Reverse Engineering Gene Networks with Microarray Data

Transcription

1 Reverse Engineering Gene Networks with Microrry Dt Roert M Mllery Avisors: Dr Steve Cox n Dr Mrk Emree August 25, 2003 Astrct We consier the question of how to solve inverse prolems of the form e At x(0) = x(t) for n unkown mtrix A, given mesurements of x(t) t ifferent time points Prolems of this form hve pplictions in reverse engineering gene networks In prticulr, we exmine the cses where A is circulnt n Toeplitz We re lso le to exten our finings of the circulnt cse to some generliztions of circulnt mtrices Introuction DNA Microrrys cn e use to etermine mesurements of cellulr gene proucts t given point in time These concentrtions of gene prouct provie clues to the overll interction of the genes in the gene network eing stuie We cn mesure the perturtions x 1,,x n of the gene mrna expression concentrtions from the stey stte, which re governe y the eqution x(t) t = Ax(t), (1) where x(t) = (x 1 (t),, x n (t)) T In other wors, the rtes of chnge of the gene prouct concentrtions re etermine y the evitions from the stey stte of ll the gene proucts present If A is unstructure, then A contins n 2 egrees of freeom, so we woul expect tht n 2 concentrtion mesurements woul suffice to etermine A uniquely However, mesuring the given gene prouct concentrtions t prticulr time is oth time consuming n expensive Gene networks cn e on the orer of 10,000 genes, so tking 10 8 mesurements is impossile Hence, we woul like to impose some structure on A tht will llow us to tke fewer gene concentrtion mesurements, enling us to etermine A more esily This will provie solution to our gene network which is esy to fin n cn provie strting point for etermining the exct structure of the gene network In prticulr, we will first stuy the cse where A is circulnt n the cse where A is Toeplitz It is expecte tht if A is circulnt, then A cn e uniquely etermine from mesurement of the n gene proucts t single point in time This seems intuitive since circulnt A contins only n egrees of freeom If A is Toeplitz, we expect A my e etermine y the mesurement of the n gene proucts 1

2 t two points in time since Toeplitz A contins 2n 1 egrees of freeom Once A is etermine, the gene prouct concentrtions for ny time t re given y x(t) = e At x 0, (2) where x 0 = x(0) is the vector of gene prouct evitions from the stey stte cuse y perturtion to the system t t = 0 Circulnt Mtrices Circulnt mtrices re those squre mtrices C of the form c 1 c 2 c 3 c n c n c 1 c 2 c n 1 C = circ(c 1, c 2,, c n ) = c n 1 c n c 1 c n 2 c 2 c 3 c 1 It is instructive to consier wht grph of four gene network represente y eqution (1) looks like when A is circulnt Let A = circ(,, c, ) Then eqution (1) ecomes x 1 (t) c x 1 (t) x 2(t) c x x 3 (t) = 2 (t) c x 3 (t) x 4 (t) c x 4 (t) Figure 1 elow represents the grph of 4 gene network Genes x 1,,x 4 re represente y the circles numere 1 through 4 Interctions etween genes re inicte y rrows rwn etween the genes, with weights to escrie the mount of effect one gene hs on nother We sy, for exmple tht gene 1 feels itself with weight of, gene 2 with weight of, gene 3 with weight of c, n gene 4 with weight of 1 c c Fig 1 Grph of four gene network with circulnt A Circulnt mtrices hve nice properties tht provie n elegnt solution to our prolem In prticulr, ll circulnt mtrices of orer n re igonlizle y the Fourier mtrix F n, where F n = 1 n 1 ω ω 2 ω n 1 1 ω 2 ω 4 ω 2(n 1), 1 ω ( n 1) ω 2(n 1) ω (n 1)(n 1) 2

3 n ω = e 2πi n [1] From now on we will simply use F to enote the Fourier mtrix of orer n Hence, if A is circulnt, we hve tht A = F ΛF, where Λ = ig(λ 1,, λ n ) n λ 1,, λ n re the eigenvlues of A repete ccoring to multiplicity Also, we hve tht F is unitry [1], ie FF = F F = I Let us consier the mtrix exponentil in (2), utilizing its Tylor series representtion We hve tht e At = I + At + 1 2! A2 t ! A3 t 3 + = F F + F ΛFt + 1 2! (F ΛF) 2 t ! (F ΛF) 3 t 3 + = F (I + Λt + 1 2! Λ2 t ! Λ3 t 3 + )F = F e Λt F (3) Now, let us suppose tht we tke mesurement of the mrna concentrtions t specific time t m > 0 This mens tht we know x(t m ) n woul like to show tht we cn solve for A in (2) Hence, from (3) we hve tht x(t m ) = e Atm x 0 = F e Λtm Fx 0 Multiplying oth sies y F, n efining ˆx(t) = F x(t) to e the Discrete Fourier Trnsform of x(t), we get ˆx(t m ) = FF e Λtm Fx 0 = e Λtm ˆx 0 (4) Since Λt m is igonl mtix, e Λtm = ig(e λ1tm,, e λntm ), n we otin tht e λ1tm e ˆx(t m ) = λ2tm 0 ˆx e λntm In other wors, ˆx j (t m ) = e λjtm ˆx j (0) for ll j {1,, n} (5) Eqution (5) cn e solve for ech λ j, n circulnt mtrix A = F ΛF stisfying x(t m ) = e Atm x 0 cn e etermine since A is circulnt for ny vlues of λ 1,,λ n We woul like to show tht A is unique since we esire to etermine A with only one mesurement Assume tht e A1tm x 0 = e A2tm x 0 for two circulnts A 1 n A 2 Suppose tht A 1 = F Λ 1 F n A 2 = F Λ 2 F for igonl mtrices Λ 1 = ig(λ 1,,λ n ) n Λ 2 = ig(µ 1,, µ n ) We get tht F e Λ1tm Fx 0 = F e Λ2tm Fx 0 Hence e Λ1tm ˆx 0 = e Λ2tmˆx 0 It follows tht e λjtm = e µjtm for ll j {1,, n} (6) Let j {1,,n} e given Let λ j t m = + i n µ j t m = c + i,,, c, From (6) we get tht e +i = e c+i This implies e e i = e c e i In other wors, e (cos + i sin) = e c (cos + i sin) Equting the rel n imginry prts, we otin e cos = e c cos n e sin = e c sin Since sin 2 + cos 2 = sin 2 + cos 2 = 1, we hve tht e 2 = e 2c Thus,, c implies = c Also, it cn e seen tht n must iffer y multiple of 2π since cos = cos, n sin = sin, ie = + 2πk j, k j Hence, µ j t m = + ( + 2πk j )i Diviing y t m, we get µ j = +i t m + 2πkj t m i We conclue tht µ j = λ j + 2πk j t m i, for some k j (7) 3

4 In other wors, solving eqution (5) for ech λ j oes not necessrily prouce unique solution Hence, one time mesurement will not suffice to etermine λ 1,, λ n ue to the perioicity inherent in the exponentil function However, in the cse tht ech e λjtm is rel vlue, we cn solve for λ j uniquely Whenever A is symmetric, we re gurntee rel eigenvlues [2] Thus, when A is symmetric circulnt, one time mesurement will suffice to etermine A uniquely This les to the following sttement Theorem 1: Let x 0 = x(0) e given When A is symmetric circulnt, e At x 0 = x(t) cn e solve uniquely for A with vlue of x(t) t single time t m In orer to ssure tht we re le to etermine λ 1,, λ n uniquely in the non-symmetric cse, it is necessry to tke secon mesurement of x(t) t lter time However, there re restrictions on when the secon time mesurement cn occur Suppose tht we tke mesurements of x(t) t ifferent times t 1 n t 2 Then we know tht ˆx j (t 1 ) = e λjt1ˆx j (0) n ˆx j (t 2 ) = e λjt2ˆx j (0), for j = 1,,n Suppose tht we fin two ifferent vlues µ n σ tht stisfy oth equtions for λ j We hve tht e µt1 = e σt1 n e µt2 = e σt2 From the results of eqution (7), we see tht µ = σ + 2πn t 1 i n µ = σ + 2πm t 2 i, for some n, m Since µ σ, m n n re nonzero integers Thus, 2πn t 1 = 2πm t 2, which implies t 2 = m, for nonzero integers m, n (8) t 1 n Hence, if t 2 /t 1 /, then µ = σ n we hve tht λ j my e solve for uniquely with vlues of x(t 1 ) n x(t 2 ) We otin the following theorem Theorem 2: Let x 0 = x(0) e given When A is generl circulnt, e At x 0 = x(t) cn e solve uniquely for A with vlues of x(t) t two istinct times t 1 n t 2, provie tht the rtio of t 2 to t 1 is n irrtionl numer The question remins whether tking more thn two time mesurements of x(t) will llow us to etermine A uniquely without tking mesurement of x(t) t n irrtionl time Suppose tht we know x(t 1 ), x(t 2 ),, x(t m ), where t 1,, t m Let A = F ΛF, where Λ = ig(λ 1,, λ n ) Let λ j {λ 1,, λ n } e given Then e λjt1ˆx j (0) = ˆx j (t 1 ), e λjt2ˆx j (0) = ˆx j (t 2 ),, e λjtm ˆx j (0) = ˆx j (t m ) Suppose tht there exist two complex numers µ n σ tht stisfy these equtions for λ j Then y eqution (7) σ = µ + 2πk1 t 1 i = µ + 2πk2 Then we hve tht σ = µ + 2πk11 1 t 2 i = = µ + 2πkm t m i, k 1,,k m Let t j = j / j, j, j +, for j = 1,,n i = µ + 2πk22 2 i = = µ + 2πkmm m i Let γ = lcm( 1,, m ) e the lest common multiple of 1,, m We hve tht σ = µ + 2πk11 γ i = µ + 2πk22 γ i = = µ + 2πkmm γ i, where k γ for = 1,,m Let δ = lcm( γ1 1,, γm m ) Then these m equlities reuce to the equlity σ = µ + 2π γ ki, where k δ is ritrry Hence, the m mesurements t rtionl times t 1,, t m o not etermine λ j uniquely Therefore, we my revise Theorem 2 Theorem 2 : Let x 0 = x(0) e given In orer to uniquely solve the inverse prolem e At x 0 = x(t) for circulnt A, it is necessry n sufficient tke mesurements of x(t) t two istinct times t 1 n t 2, such tht the rtio of t 2 to t 1 is n irrtionl numer Thus, y tking one time mesurement in the cse tht circulnt A is symmetric or two time mesurements in the generl cse, we re le to etermine A uniquely As si efore, this solution to eqution (1) for A circulnt cn provie strting point from which to etermine etter moel for the gene network Block Circulnt Mtrices with Circulnt Blocks An interesting n similr result to our solution for A circulnt hols when A is ssume to e lock 4

5 circulnt mtrix with circulnt locks These re mtrices of the form C 1 C 2 C m C m C 1 C m 1 A m,n =, C 2 C 3 C 1 where ech C j, j = 1,,m, is circulnt mtrix of orer n For nottionl ese we will sy tht n m m lock circulnt mtrix with n n circulnt locks is of the clss BCCB m,n [1] It is importnt to note tht mtrix in BCCB m,n is not necessrily circulnt, n the topology of generl four gene network with A BCCB m,n iffers from the topology of the circulnt cse in Figure 1 Consier the mtrix c c A = BCCB c 2,2 c Figure 2 elow shows the grph of the four gene network etermine y A from eqution (1) 1 2 c c 4 3 Fig 2 Grph of four gene network with A in BCCB 2,2 Also, suppose tht mtrix A hs orer p When solving eqution (2) for mtrix A tht is lock circulnt with circulnt locks, we must tke into consiertion oth the numer of locks n the size of the circulnt locks Suppose tht p = mn n p = qr, m, n, q, r + We cn fin ifferent solutions for A se on whether we tke A BCCB m,n or A BCCB q,r For exmple, the mtrices c e f c e f c f e c f e e f c c e f A 1 = BCCB f e c 3,2 n A 2 = BCCB e f c 2,3 c e f f e c c f e e f c re oth 6 6 lock circulnt mtrices with circulnt locks However, they re clerly not equivlent n o not represent the sme gene network This fct my prove useful when reverse engineering the gene network using numer of mesurements of the gene prouct concentrtions One type of BCCB mtrix my serve s etter moel to strt from thn nother when reverse engineering the gene network Let us suppose tht A BCCB m,n Thus, A is mtrix of orer mn Then, we know tht A is in 5

6 BCCB m,n if n only if it my e igonlize y the unitry mtrix F m F n = 1 n F n F n F n F n F n ωf n ω 2 F n ω m 1 F n F n ω 2 F n ω 4 F n ω 2(m 1) F n, F n ω (m 1) F n ω 2(m 1) F n ω (m 1)(m 1) F n where F m n F n re the Fourier mtrices of orer m n n respectively, F m F n is their irect or Kronecker prouct, n ω = e 2πi n s efine ove [1] Hence, we hve tht A = (F m F n ) Λ(F m F n ) where Λ = ig(λ 1, λ 2,, λ mn ) We cn now otin n nlog to eqution (3) for A BCCB m,n, nmely e At = (F m F n ) e Λt (F m F n ) (9) Now, consier the eqution x(t) = e At x 0 When A BCCB m,n we hve x(t m ) = (F m F n ) e Λt (F m F n ) Left multiplying oth sies y (F m F n ) n efining x(t) = (F m F n )x(t), we get (F m F n ) x(t) = (F m F n )(F m F n ) e Λt (F m F n )x 0 Since F m F n is unitry, x(t) = e Λt x 0, n (10) x j (t) = e λjt x j (0) for ll j {1,,mn} (11) By the sme rgument s in the cse where A is circulnt, we hve the following two corollries to Theorem 1 n Theorem 2 Corollry 1: Let x 0 = x(0) e given When A is symmetric lock circulnt mtrix with circulnt locks, e At x 0 = x(t) cn e solve uniquely for A with vlue of x(t) t single time t m Corollry 2: Let x 0 = x(0) e given Suppose tht A is lock circulnt mtrix with circulnt locks In orer to uniquely solve the inverse prolem e At x 0 = x(t) for A, it is necessry n sufficient tke mesurements of x(t) t two istinct times t 1 n t 2, such tht the rtio of t 2 to t 1 is n irrtionl numer Level-N Circulnts We cn mke further extension of our circulnt cse, consiering mtrices A tht re level-n circulnts A level-1 circulnt is just n orinry circulnt A level-2 circulnt is one tht is in BCCB A level-3 circulnt is lock circulnt whose locks re level-2 circulnts In generl, level-n circulnt is lock circulnt whose locks re level-(n 1) circulnts The following mtrix C is n exmple of the smllest level-3 circulnt c e f g h c f e h g c g h e f c h g f e C = e f g h c f e h g c g h e f c h g f e c 6

7 g g 5 6 e h 8 e f c c h f e f h 1 2 c h c 7 f e g 4 3 g Fig 1 Grph of the eight gene network represente y C ove, the smllest level-3 circulnt We will use C(N) m1,,m N to enote the clss of level-n circulnts consisting of n m N m N lock circulnt whose locks re m (N 1) m (N 1) lock level-(n-1) circulnts n elong to C(N 1) m1,,m (N 1) For exmple, the mtrix C ove elongs to the clss of level-n circulnts enote y C(3) 2,2,2 Suppose tht mtrix A in eqution (1) elongs to C(N) m1,,m N Similr to the circulnt n level-2 circulnt cse, A my e igonlize y the mtrix F mn F m(n 1) F m1 [1] Hence, we hve tht Defining x(t) = (F mn F m1 )x(t), we see tht A = (F mn F m1 ) Λ(F mn F m1 ) (12) x(t) = e Λt x 0 n hence, (13) x j (t) = e λjt x j (0) (14) Hence we hve the following generliztions for Theroem 1 n Theorem 2 Theorem 1 : Let x 0 = x(0) n positive integer N e given When A is symmetric level-n circulnt mtrix, e At x 0 = x(t) cn e solve uniquely for A with vlue of x(t) t single time t m Theorem 2 : Let x 0 = x(0) n positive integer N e given Suppose tht A is level-n circulnt mtrix In orer to uniquely solve the inverse prolem e At x 0 = x(t) for A, it is necessry n sufficient tke mesurements of x(t) t two istinct times t 1 n t 2, such tht the rtio of t 2 to t 1 is n irrtionl numer When reverse engineering gene network with n genes, it will prove useful to consier the vrious level-n circulnts which hve the orer n Suppose tht we hve gene network with 100 genes Then, we cn solve our inverse prolem for level-3 circulnts elonging to vrious clsses, incluing, C(3) 2,2,25,C(3) 2,25,2, C(3) 5,5,4, n C(3) 2,5,10 We coul lso solve our prolem for level-4 circulnt of the clss C(4) 5,5,2,2 Also, suppose tht n hs prime fctoriztion given y n = p 1 1 p 2 2 p l l We see tht the lrgest numer of levels our level-n circulnt coul hve woul e N = l In the cse where n = 100, 4 is the 7

8 lrgest possile vlue for N Toeplitz Mtrices Toeplitz mtrices re those m n mtrices T of the form n (n 1) T = (n 2) m (m 1) (m 2) 0 Before preceing to the generl Toeplitz cse, we will first consier the solution to eqution (2) when A is n n n upper tringulr Toeplitz mtrix Define A = utoeplitz( 1, 2,, n ) to e the upper tringulr Toeplitz mtrix given y A = 1 2 n 0 1 n We ssert tht e At is lso n upper tringulr Toeplitz mtrix Consier the expnsion e At = I + At + 1 2! A2 t ! A3 t 3 + Given two upper tringulr Toeplitz mtrices, B = utoeplitz( 1,, n ) n G = utoeplitz(g 1,, g n ), their prouct BG is the upper tringulr Toeplitz n given y BG = utoeplitz( 1 g 1, 1 g g 1,, n g 1 + n 1 g g n ) Hence A j is upper tringulr Toeplitz for ll positive integers j The sum of upper tringulr Toeplitz mtrices is gin upper tringulr Toeplitz This implies e At is n upper tringulr Toeplitz mtrix Let iscrete time t m > 0 e given Then, let T = e Atm e the upper tringulr Toeplitz mtrix T = utoeplitz(u 1, u 2,,u n ) Assuming we know x(t) t t = 0 n t = t m, we wish to solve the inverse prolem e Atm x(0) = x(t m ) for A We cn esily solve the eqution Tx(0) = x(t m ) for T = e Atm Using T we will e le to clculte e At t ny time t m We hve u 1 u 2 u n x 1 (0) x 1 (t m ) 0 u 1 u n 1 x 2 (0) Tx(0) = = x 2 (t m ), n hence 0 0 u 1 x n (0) x n (t m ) u 1 x 1 (0) + u 2 x 2 (0) + + u n x n (0) u 1 x 2 (0) + u 2 x 3 (0) + + u n 1 x n (0) = u 1 x n 1 (0) + u 2 x n (t m ) u 1 x n (0) x 1 (t m ) x 2 (t m ) x n 1 (t m ) x n (t m ) (15) It is esily seen tht u 1 = xn(tm) x Thus, we cn solve for u n(0) 1 whenever x n (0) 0 Suppose x n (0) is nonzero Once we otin u 1, we re le to clculte u 2 = xn 1(tm) u1xn 1(0) x n(0) In generl, u 1 = x n(t m ) x n (0) u j = x n j+1(0) x n (0) n 1 j 1 x n (0) k=1 u k x n j+1 (16) 8

9 whenever x n (0) 0 Thus, we cn otin the full mtrix T = e Atm We woul like to show tht there exists unique A such tht e Atm = T n provie metho for etermining A Suppose tht e Atm = T where A = utoeplitz( 1, 2,, n ) n T = utoeplitz(u 1, u 2,, u n ) Let V = At m = utoeplitz(v 1, v 2,, v n ) We woul like to solve the prolem e V = T for V We hve tht I + V + 1 2! V ! V 3 + = T Consier the element u 1 of T We see tht 1 + v ! v ! v3 1 + = u 1, so u 1 = e v1 Since u 1 = xn(tm) x n(0) is rel, we my solve for v 1 uniquely We will nee the following lemm Lemm 1: Let V e the n n upper tringulr Toeplitz mtrix given y V = utoepliz(v 1, v 2,,v n ) Define V k to e the sumtrix of V given y V k = utoeplitz(v 1, v 2,, v k ), where k n Then, [e V k ] i,j = [e V ] i,j for i, j {1,,k} Proof: We hve tht v 1 v k v k+1 v n 0 ( V = v1 v 2 v k+1 Vk W 0 0 v 1 v = n k 0 V n k v 1 For j +, ( V j V j = k Ψ(V k, V n k, W) 0 V j n k where Ψ is some function of V k, V n k, n W Thus, we see tht ), where W = ), e V = I + V + 1 ( 2! V 2 e V k Ω(V + = k, V n k, W) 0 e V n k v k+1 v n v 2 v k+1 where Ω is some function of V k, V n k, n W Thus, for 1 k n, [e V ] i,j = [e V k ] i,j for i, j {1,,k} We now return to our question of etermining v 1, v 2,,v n in mtrix V Consier e V = T From Lemm 1 we see tht e V k = (e V ) k, for 1 k n In other wors, e V k = T k = utoeplitz(u 1,,u k ) We woul like to show tht u k = f(v 1,, v k 1 ) + v k e v1, where f is some function of v 1,, v k 1 v 1 v k T k = e V k = I ! v 1 v k 2 ), + 1 v 1 v k 3! v 1 Let us consier only the vlue of u k = [T k ] 1,k 0 v 1 0 v 1 u k = v k + 1 2! (2v 1v k + f 1 (v 1,,v k 1 )) + 1 3! (3v 1 2 v k + f 2 (v 1,, v k 1 )) + 1 4! (4v 1 3 v k + f 3 (v 1,, v k 1 )) + = v k (1 + v ! v ! v3 1 + ) + f(v 1,, v k 1 ) = f(v 1,, v k 1 ) + v k e v1 (17) References [1] P J Dvis Circulnt Mtrices Chelse Pulishing: New York, 1994 [2] S H Frieerg Liner Alger, 3r e Prentice Hll: Upper Sle River, NJ,

10 [3] G Heinig Inverse Prolems for Hnkel n Toeplitz Mtrices Liner Alger Appl 165 (1992), 1 23 [4] R M Reheffer Differentil Equtions Jones n Brtlett Pulishers: Boston,