429-2009-HW6key

429-2009-HW6key - SB3 HW6 Key_v2(corrected 2.2(b 3 7(e bonus by An-Chi Wei 1 Q2.1 dY = 2 Y Y(0 = 1 dt sY s Y(0 = Y(s = 2 s Y(s 1 1 s(s(s 2 t Y(t =

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SB3 HW6 Key_v2 (corrected 2.2(b), 3, 7(e), bonus) by An-Chi Wei Oct 28, 2009 1. Q2.1 1/2 1 2 2 21 22 1 2 212 ,( 0 ) () (0 ) 1 () ( ) ln 2 2 tt t t dY YY dt sY s Y Y s s Ys ss s Yt e e e e t αα α β βα ββ βββ −− =− = −= =+ ++ + = + = ±± ± 0 50 100 150 200 250 300 350 400 450 500 0 10 20 30 40 50 60 time(min) Y(t) beta1=1/hr, beta2=1/min, alpha=1/hr
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mm 2. Q2.2 (a)The dynamic equation for the concentration of mRNA of gene Y: - β : mRNA production rate, α : mRNA degradation rate The dynamic equation for the concentration of protein due to produ m m dY Y dt βα = ction of p copies per mRNA and degradation rate - m dY pY Y dt α = m () (i)With fast RNA dynamics ( >> ), the quasi steady state assumption (mRNA reaches steady state with respect to slower process) can be used. Thus, 0 -- effective m m m m m b dY Y dt dY pY Y p Y Y dt αα β =⇒ = == = m protein production rate (ii) with slow RNA dynamics( α << α ) , , () ( ) 1 ( ) ()() a m m m m m m m t t m m m p sY s Y s Y s ss s pY s p sY s pY s Y s Y s s s ee Yt p ββ α α = =− = + = = ++ + =+ + −−−− ±± ± ± ± ± 1/2 m s α << α , 1 ( ) (1 ) ( 1 ) 2 ln 2 m m m t t m m m t m p e p e pp e t ≈+ = ⇒== ⇒=
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0 0 0 1 2 12 1 3. Q2.3 assume X >K and / >K (set =1/min, =1/hr, K=10, X =50) () ( ) ( ) let t and t represent the time when X and Y reach threshold K X(t )=K, Y(t )=K ( t Xt X t Yt X K Y Zt Y K Z t βα β α =Θ > − =Θ> ± ± ± 11 1 1 2 1 1 ) ( t) , Y(t)= 2i ( ) ( ) (i) t>t , close in left half plane (poles at 0, - ) (ii)t<t , close in rig st st ii st st t st Y t Z ee sY s Y s Y s ss s ds e ds e ββ πα −− ∞∞ −∞ =Θ> − =− = + = ++ ∫∫ ± ²² ² 1 22 2 β - α (t-t ) 1 α 1 β - α (t-t ) 2 α 2 ht half plane (no ploles) (1-e ) ,t t Thus, Y(t)= 0 , t<t Similarly, , (1-e ) ,t t z(t)= 0 , t<t To reach thr st st sZ s Z s Z s s = + ² 21 1/2 2 101 β - α (t -t ) α 1/2 - α (t -t ) 1/2 2 αα 1/2 2 / ln2 1/2 α eshold K, X(t )=K=X t =0 -1 -1 Y(t )=K= (1-e ) t = ln(1 ) // response time t for Z Z(t )= (1-e ) t ln 2 (ln 2 ln(1 )) (as / >K, t ) K K KK K t αβ ⇒− = = ⇒= + = ≈+
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0 50 100 150 200 250 300 350 400 0 10 20 30 40 50 60 concentration Time(min) Y(t) Z(t)
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12 1 2 11 1 22 2 1 2 1 2 2 2 4. Q2.5 assume / >K , K , X(0)=Y (0)=Y (0)=0 1 2 () ( ) ( ) let t and t be the time when X reach threshold K and K X(t )=K , X(t )=K For X, ( XY Y Xt t X Yt X K Y Y sX s βα =Θ − =Θ > ± ±
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This note was uploaded on 03/30/2010 for the course SBE 580.429 taught by Professor Joelbader during the Fall '09 term at Johns Hopkins.

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429-2009-HW6key - SB3 HW6 Key_v2(corrected 2.2(b 3 7(e bonus by An-Chi Wei 1 Q2.1 dY = 2 Y Y(0 = 1 dt sY s Y(0 = Y(s = 2 s Y(s 1 1 s(s(s 2 t Y(t =

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