Dear Troels,
I send you an updated version of the response. I think that a hybrid
approach is probably the best solution:
- Input / Starting values = free: ATP, ADP, Crea, CreaP, lactate,
inorganic phosphate;
- Output: diff(Total, given total value);
- I assume that the pH is given;
- I did NOT check all individual eqs;
library(rootSolve)
solve.AcidSpecies = function(x, H, Mg=0.0006, K = 0.12) {
??? # ... the eqs: ...;
??? ATPTotal = KaATPH * ATP * H + KaATPH2 * KaATPH * ATP * H^2 +
??? + KaATPH3 * KaATPH2 * KaATPH * ATP * H^3 + KdATPMg * ATP * Mg +
??? + KdATPHMg * ATP * H * Mg + KdATPMg2 * ATP * Mg^2 + KdATPK * ATP * K;
??? ### Output:
??? total = c(
??? ??? ATPTotal - 0.008,
??? ??? ADPTotal - 1E-5,
??? ??? CreaTotal - 0.004,
??? ??? CreaPTotal - 0.042,
??? ??? PiTotal - 0.003,
??? ??? LactateTotal - 0.005);
??? return(total);
}
KaATPH = 10^6.494; # ...
x0 = c(ATP = 0.008, ADP = 1E-5,
??? Crea = 0.004, CreaP = 0.042, Pi = 0.003, Lactate = 0.005) / 2;
x = multiroot(solve.AcidSpecies, x0, H = 4E-8);
print(x)
I think that it is possible to use the eqs directly as provided
initially (with some minor corrections). You only need to output the
known totals (as a diff), see full code below.
Sincerely,
Leonard
library(rootSolve)
solve.AcidSpecies = function(x, H, Mg=0.0006, k = 0.12) {
?? ?# with(list(x), { seems NOT to work with multiroot });
?? ?atp = x[1]; adp = x[2]; pi = x[3];
?? ?pcr = x[4]; cr = x[5]; lactate = x[6];
###
hatp <- 10^6.494*H*atp
hhatp <- 10^3.944*H*hatp
hhhatp <- 10^1.9*H*hhatp
hhhhatp? <- 10*H*hhhatp
mgatp <- 10^4.363*atp*Mg
mghatp <- 10^2.299*hatp*Mg
mg2atp <- 10^1-7*Mg*mgatp
katp <- 10^0.959*atp*k
hadp <- 10^6.349*adp*H
hhadp <- 10^3.819*hadp*H
hhhadp <- 10*H*hhadp
mgadp <- 10^3.294*Mg*adp
mghadp <- 10^1.61*Mg*hadp
mg2adp <- 10*Mg*mgadp
kadp <- 10^0.82*k*adp
hpi <- 10^11.616*H*pi
hhpi <- 10^6.7*H*hpi
hhhpi <- 10^1.962*H*hhpi
mgpi <- 10^3.4*Mg*pi
mghpi <- 10^1.946*Mg*hpi
mghhpi <- 10^1.19*Mg*hhpi
kpi <- 10^0.6*k*pi
khpi <- 10^1.218*k*hpi
khhpi <- 10^-0.2*k*hhpi
hpcr <- 10^14.3*H*pcr
hhpcr <- 10^4.5*H*hpcr
hhhpcr <- 10^2.7*H*hhpcr
hhhhpcr <- 100*H*hhhpcr
mghpcr <- 10^1.6*Mg*hpcr
kpcr <- 10^0.74*k*pcr
khpcr <- 10^0.31*k*hpcr
khhpcr <- 10^-0.13*k*hhpcr
hcr <- 10^14.3*H*cr
hhcr <- 10^2.512*H*hcr
hlactate <- 10^3.66*H*lactate
mglactate <- 10^0.93*Mg*lactate
tatp <- atp + hatp + hhatp + hhhatp + mgatp + mghatp + mg2atp + katp
tadp <- adp + hadp + hhadp + hhhadp + mghadp + mgadp + mg2adp + kadp
tpi? <- pi + hpi + hhpi + hhhpi + mgpi + mghpi + mghhpi + kpi + khpi + khhpi
tpcr <- pcr + hpcr + hhpcr + hhhpcr + hhhhpcr + mghpcr + kpcr + khpcr +
khhpcr
tcr? <- cr + hcr + hhcr
tlactate <- lactate + hlactate + mglactate
# tmg <- Mg + mgatp + mghatp + mg2atp + mgadp + mghadp + mg2adp + mgpi +
#??? kghpi + mghhpi + mghpcr + mglactate
# tk <- k + katp + kadp + kpi + khpi + khhpi + kpcr + khpcr + khhpcr
total = c(
?? ?tatp - 0.008,
?? ?tadp - 0.00001,
?? ?tpi - 0.003,
?? ?tpcr - 0.042,
?? ?tcr - 0.004,
?? ?tlactate - 0.005)
return(total);
# })
}
# conditions
x0 = c(
?? ?atp = 0.008,
?? ?adp = 0.00001,
?? ?pi = 0.003,
?? ?pcr = 0.042,
?? ?cr = 0.004,
?? ?lactate = 0.005
) / 3;
# tricky to get a positive value !!!
x0[1] = 0.001; # still NOT positive;
x = multiroot(solve.AcidSpecies, x0, H = 4E-8)
On 1/23/2023 12:37 AM, Leonard Mada wrote:
Dear Troels, The system that you mentioned needs to be transformed first. The equations are standard acid-base equilibria-type equations in analytic chemistry. ATP + H <-> ATPH ATPH + H <-> ATPH2 ATPH2 + H <-> ATPH3 [...] The total amount of [ATP] is provided, while the concentration of the intermediates are unknown. Q.) It was unclear from your description: Do you know the pH? Or is the pH also unknown? I believe that the system is exactly solvable. The "multivariable" system/solution may be easier to write down: but is uglier to solve, as the "system" is under-determined. You can use optim in such cases, see eg. an example were I use it: https://github.com/discoleo/R/blob/master/Stat/Polygons.Examples.R a2 = optim(c(0.9, 0.5), polygonOptim, d=x) # where the function polygonOptim() computes the distance between the starting-point & ending point of the polygon; # (the polygon is defined only by the side lengths and optim() tries to compute the angles); # optimal distance = 0, when the polygon is closed; # Note: it is possible to use more than 2 starting values in the example above (the version with optim) works quit well; # - but you need to "design" the function that is optimized for your particular system, e.g. #?? by returning: c((ATPTotal - value)^2, (ADPTotal - value)^2, ...); S.1.) Exact Solution ATP system: You can express all components as eqs of free ATP, [ATP], and [H], [Mg], [K]. ATPH = KaATPH * ATP * H; ATPH2 = KaATPH2 * ATPH * H = KaATPH2 * KaATPH * ATP * H^2; [...] Then you substitute these into the total-eqs. It is a bit uglier, as you have 5 coupled systems: ATP, ADP, Creatinine, CreaP and Lactate. This is why I thought that it may be easier to do it, at least partially/hybrid, in a "multivariate" way. # Total ATP: ATPTotal = KaATPH * ATP * H + KaATPH2 * KaATPH * ATP * H^2 + ??? + KaATPH3 * KaATPH2 * KaATPH * ATP * H^3 + ...; Note: - the system above is solvable, if the pH is known; it may be undetermined if the pH is NOT known (but I am not very certain: easier to spot only when all the eqs are written down nicely); - the eqs for total K & total Mg: are not useful to solve the system, as there are NO constraints on the total values; they will yield only some numerical values (which may be interesting for the analysis); - total K & total Mg can be computed after solving the system; Please let me know if you need further assistance. Sincerely, Leonard