Solution Manual Dynamic Systems Biology Modeling 1st Joseph Distefano (SOLUTIONS or ANSWERS to SELEC

SOLUTION MANUAL

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

SOLUTIONS or ANSWERS to SELECTED EXERCISES in CHAPTERS 1 – 17 Chapter 1 Has no Exercises.

Chapter 2 2.1 is a discussion question.

S2.2: Substitute et for y, differentiate four times and substitute in the homogeneous equation: ( 4   3  3 2  5  2)e  t  0 Factoring: (  1)3 (  2)e t  0 Two distinct solutions are λ = 1 and λ = −2, or y(t) = et and e−2t, which are linearly independent because the only c1 and c2 that satisfy c1et  c2 e2t  0 for all t are c1  c2  0. Since one of the roots has multiplicity 3, the overall solution of the ODE is y (t )  c1et  c2 e 2t  c3tet  c4t 2 et  (c1  c3t  c4t 2 )et  c2 e 2 t

Since the only c1, …, c4 that satisfy this equation for all t are c1 = c2 = c3 = c4 = 0, the four solutions are independent.

S2.3: Let

x1  y x2  y  x1

x3  y  x 2 x4   y  x3  y  x 4

From the equation, x4  2 x1  5 x2  3 x3  x4  u. Therefore, we have:

 x1   0 1  x   0 0 x   2     x3   0 0     x 4   2 5

0 0 0   0  1 0  x u 0  0 1    3 1 1 

1

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

S2.4: Following the hints, let x1 (k )  y2 (k ), the output. Then x1 (k  1)  y2 (k  1)  x2 (k ) x2 (k  1)  y2 (k  2)  x3 (k ) x3 (k  1)  y2 (k  3)  x4 (k ) Now substitute xi into the second equation: x4 (k ) 

y1 (k  2)  x2 (k )  0 2

and solve for y1:

y1 (k  2)  2  x2 (k )  x4 (k )  Now substitute y1 into the first equation: 2  x2 (k )  x4 (k )  

5  x2 (k  1)  x4 (k  1)  3



 x2 (k  2)  x4 (k  2)   u (k  1) 3

Then simplify and substitute x1 (k )  x2 (k  1), etc.

5 5 1 1 2 x2 (k )  2 x4 (k )  x1 (k )  x3 (k )  x1 (k  1)  x2 (k )  u (k  1) 3 3 3 3

Which is equivalent to 5 5 1 1 2 x2 (k  1)  2 x4 (k  1)  x1 (k  1)  x3 (k  1)  x1 (k )  x2 (k  1)  u (k ) 3 3 3 3 and 5 5 1 1 2 x3 (k )  2 x4 (k  1)  x2 (k )  x4 (k )  x1 (k )  x3 (k )  u (k ) 3 3 3 3

Therefore, the difference equation for x4 is: 1 5 7 5 1 x4 (k  1)   x1 (k )  x2 (k )  x3 (k )  x4 (k )  u (k ) 6 6 6 6 2

which, in matrix form, is:

2

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

 0  x1 (k  1)    x (k  1)   0  0 x (k  1)   2  x3 (k  1)     1  x4 (k  1)     6

1 0

0 1

0

0

5 6





7 6

0   0    0  0    1  x (k )   0  u (k )    5  1    6   2 

S2.5: (a) The question implies finding the impulse response h(t), the inverse transform of H(s). (b) One way is to solve by using the fact that a step is the integral of an impulse, etc.

Another is to convert into state variable form and find the Transfer Function (TF) for the 2x2 system of ODEs (which is what is done in Example 2.32). Another approach is to convert the equation directly into the Laplace domain and find the TF that way, i.e.

L   y  3 y  2 y   L 5  1(t ) 

L   y   3L  y   2L  y   5L 1(t ) 

s ( sY ( s )  y (0))  y (0)  3( sY ( s )  y (0))  2Y ( s )  5 / s

Substituting in the ICs and solving for Y(s):

Y (s) 

5 s ( s  3s  2) 2

Now we do PFE: Y (s) 

c c  5 c  5  11  21  31  s ( s  1)( s  2)  s s 1 s  2 

c11  sY ( s ) s  0 

1 1  (0  1)(0  2) 2

c21  ( s  1)Y ( s ) s 1 

1  1 (1)(1  2)

c31  ( s  2)Y ( s ) s 2 

1 1  (2)(2  1) 2

Thus  1 1 1  Y (s)  5      2 s s  1 2( s  2)  Taking the inverse Laplace transform:

3

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

y (t )  L1 Y ( s )  5e  t 

5 2 t  e  1 2

S2.6: We are given the impulse response in the Laplace domain: Y (s) 

1

 s  1  s  2  2

To find the time domain solution to this response, we first do a partial fraction expansion (PFE) on Y(s) and then take the inverse transform.

Y (s) 

1

 s  1  s  2  2



c11 c12 c   21 2 s  1  s  1 s  2

The first subscript corresponds to “root group” and the second to the order, so c12 is the residue for the first group of second order.

For non-distinct roots: 1 d ni  k ( s  pi ) ni Y ( s )  cik  ni  k  s  pi (ni  k )! ds ni is the highest order for that root grouping, k is the order for that coefficient. So for c11 , ni  2 , k  1 , pi  1 , and: c11 

1 d 21 d 1  ( s  1)2 Y ( s )    1 2 1  s  1 (2  1)! ds ds ( s  2) s 1

Similarly for c12 , ni  2 , k  2 , pi  1 , c12 

1 d 22 1 ( s  1)2 Y ( s )   1 22  s  1 (2  2)! ds ( s  2) s 1

c21 Corresponds to a distinct root or ( ni  1 , k  1 , pi  2 ) c21 

1

 s  1 s 2 2

1

So our expanded Laplace solution is:

4

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

Y (s) 

1

 s  1  s  2  2



1 1 1   2 s  1  s  1 s  2

Taking the inverse: y (t )  L1 Y ( s )  e  t  te  t  e 2t .

S2.7: With all ICs equal zero, the output transform is 20   L[ y (t )]  Y ( s )   3 U ( s) 2  s  12 s  22 s  20 

The TF Y(s)/U(s) can be factored and rewritten as

H ( s) 

10  1 s 1  80    2    2  41  s  10 s  2 s  2  41  s  2 s  2 

The constant factor 80/41 of the second term is 8 times the constant factor 10/41 of the first term. Therefore, the output y(t) is dominated by the time function

80 1  U ( s )  L 41  s 2  2 s  2  The output transform Y(s) can thus be approximated by this second term, i.e.

Y (s) 

80  U ( s )   2   2  2 U (s) 41  s  2 s  2   s  2s  2 

The third-order original model can then be approximated by

d2y dy  2  2 y  2u.

2 dt dt

S2.9: 10t is the integral of constant 10. So, with zero ICs, superposition applies and the linear system model solution is 10 times the integral of the given step response t

t

0

0

y1 (t )  1  et for t  0 . This is: 10 (1  e  )d  10(t  e  t

)  10(t  e  t  1) .

Chapter 4 S4.8: For n = 8, k = 3 and kelim = 0.16/min.

S4.13: Percent absorption of T4 is computed from model parameter values as 49.3%.

5

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

S4.20: See figure below for the answer. Other configurations are possible, but this one is a most likely interpretation. Note the additional measurement port in feces.

S4.25: Deadly Diseases. The following alternative derivation of R0 is instructive. For an epidemic to take off, we need q I  k IS qI qS  k RI qI  0 from the start, i.e. q I (0)  (k IS qS (0)  k RI )qI (0)  0 and this implies k IS qS (0)  k RI  0 because qS (0)  0 . Then k IS qS (0)/kRI  1 and as qS (0)  N we have R0  k IS N /k RI  1 .

Thus, in a completely susceptible population, introduction of a small number of infectives causes the epidemic to spread if R0 > 1. y1 u1

X1 in BLOOD

METABOLISM PATHWAY

X1 in LIVER

BILIARY PATHWAY

X2 in LIVER

X2 in GUT

BILIARY PATHWAY

EXCHANGE PATHWAYS

X1 in FECES

X1 in GUT

y2 X2 in FECES

EXCRETORY PATHWAY

y3

EXCRETORY PATHWAY

MC model with additional enterohepatic pathways

S4.27: Time-Domain Solutions of the 2-Compartment Model – First way: Consider the model with only one input u1: q1  k11q1  k12 q2  u1 q2  k 21q1  k 22 q2 We convert these two first-order equations into a single 2nd-order equation. q  k q From the second equation, we solve for q1  2 22 2 . Then, differentiating k21 q  k q once, we have q1  2 22 2 , or q2  k22 q2  k21 (k11q1  k12 q2 )  k21u1 . This k21 second-order constant coefficient inhomogeneous ODE can be solved by classical methods, e.g. using fundamental sets of solutions. The complete solution is the “free” plus the “forced” responses or, equivalently, the “homogeneous” plus the “particular” solutions, as developed in Chapter 2. For n-compartment models, we follow the same successive substitution and differentiation procedure n  1 times.

6

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

Second way. Solve the set of 1st-order ODEs directly. The solution always has the same form for any number of 1st-order linear equations. In Chapter 2, Eq. (2.22), the complete solution of the constant-coefficient vector-matrix ODE: q  Kq  Bu is t

q(t )  e K ( t t0 )q(t0 )   e K ( t  ) Bu( )d 

(4.90)

t0

where eKt  F(t) are the fundamental solutions of the matrix ODE, F  KF , with initial conditions F(0) = I.1 The integral term in Eq. (4.91) is the convolution of the exponential impulse response matrix with the input (see Chapter 2), often abbreviated: e Kt  Bu(t ) .

S4.28: (a) intravenous (IV) input; (b) an oral input; (c) an intramuscular (IM) input and (d) a cutaneous input via a skin patch that include a slow release drug.

Compartment 1 is blood, and 2 is kidney tissue.

STEP INPUT

k21 V1

k12

V2 k02

C K  q1

SKIN PATCH

MUSCLE

GUT

k21 V1

k12

C K  q1

k21

k21 V2

V1 k02

k12

C K  q1

V1

V2 k02

V2

k12

k02

C K  q1

Chapter 5 S5.5: (a) Impulse input into node 3; (b) impulse into node 2 or exponential input input into node 1; (c) impulse into 5, exponential into 2, or biexponential into 4; (d) nonzero initial condition on node 2 and any value (including zero) on all others; (e) nonzero initial conditions on nodes 1 or 2 and any values (including zero) on all others.

S5.7: First, let’s try zeroing out each of the rate constants connecting compartments 1 and 2 separately. Inspection of Eq. (5.9) indicates that k12 = 0 or k21 = 0 also yields a pole-zero cancellation, i.e.

1

Appendix C can be helpful here. It provides a synopsis of linear system theory pertinent to state space modeling, including solution of this equation.

7

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

H11 k 0 or k  0  12

21

( s  k22 )( s  k33 ) ( s  k22 )  ( s  k11 )( s  k33 )  k13k31 

s  k33  ( s  k11 )( s  k33 )  k13 k31

(5.46)

We can also glean this graphically, without the math, by deleting arrows: when the k12 (= 0) arrow is removed, compartment 2 is only a “sink” for compartment 1 (Fig. 5.26a) and therefore the mode for compartment 2 has no path to 1, so it’s not visible in the output of 1. And, when the k21 (= 0) arrow is removed, inputs into compartment 1 never reach compartment 2 and therefore “state” 2 is effectively uncoupled, again yielding a lower order model output ((Fig. 5.27b). By symmetry, we would also have a model reducible to second-order if k13 = 0 or k31 = 0, and a model reducible to first-order if k12 = 0 or k21 = 0 and k13 = 0 or k31 = 0, i.e. if k12k21 = 0 and k13k31 = 0.

In other words, the 3-compartment models inFig. 5.27 have only 2 visible modes, established by inspection – using our growing intuition – so they both reduce to the 2-compartment models on the right. The models on the right are called inputoutput equivalent to those on the left.

Note that zeroing out parameters, or removing links – as we did above – is not really state variable aggregation, although it reduced the visible dimensionality of the model, as viewed from its output. Parameter space aggregation, i.e. combining parameters to simplify a model, is discussed along with model identifiability analysis in Chapter 10.

S5.9: Linear cyclic models can have complex roots in their modal solutions for some model parameter values. For other values, the roots can be real, negative and distinct, or real and repeated. This means that choosing an appropriate model structure can be difficult, because cyclic models can be fitted to a variety of kinetic data types. For example, if the data exhibits no oscillatory behavior, they would be difficult to distinguish from model structures with no cycles possible.

Chapter 6 S6.8: Effects of Temperature on Reaction Rate: Arrhenius vs. Square-Root For a reaction with a very high energy barrier, consider the change in rate constant for an approximately 2% increase in temperature from 298K to 303K, first using the Arrhenius’ formula for the ratio of rate constants:  Ea  1 1    25000 cal mol-1  1 k (T2 ) Ae Ea / RT2 1    exp        exp       2.01  Ea / RT1 -1 -1  k (T1 ) Ae  1.99 cal mol K  303 K 298 K    R  T2 T1  

8

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

This is an approximate doubling of the rate constant (~100% change). Using the square-root law, the percent change is: 100(k wV0,Tx  kwV0,Tx0 ) / k wV0,Tx0  100  303  1   0.8% .

298   This estimate is > 100-fold smaller than the Arrhenius estimate!

S6.9: Let v ≡ kwnxm. Then:

 w   n   x     m  v      z   1 

S6.10: Taking logarithms, we have log10 cH  log10 K a  log10

 log10 cH   log10 K a  log10

S6.11: From Eq.(6.103), log10

(6.104) c AH , or cA

cA . Then, rearranging, we obtain(6.103). c AH

cA  1.5  3.5  2 and cAH

log10

cA c AH

cA 1   0.01. For an acid, the proton donor is c AH 100 cA 1 the unionized form, i.e. AH. Therefore cAH = 100cA, c AH i.e. > 99% of aspirin in gut is unionized. In other words, over 99% of aspirin is absorbed by the gut into Fig. 6.17 the bloodstream.

Buffers: A substance that counteracts changes in H+ concentration must be a proton-donor/proton-acceptor pair, e.g. HCO3 , H 2CO3 , CO 2 , the bicarbonate-carbonic acid system. Such a pair acts as a buffer to changes in pH when its pH  pKa, i.e., cA ~ cAH (50% ionized), at the crossover point, 1. Fig. 6.17, a plot of the H-H equation, illustrates this.

Chapter 7 S7.3: Linearized Mass-Action-Based Kinetics: Applying Eqs. 4.112 and 4.113 (Chapter 4) to Eqs. (7.40) through (7.42) yields: ַ◌  f   f   f  x1   1N  x1   1N  x2   1N  u  2c2 x1 N x1  c1u N x2  c1 x2 N u  u   x1   x2  ַ◌  f   f  c c x  f  x2   2 N  x1   2 N  x2   2 N  u  3 4 1 2 (c4  x1 N )  u   x1   x2 

(7.47) (7.48)

y  c6 x1

(7.49)

9

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

If the nominal input uN and nominal solutions x1N and x2N are constant or timevarying, we may formally write:  2c2 x1N x   c3c4  2  ( c4  x1N ) ַ◌

c1uN   x   c1 x2 N  u  0  0    

y   c6

(7.50)

0 x

(7.51)

We note that the “zero-order” term, parameter c5, in the nonlinear Eq. (7.41) does not appear in the linearized system, Eq. (7.50) or(7.51). Thus, although we have reduced the nonlinear problem to a linear one, we cannot learn anything about c5 from the perturbation experiment model. On the other hand, information about c5 can still be obtained if u = u is constant and the endogenous steady state variable x1 is also measured, i.e. solving Eq. (7.41) for c5 when x2  0 gives cx c5  3 1 . c4  x1 S7.4: The stoichiometric matrix for this reaction is:

S7.5: Ka  r

s

i 1

i 1

 1 1   1 1   N   1 1    1 1

(7.55)

k1 y m1 y m2  ykms  1n1 n22 k1 x1 x2   xmnr

(7.52)

S7.6: Let v = v+ - v- = k1  xini  k1  yimi . Then:   n1   n   2      N    nr   m1        ms 

(7.53)

S7.7: Kd = p1∞p2∞/d∞. The ODEs are:

10

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

 1 1  N   1 1     1 1

p 1  p 2   d  k1 p1 p2  k 1d  v1  v1  v,

S7.10: The equations and two graphical versions of these reactions are given in Eqs. (7.54) and Fig. 7.14 and Fig. 7.15. In the compartmental diagram 5.z, the usual convention of the parameter arrow symbol multiplying the mass at the tail end of the arrow is assumed (not shown), so the q1 and q2 multipliers in Eqs. (7.54) are not shown in the diagram; they are implied by the convention.

Aq1 Aq1 Aq1 q1    k12 q2  u1    (1   )  v12  u1  vtot  v12  u1 q1  B q1  B q1  B Aq1 q B    k12 q2  k02 q2  v21  v12  v02 , q1  B

y1(t)

u1(t)

A q1  B

1

k12 (1   )

A q1  B

y1  q1 / V1

v21

u1

2

(7.54)

● v 12

● v0

v01 k02

Fig. 7.14

Fig. 7.15

Chapter 8 S8.5: The ODEs for the four 1-compartment flow-limited organs (plasma + 3 organ compartments) of the PBPK model shown are established by balancing the fluxes for all compartments.

S8.7: Compartment 1 and 2 in Fig. 8.20 each have 2 modes, because each contributes its own to the other. Compartment 3 then has 3 modes, and 4 has 4 modes, each receiving the accumulation of modes from 1 and 2, and 1 and 2 and 3.

Compartment 5 gets 4 modes from compartment 4, generates its own – making 5 so far; and several more from compartment 7.

To find this number, we see that 6, 7, 9 and 10 share the 4 modes generated in these compartments. These 4 are transmitted back to compartment 5, yielding a total of 9 for compartment 5.

Compartment 6 then receives 5+4 = 9 from compartment 5 – and we note that 4 of these

11

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

duplicate the 4 received from 7. So, 6, 7, 9 and 10 receive 9 modes total – 4 of which are duplicates. Finally, compartment 8 receives all 9 (distinct) modes from 7 and, adding one for itself, 10 modes can be measured in compartment 8.

We remark that it is unlikely that all the modes theoretically possible (structurally identifiable, SI) in each of the 10 compartments would be observably distinguishable if the model were fitted to real data measured in these compartments. Some of these modes would be hidden (numerically unidentifiable, albeit SI) or otherwise obscured in the “noisy” numerical measurement data. This reinforces the point that accurate biostructural information is needed – and generously supplied in physiologically based models – to precisely quantify and validate them from kinetic data in one or more of these compartments.

Chapter 9 S9.1: With all ki  0 , only the third one is stable (poles in the LHP).

S9.2: Let x  x1 , y  x 2 , p5  p1 p4 . Then: x1  p1 x1  p2 x1 x2 x2   p3 x2  p5 x1 x2 (1) When x1  x2  0, x1e  p1  p2 x2 e   0 , x2e  p3  p5 x1e   0 .

Therefore, x1e  x2e  0 is a trivial, stationery steady-state solution. x2e  p1 / p2 , x1e  p3 / p5 is a stationary steady-state solution, as are the other two possible combinations of these four steady states.

(2) Linearize: Let x  x  xe . 

 x1   p1  p 2 x 2 e x1  p 2 x1e x 2 

p 2 p3 x 2  ax 2 p1 p 4

 p2 p  x 2   p5 x2 e x1   p3  p5 x1e x2  1 4 x1  bx1 p2

(3) Solve by separation of variables:

d x1  d x1   ax2  dt  dt ax 2 d x2  dt

 bx1  dt 

dx 2  bx1 12

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

Then

d x1  d x2   or bx1d x1   ax2 d x2 . ax2 bx1

Now integrate both sides: c1 

bx12 a 2  x2  c 2 2 2 2

2

bx1  ax2  c , or

x12 x22  c A B

representing ellipses dependent on c, and c is determined from IC’s, x1 (0)  x10 x2 (0)  x 20 : (x10  x1e )2 (x 20  x 2e )2  c A B

x(t)  xmax cos[K(t  t 0 )] y(t)  y max sin[K(t  t 0 )]

The trajectories are ellipses centered at xe : x2 x2e

•

x20 x1e

x10

x1

xe is not a stable equilibrium point, which would look like a spiral converging on xe .

Instead, limit cycle oscillations in x and y continue forever, with same amplitude and frequency, if perturbed about equilibrium. As:  host   prey …  prey + host …  host, etc.

Chapter 10 S10.4: (a) Only k21, c1k31 and k11 = - (k01 +k21 + k31) are uniquely identifiable. Others (and model) are unidentifiable. (b) Model, k21 and k31 are unidentifiable, but combo k21 + k31 is uniquely identifiable.

13

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

S10.7: (a) V, p1, p2, p3 and p4p5 are uniquely identifiable; p4 and p5 are unidentifiable. (b) YES, if –p1 ≥ p3 ≥ 0 and –p4p5 ≥ p2 ≥ 0 and V ≥ 0.

S10.11: The TF:

Y2 ( s ) 

k12U1 ( s )  ( s  k22 )U 2 ( s ) s  (k11  k22 ) s  k11k22  k12 k21 2

is for the model: dq / dt = Kq+ u and y = [0 1]q = q2. The unknown parameters are k11, k12, k21, k22. Input u1(t) is a unit impulse, so U1(s) = 1. u2(t) is a unit step, so U2(s) = 1/s. Both are applied at time t = 0. To get identifiable parameters, look at structural invariants of the TF:

Y2 ( s ) 

k12  ( s  k22 ) / s s (k12  1)  k22  3 s  (k11  k22 ) s  k11k22  k12 k21 s  ( k11  k22 ) s 2  ( k11k22  k12 k21 ) s 2

The numerator gives k12 and k22. The denominator then gives k11 from the term in s2; and k21 from the last term. All parameters are thus uniquely identifiable.

Chapter 11 S11.2: 2nd-Order Sensitivity Functions. We used a Taylor series expansion of the state vector x(t,p) about a nominal value p0 of p, i.e. Eq. (11.7), to obtain the 1st-order local variational Eq. (11.10) for changes x in x due to changes p about p: x  t , p   ( x /  p )p  V  t, p  p . A more accurate prediction for the state vector perturbation caused by small parameter variations can be obtained from the 2nd-order Taylor series approximation, from Eq. (11.8) and (11.14). To avoid complex notation, we evaluate this for the kth component of x:

xk  t, p  

  2 xk   xk 1 p  pT  p 2  p 2  p 

(11.43)

The 2nd-order sensitivity functions for xk with respect to pi and pj are the elements of the matrix:  2 xk   2 xk    2   2 0     vkij   Vk  t , p  2 p  p  p  i j 

(11.44)

T

T T  2 T  with columns vij 2    v1ij2  v2 2ij  vnij In these terms, (11.43) becomes:   .

xk  t , p   V  t , p  p 

P 1 T 2 x 1  2 xk p Vk  t , p  p   k pi   pi p j 2 2 i j  pi p j i 1  pi

14

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

We assume the xi are suitably well behaved so we can exchange the order of total and 2 partial differentiation. The ODE for vij  and solution for model x  Ax  Bu are: vij 2  t , p  

d   2x   2 x  2 ( Ax  Bu)     dt   pi p j   pi p j  pi p j

 Avij   2

(11.45)

A A  2A  2B vj  vi  x u  pi  pj  pi p j  pi p j

, with ICs: vij 2  t0 , p  

 2 x0 . The 2nd-order output sensitivity functions are then:  pi p j T

  2 y1  2y  2 ym  C C  2C 2   vj  vi  x   Cvij   pi p j   pi p j  pi p j   pi  pj  pi  p j

(11.46)

Note that n(P + 1) ODEs need to be solved to obtain V(t,p0) and x(t,p0). For 2nd-order sensitivities, Eqs. (11.45) is nP2 dimensional and (11.46) is mP2 dimensional.

2 However, each Vk  and each 2yk/pipj are symmetric and thus only nP(P + 1)/2 and mP(P + 1)/2 additional equations (at most) require solution to obtain the 2nd-order state and output sensitivity functions.

Second-order sensitivity functions are components of 2nd-order gradient, Newton-Raphson, quasilinearization and other optimization algorithms. They are often used in obtaining measures of parameter covariances.

S11.3: Sensitivities for constant-coefficient models. Eqs. (11.1), (11.2) and (11.31) become:

x (t,p)  A(p)x (t, p)  B(p)u(t) x (t 0 , p)  x0 (p)

y (t ,p)  C  p  x  t ,p   D  p  u(t )

(11.47) (11.48) (11.49)

We assume, for generality, that the initial state is also a function of p and the input is also measured. If we give a special partitioned matrix interpretation to the operation of differentiation of a matrix with respect to a vector, we can rewrite Eqs. (11.19), (11.34), and (11.49) as:

A B V  AV  x u p p  x  p V  t0 , p   0 p

(11.50) (11.51)

15

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

  yi  C D x u    CV  p p   p j 

(11.52)

However, this new formalism can be easily avoided by writing (11.47) and (11.49) as:

 (Ax )  (Bu ) V  AV   p p   yi   (Cx )  (Du )     CV  p p   p j 

(11.53) (11.54)

where Ax, Bx, Cx and Du are vectors, and differentiation of a vector with respect to a vector is a matrix (see Appendix B).

In (11.50) and (11.52), only A and C are functions of p, and thus:

  Ax    A  p  x(t)  , p p

  Cx   C  p  x(t)  p p

(11.55)

To see this, we derive (11.50) from the scalar equation (11.15): n  f   fi vij    i  vkj   pj k 1   xk 

(11.56)

From (11.59), we have: n

n

s 1

s 1

xi   ais xs   bis us  f i

(11.57)

Then (11.54) becomes: n        vij    ais xs   bis us  vkj  aik ( p) xk   bik ( p)uk      p j  k k 1  xk  s s k   n n a b    (ais xs )vkj   ik xk   ik uk k 1 s 1 xk k pk k p j

(11.58)

n   r b a    aik vkj  ik xk    ik uk  p j  k 1 p j k 1 

Equation (11.58) yields the ijth component of (11.52) or (11.54). The reason x is not considered a function of p in (11.57) and (11.52) is that the first term on the righthand-side (RHS) of (11.56) (and (11.15)) represents the contribution of x as a function of p in the derivation of Eqs. (11.17) and (11.56) by the chain-rule. The

16

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

second term on the RHS of (11.56) then is the partial of fi[x(p), p] with respect to p, with x held constant.

S11.4: Initial State Sensitivities. Suppressing p in Eq. (11.5), and defining xj0  xj(t0), we have

 fi  x   fi  x (t , x0 )    xi (t , x0 )   x j0  x j0  x  xj0

(11.59)

This yields the homogeneous equation: f i x   xi     t  x j 0  x x j 0

with initial condition

(11.60)

xi (t0 ) xi 0  . In matrix form, this becomes: x j 0 x j 0

  xi (t )    f    x        x j 0    x    x0 

(11.61)

 x  with initial conditions  i 0   I .  x j 0 

Chapter 12 S12.3: Jacobian and Hessian of WLS(p). Suppressing x, we write: WLS(p) ≡ L(p) = ET(p)WE(p)  L[E(p)]. Then2 Jacobian L ≡

L L E E Y   2 E TW  2 E TW  p E  p p p

(12.61)

Then, writing Y / p  Y ( p), L / p  L( p) and (12.61) as L(p)  L[E(p), Y(p)], we again apply the chain-rule: Hessian  H   2 L  p 2 

Now 2

  L   E   L    Y    L  p    p  E  p   Y   p

  L     T  2 E TW Y   2 W Y  E   2Y T W T  2Y T W ,    E E E

See Appendix B for help with the matrix calculus.

17

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

  L    Y   2Y , where we also have taken advantage of the  2 E TW and    Y  p  p2 symmetry of both H and W. Since E = Y – Z, ∂E/∂p = Y / p . Then, gathering terms and redefining symbols in their partial derivative terms,  Y T  Y   2Y  T H  2   W   E  p W 2  p   p   p 

(12.62)

Note that H is a function of both first- and second-order sensitivity matrices, with: T Y  (mN × P)   y1  t1  y2  t1   ym  t N     yi  p j  p p  2Y The 2nd-order sensitivity matrix is written term-by-term in a similar manner. p 2 Note that, whereas L in (12.61) is a 1 × P (row) vector, H is a symmetric P × P matrix.

S12.4: Gauss-Newton Algorithm. At a point p̂ sufficiently near the minimum of L, Y  pˆ   Z , i.e. E  0. In this case, the second term in Eq. (12.62) above is small compared with the first. At the ith iteration we can then approximate Hi by: T

 Y   Y  Hˆ ( i )  2  ( i )  W  ( i )   p   p 

(12.63)

This approximation leads to the Gauss-Newton iterative algorithm:

pˆ ( i 1)  pˆ ( i )  ( Hˆ 1 )( i )  L( i ) 

T

(12.64)

With this algorithm, the search proceeds in the gradient direction, modified by the curvature of the loss function L, approximated with the help of Ĥ .

S12.5: Newton-Raphson Algorithm. Start by expanding the loss function L in a Taylor series about p̂ up to the 2nd-order term: 2 L 1 T  L L  p   L  pˆ    p  pˆ    p  pˆ   p  pˆ   h.o.t.  pˆ 2  pˆ 2 1 T  L  pˆ   L  p  pˆ    p  pˆ  H  p  pˆ   h.o.t.

2 with L   L  pˆ   L  p  p  pˆ and H   2 L  pˆ 2   2 L  p 2 

p  pˆ

(12.65)

.

18

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

Now, if L(p) has a minimum at p, differentiating (12.65) with respect to p gives:

L 1 T  L   p  pˆ  H  0 p 2 Solving for p, pˆ  pˆ  H 1LT

(12.66)

Next, we give approximation (12.66) an iterative interpretation. Let pˆ ( i ) be the current (ith) estimate of p; then the next best estimate of p is pˆ ( i1) , and H(i) L(i) are the (exact) Hessian and gradient of L with respect to p, evaluated at pˆ ( i ) . In these terms, (12.66) becomes the Newton-Raphson iterative algorithm for estimating p: pˆ ( i 1)  pˆ ( i )  ( H 1 )( i )  L( i ) 

T

(12.67)

S12.6: ML Estimator for Multiexponential Model Parameters. We show that optimized parameter vector pML that maximizes the log-likelihood LL function minimizes the weighted least square error WLS. We seek ML estimates of the parameters p  [A1 1 … An n]T of the multiexponential n

model y  t    Ak e k t , from data time series z1, z2,..., zN , using the measurement k 1

model zi = yi + ei, i =1,, N. If the data zi are Gaussian distributed, the marginal density of zi, by definition (Papoulis 1965), has the form:

f  zi  

 1 ( zi  i ) 2  1 exp    2  i 2  2  

(12.68)

where i  E(zi) is the population mean or expected value of zi and  i2  E[( zi  i ) 2 )] is its variance. Given that the ei are independent measurement errors with zero mean n

values, i.e. E(ei)  0, then E  zi   E  yi  ei   E  yi    Ak e k ti k 1

n

and E  zi p   E  yi p    Ak ek ti conditioned on p. Then, substituting E  zi p  k 1 n

for f  zi  and E  zi p  =  Ak ek ti for i in (12.68) yields the desired marginal k 1

density: 2 n  1   1 k ti  f  zi p   exp   2  zi   Ak e    i 2 k 1    2 i 

(12.69)

19

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

Now take the natural log (ln) of the product of these marginal densities: 2 N N N  1   1 k ti  ln  f  zi p   ln  exp   2 z i   Ake   i 1 i 1  i 2 k 1    2 i  N 2   N  1   1  1  2 k ti   ln   exp   2  zi   Ak e     k 1     2 i   2   i  N  N 2

N

N

2

(12.70)

 1  1    1   ln   ln exp   2  zi   Ak e k ti     2   i  N k 1    2 i  2

N 1 1    1 2  ln   z  Ak e k ti    ( i )  WLS ( p) / 2   2  i  2   i  N 2 i  k 1 

The first term  ( i ) in the last equation above is not a function of p and the ½ doesn’t matter in optimizing WLS. Thus, finding the pML that maximizes the LL function when the residual errors are independent, Gaussian distributed and with zero means is equivalent to minimizing the weighted least square error, WLS. Remark: if the variances i2 are also unknowns in this problem, i.e. we can write p ≡ [A1 1 A2  An n 12 N2]T and both terms in the last equation in Eq. (12.70) are optimized, w.r.t. the variances and the multiexponential parameters. In this case, ML and WLS results are not same, because the first term  ( i ) in the last equation above is a function of p.

Chapter 13 S13.4: The variances VAR ( cˆ ) of the structural invariants are established by the output data. When parameter values are fixed (known), the variances of their parameter estimates are zero by definition. VAR( cˆ ) must nevertheless remain the same. Therefore, other parameter bound variances might increase, to reflect this conservative feature of the data. For example, suppose estimates of k02max and k12min are approximately uncorrelated ( COV (k02max , k12min ) ≈ 0) . Then, for k22  k02max  k12min , VAR( k22 )  VAR(k02max )  VAR(k12min ). Also, let VAR (k22 )  2 and VAR (k02max )  VAR (k12min )  1 in the unconstrained case. Then, if k02 is known,

VAR (k02max )  0 and thus VAR (k12min ) increases from 1 to 2. If k02max and k12min estimates were correlated, the variance estimate increase would possibly be smaller or nonexistent.

In (Vicini 2000), COV (bˆ) are computed two additional ways for these models: by Monte Carlo (MC) simulations, with 1,000 randomized runs of each model (Chapter 12); 20

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

and by the uniquely identifiable joint submodel parameters approach (Chapter 10), with statistics evaluated using SAAMII. Results compared well across the methods.

Chapter 14 S14.5: Plant Metabolism Modeling and Analysis The Problem. Consider the simple plant metabolism model in Fig. 7.6 of Chapter 7. (a) Implement the model Matlab or equivalent software. (b) Simulate its metabolite dynamics with parameter values. (c) Explore the model, by simulation, varying the parameter values, to see how they affect metabolite concentrations. (d) Do a metabolic control analysis (MCA) of this model. Compute the sensitivity functions of MCA that you can compute, with Matlab, using the model augmented with sensitivity equations.

Also compute the length of the sensitivity vectors over the time course 0 to 12 minutes.

Plot all results on the same graph, along with your simulated metabolite concentrations.

Discuss your results. (e) Discuss how you might increase the production of metabolite B (=2) or C (=3). The idea here is that B or C might be biofuel.

Part I. Implementation of the Simple Plant Metabolism Model

Figure 1. Metabolite dynamics of the simple plant metabolism model.

Part II. Exploration of Parameter Values and Effects on Metabolite Concentrations Varying the parameter values of the model affects the metabolite concentrations in the following ways.

As shown in Figure 2 with the solid lines representing the model with the original parameter values, decreasing Km1 will cause the reaction velocity, v1, to increase which makes the concentration of metabolite A decrease faster. At steady-state there is more concentration of metabolite C and less of metabolite B compared to the original.

Similarly, increasing Km1 will decrease v1 and cause A to decrease slower. At steadystate, there is less C and more B.

21

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

Figure 2. Effects of decreasing Km1 on the simple plant metabolism model.

Decreasing Km2 causes the reaction rate v2 to increase which causes the concentration of metabolite A to decrease slightly faster than the original (shown with a solid line), the concentration of metabolite B to increase faster to a higher steady-state concentration, and the concentration of metabolite C to reach a lower steady-state value. Similarly, the opposite occurs when Km2 is increased. These effects are shown in Figure 3.

Figure 3. Effects of decreasing Km2 on the simple plant metabolism model.

Shown in Figure 4, decreasing Km3 causes no appreciable difference in the concentrations of metabolites A, B, or C and no difference in the reaction rates. This is because vmax3 is very small compared to vmax1 and vmax2. Km3 is also very small compared to Km1 and Km2.

Figure 4. Effects of decreasing Km3 on the simple plant metabolism model.

Part III. Metabolic Control Analysis (MCA) Compute the sensitivity functions of MCA MCA is a method for analyzing how control over flux and metabolite concentration is distributed among the enzymes that constitute a metabolic pathway (or network of pathways).

22

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

Flux control coefficient The Flux Control Coefficient represents the sensitivity of the flux to the velocities, which may tell us how we can alter the system to make more of a certain product.

Flux J2 = Cb2 = C(v2 + v3) = C((vmaxEnz2A / (A+KmEnz2)+vmaxEnz3B / (B+KmEnz3))) dJ2/dvmaxEnz1 = 0 dJ2/dvmaxEnz2 = CA / (A + KmEnz2) dJ2/dvmaxEnz3 = CB / (B + KmEnz3) Flux J3 = Bb3 = B(v1 - v3) = B((vmaxEnz1A / (A + KmEnz1) - vmaxEnz3B / (B + KmEnz3))) dJ3 / dvmaxEnz1 = BA / (A + KmEnz1) dJ3 / dvmaxEnz2 = 0 dJ3/ dvmaxEnz3 = - B2 / (B + KmEnz3) The plots of the Flux Control Coefficients are shown in Figure 5. The Flux Control Coefficients for J3 with respect to dvmaxEnz1 and J2 with respect to dvmaxEnz2 are zero. As shown in the graph, the Flux Control Coefficients vary with time.

We notice that the partial derivatives of both flux with respect to V3 reach steady state while the flux with respect to V2 and V1 reach zero over time.

The velocities seem to have more impact on the J2 system than the J3 system, which means the velocity has more influence on J2. The enzyme with velocity vmaxEnz3 exerts more control on J2. This allows us to theoretically alter the system to make more product C because flux J2 is Cxb2.

Figure 5. Flux control coefficients.

Concentration control coefficient: The concentration control coefficient quantifies the relative change in steady state concentration of metabolite or enzyme to a small relative change in a vi -From the model we have dA/dt = b1Aext-A(∞)(v2+v1), for steady state dA/dt = 0, therefore, b1Aext = A(∞)(v2+v1). The concentration of A(∞) = b1Aext / (v2+v1). Since we are assuming b1 is negligible, then concentration A(∞) would actually go to zero in this situation.

23

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

- dB/dt = B(∞)v3 + A(∞)v1 - B(∞)b3 where b3 = v1 - v3 and dB/dt = 0 (steady state condition).

- dC/dt = B(∞)v3+A(∞)V2 - C(∞)b2 where b2 = v2 + v3 and dC/dt = 0 (steady state condition). → solving for B(∞), A(∞), C(∞), we have two equations and three unknowns...

-if we use A(∞) = 0 situation, then we would end up with B(∞) = 1 and C(∞) = 1 where B(∞)b3 = C(∞)b2.

-Without equations for concentration, computation and graphing of the concentration control coefficients proved impossible.

Discussion The goal is to increase the production of metabolites B or C. Because flux J2 is influenced greatly by enzyme 3, we can use this to advantage in devising a scheme to produce more metabolite B or C. Increasing enzyme concentration will increase vmax. So the corresponding velocity v will also increase. If we want to increase production of metabolite C, the concentration of enzymes 2 and 3 should increase while the concentration of enzyme 1 should decrease.

As an alternative, a different enzyme can be used in place of enzyme 3 which has the same properties (Km) but a faster vmax. This will cause the flux J2 to increase, due to its sensitivity to enzyme 3, which will produce more metabolite C. From earlier analysis of parameter variation effects on metabolite concentration, we saw that decreasing Km2 will cause an increased metabolite B concentration and decreasing Km2 will increase metabolite C and decrease metabolite C. Depending on which metabolite, B or C, we are trying to produce more of, we can exploit these differences to alter the system.

Chapter 15 S15.2: Full Wagner Model Sim, eqns and simulink diagram (287 Janay):

24

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

Reduced (simplified) Wagner Model eqns and Simulink Diagram:

25

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

26

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

Chapter 16

u1(t)

k21

01

12

u2(t)

1 2 S16.3: Experimental TF Matrix and V1 V2 Identifiability Analysis for the General 3k k k Compartment Model k This model has 12 intrinsic parameters: k01, k02, k true true true k k03, k21, k12, k23, k32, k13, k31, V1 , V2 and V3 ; k a maximum of three possible input variables: u1(t), u2(t), and u3(t); and a maximum of three 3 V3 independent measurement (output) variables: y1(t), y2(t), and y3(t), in compartments 1, 2, and 3, respectively. Define the k's as (fractional) k u3(t) rate constants, with dimensions of time-1; the V's are volumes; the u's are mass flux rates (mass/time); and the y's are concentrations (mass/vol) or amounts (mass).

31

13

02

23

32

03

The most general model may be written by choosing the three state variables as amounts in compartments 1, 2 and 3, q1(t), q2(t), and q3(t), and the measurements as concentrations: y1(t) = q1(t) / V1true , y2(t) = q2(t) / V2true and y3(t) = q3(t) / V3true . Thus, we have:

q1 = (k01 + k21 + k31)q1 + k12 q2 + k13 q3 + u1

q2 = k21 q1 - (k02 + k12 + k32 )q2 + k23 q3 + u2

q3 = k31 q1 + k32 q2 - (k03 + k13 + k23) q3 + u3 q (t ) y1 (t )  1 true V1 q (t ) y2 (t )  2 true V2 q (t ) y3 (t )  3true V3 (k  k  k )  k 12 k 13 31 31 1 0 0  01    q  (k  k  k 32) q  0 1 0 u k 21 k 23 02    12  0 0 1    (k  k  k 23) k 31 k 32 03   13 1 V1true 0 0    y  0 1 V2true 0  q  0 0 1 V3true  

(16.42)

(16.43) (16.44)

(16.45) (16.46) (16.47)

(16.48)

(16.49)

27

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

The Experimental Transfer Matrix: Hn(s) = Cn[sI - K] -1Bn. With Bn = I,

0 0  1 V 1true   1 true 1V2 0  H n ( s )  C n  sI - K    0  0 0 1 V 3true  -k12 -k 13 ( s  k 01  k 21  k 31)    -k 21 ( s  k 01  k 21  k 31) -k 23   -k 31 -k 32 ( s  k 01  k 21  k 31)   k11 Let K   k21  k31 i.e.

k12 k22 k32

1

(16.50)

k13   (k01  k21  k31 ) k12 k13     (16.51) k23    k21 (k02  k12  k32 ) k23  k33   k31 k32 (k03  k13  k23 )  k11  - (k01 + k21 + k31) k22  - (k02 + k12 + k32) k33  - (k03 + k13 + k23)

Then:

0 0  ( s - k 11) -k 12 -k 13  1 V 1true    true H ( s)   0 1 V2 0  ( s - k 22) -k 23   k 21   0 0 1 V 3true   -k 31 -k 32 ( s - k 33) 

(16.52) 1

(16.53)

Now let D  s3 - (k11 + k22 + k22) s2 + (k11k22 + k22k33 + k11k33 - k12k21- k23k32- k13k31)s + (k11k23k32 + k22k13k31 + k33k12k21 - k11k22k33 - k12k23k31 - k13k21k32)

(16.54)

The complete Experimental Transfer Function Matrix is therefore given by:  s2  (k22  k33 )s  (k22  k33  k23  k32 )  k12s  (k13k32  k12k33 ) k13s  (k12k23  k13k22 )   true true true V1 D V1 D V1 D   (16.55) 2   k21s  (k31k23  k21k33 ) s  (k11  k33 )s  (k11k33  k13  k31 ) k23s  (k13k21  k11k23 ) Hn(s)    true true true V D V D V D 2 2 2    k31s  (k21k32  k31k22 ) k32s  (k12k31  k11k32 ) s2  (k11  k22 )s  (k11  k22  k21  k21)    V3trueD V3true D V3true D  

Thus, 1 H ( s)  V 11

true 1

 s 2 - (k22  k33 ) s  ( k22k33  k 23k32 )  D

(16.56)

28

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

1 true

H12 ( s )  V 1

1 true

H13 ( s )  V 1

1 true

H 21 ( s )  V 2

1 H 22 ( s )  V

true 2

D

D

true

1 true

H 31 ( s )  V 3

1 true

H 32 ( s )  V 3 1 true 3

(16.58)

k21s  (k31k23  k21k33 ) D

D

H 23 ( s )  V 2

(16.57)

k13s  (k12k23  k13k22 )

 s 2 - (k11  k33 ) s  ( k11k33  k13k31 )  1

H 33 ( s )  V

k12 s  (k13k32  k12k33 )

(16.59)

(16.60)

k23s  (k13k21  k11k23 ) D

(16.61)

k31s  (k21k32  k31k22 ) D

(16.62)

k32 s  (k12 k31  k11k32 ) D

 s 2 - (k11  k22 ) s  (k11k 22  k 21k12 )  D

(16.63)

(16.64)

and D is given by Eq. (16.64).

Algebraic Identifiability Analysis: Consider H11 first, which can be written: b (s 2  z1s +z2 ) H11 (s )= 2 1 s  a3s 2  a2 s +a1

(16.65)

Noise-free input-output data will yield: 1, z1, z2, 2, 2, and, i.e., all coefficients of numerator and denominator polynomials.

Comparing Eqs. (16.56) and (16.65) yields the identities:

1 

1

V1true z1 = - (k22 + k33) z2 = (k22k33 - k23k32) 3 = - (k11 + k22 + k33)

(16.66) (16.67) (16.68) (16.69) 29

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

2 = (k11k22 + k22k33 + k11k33 - k12k21 - k23k32 + k13k31) 1= (k11k23k32 + k22k13k31 + k33k12k21 - k11k22k33 - k12k23k31 + k13k21k32)

(16.70) (16.71)

Therefore, from H11(s), we can solve uniquely only for V1true and k11. If we perform the same analysis for H12(s), H13(s), H21(s), ..., H33(s), we obtain Table 16.3: Table 16.3 – Parameters From Single TFs (Experiments) for the Most General 3-Compartment Model TRANSFER FUNCTION MATRIX EQUIVALENT INTRINSIC (EXPIREMENT) PARAMETERS PARAMETERS true H11 V1 , k11 V1true H12 None None H13 None None H21 None None H22 true V2 , k22 V2true H23 None None H31 None None H32 None None H33 true V3 , k33 V3true

Combining experiments in pairs, triplets, etc. gives Table 16.4: Table 16.4- Parameters From Combos of TFs (Experiments) for the Most General 3-Compartment Model TRANSFER FUNCTIONS (EXPERIMENTS)

MATRIX PARAMETERS

EQUIVALENT INTRINSIC PARAMETERS

V1true , V2true , k11 , k 22 , k33

V1true ,V2true

H11, H33

V1true , V3true , k11 , k 22 , k33

V1true ,V3true

H22, H33

V2true , V3true , k11 , k 22 , k33

V2true ,V3true

H11, H12

V1true , k12 , k11

V1true , k12

V1true , k13 , k11

V1true , k13

V2true , k21 , k 22

V2true , k21

V2true , k23 , k22

V2true , k23

H11, H22

H11, H13 H22, H21 H22, H23

30

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

Table 16.4 - continued H33, H31

V3true , k31 , k33

V3true , k 31

H33, H32

V3true , k 32 , k 33

V3true , k 32

H11, H21

V1true , k11

H11, H31

V1true , k11

V1true

V2true , k 22

V2true

H22, H12 H22, H32 H33, H13 H33, H23 H12, H21 H12, H23

V2true , k 22 V3true , k 33 true , k 33 3

V

None None None

H12, H13

V1true

V2true V3true V3true None None None

all the rows should line up with each other above....

31

Dynamic Systems Biology Modeling & Simulation 1-26-2014 2014 by JJ DiStefano III All rights reserved.

Table 16.4 - continued TRANSFER FUNCTIONS (EXPERIMENTS)

MATRIX PARAMETERS

EQUIVALENT INTRINSIC PARAMETERS

H21, H23 H31, H32 H21, H31 H12, H32 H13, H23 H13, H31

None None None None

None None None None

None

None

H23, H32

None

None

H11, H22, H33

None V1true , V2true , V3true , k11 , k22 , k33

V1true ,V2true ,V3true

V1true , V2true , k11 , k22 , k33 , k12 , k31

V1true ,V2true , k12 , k21

V1true , V2true , k11 , k22 , k33 , k13 , k31

V1true , V2true , k13 , k31

V1true , k11 , k22 , k33, k12 , k13 ,

V1true , k12 , k13 , k23 , k32 , k21,

k23 , k32 , k21 , k31

k31 , k01 , k02 , k03

H11, H22, H31

V1true ,V2true , k11 , k22, k33

V1true ,V2true

H11, H21, H31

V1true , k11

V1true V1true , k13

H11, H21, H13

V1true , k11 , k13

V1true , V2true , k11 , k22, k33 , k12 ,

V1true , V2true , k12 , k13, k31 , k32 ,

H11, H12, H13, H22 H11, H22, H12, H31 H11, H22, H12, H32 H11, H12, H13, H22, H33

k13, k31 , k32 , k23, k21

k23, k21 , k01 , k02, k03

ALL ALL ALL

ALL ALL ALL

all the rows should line up with each other in this table.....

H11, H22, H12

H11, H22, H13

None

H11, H12, H13

In each of the last three TF combos of Table 16.4, all 12 intrinsic parameters can be determined. One of these three cases needs a combination of five Hijs, while each of the remaining two cases needs a combination of only four Hijs. A combination of at least four Hijs is required to uniquely determine all 12 intrinsic parameters of the 3compartment model.

32