ( ), ( ) ad ( ) c y c y = 0 ( ) 0 c y 0 ( 4 ) ( 4 )( 4 ) c4 y4 respectively, we obtai 4 yi yj where yij =. The last matri equatio provides a immediate solutio for the i j coefficiet c. The same type of process ca be repeated to determie other coefficiets. This brigs us to a large set of problems of solvig liear equalities of the type A = b where is a vector of ukow of legth, b is a vector of parameters of the same legth ad A is a matri. A simple case to star with is of triagular matrices (this icludes the case which we just studied). Triagular problems Eample a 0 0 b a a 0 = b a a a b which is rather easy to solve. We ca immediately write = b a. Usig the (ow) kow value of we ca write for, = ( b a ) a. Similarly we ca write for, = ( ) b a a a For the geeral case we ca write a implicit solutio (i terms of the earlier j =,..., i ) j i = b a a i i ij j ii j= Note that a similar procedure applied to the upper triagular matri

5 a a a b 0 a a = b 0 0 a b To solve a geeral liear problem we search for a way of trasformig the matri to a triagular form (which we kow already how to solve). Formally, we seek the so-called LU decompositio i which the geeral A matri is decomposed ito a lower triagular matri L, ad a upper triagular matri U ( A = LU ). Note that if such a decompositio is kow, we ca solved the liear problem i two steps Step. A = b LU = b LU ( ) = b Ly = b fid y usig the lower triagular matri L Step. U = y fid usig the upper triagular matri U A way of implemetig the above idea i practice is usig Gaussia elimiatio. Gaussia elimiatio is a actio that leads to a LU decompositio discussed above, eve if the aalogy is ot obvious. I this course we will ot prove the equivalece. Gaussia elimiatio Cosider the followig system of liear equatios = where deotes ay umber differet from zero. We ca elimiate the ukow from rows to (the geeral matri is of size ). We multiply the first row by ai a ad subtract the result from row i. By repeatig the process times we obtai the followig (adjusted) set of liear equatios that has the same solutio 0 0 = 0 0

6 We ca work o the ewly obtaied matri i a similar way to elimiate from row to. This we do by multiplyig the secod row by ai a ad subtract the results from rows to. The (yet aother) ew matri ad liear equatios will be of the form = It should be obvious how to proceed with the elimiatio ad to create (a upper) triagular matri that we kow by ow how to solve.