MICHAELIS-MENTEN AND BRIGGS-HALDANE MECHANISMS:

For both the MM and HALDANE models (obviously, they are the same, with the
difference being the ratio k-1/k2), from "Solving ..." eqns 12a,b with k-2=0:
	tau1^-1 = k1*s + k-1 + k2
	tau2^-1 = tau1 * k1*s * k2
Which step is identified with tau1 (or tau2) depends on the physics of
the situation.
Note that the subscript identifiers used for the step constants in 
xrun are different from those used here and in "Solving...".
Note also that the conditions of [E]0 = [S]0 and far from equilibrium
are not be taken into account below.

MM:	
	bring xrun into a working directory
		cd tmp
		cp ~jrupley/kinetics/run.template/xrun .

	edit xrun
		ki:	1.e8	1.e2	1.e0
		e,s,p:	1.e-4	1.e-4	0

	xrun -> xgobi display
	
	select display:
		concn vs ln time
		XY

	note 	fast buildup of steady-state concn of ES
		tau^-1 ~ k1*s = 1.e4
		log10 tau ~ -4
		ln tau ~ -4*2.3 = -9.2

	note 	slow appearance of product = disappearance of ES
		tau^-1 ~ k2
		ln tau ~ 0


	select display:
		ln concn vs ln time
		XY
		identify species

	note 	buildup of ES complete before appreciable P
		break in slope of line of ln(p) vs ln(time)
		== change in rate of increase in p when reach steady state
		at ln tau = -9.2
		

HALDANE:

	edit xrun
		ki:	1.e8	1.e0	1.e2   #note switch in values of k2, k3
		e,s,p:	1.e-4	1.e-4	0

	xrun -> xgobi display:
	
	select display for both MM and HALDANE results:
		concn vs ln time
		XY

	note 	about same maximum value of steady-state concentration of ES
		Km(MM) = Km(HALDANE)

	note 	fast buildup of steady-state concn of ES
		tau^-1 = k1*s = 1.e4
		ln tau ~ -4*2.3 = -9.2
	
	note	for appearance of product,
		tau^-1 ~ k2
		ln tau(HALDANE) ~ -2*2.3 = -4.6 << ln tau(MM)


BOTH MM AND HALDANE:

	edit xrun and xrun -> xgobi display:
		e,s,p:	1.e-7	1.e-7	0
	
	select display for both MM and HALDANE results:
		concn vs ln time
		XY

	look at tau1 and tau2 for each mechanism:

		tau^-1(MM) ~ k-1
		ln tau1(MM) ~ -2.3 * log (1.e2) = -4.6
		tau2^-1(MM) ~ k1*s*k2/k-1
		ln tau2(MM) ~ 2.3

		note: for MM model at low S, tau1 is determined by the
		back reaction of ES -> E + S, and tau2 is longer than
		k1*s or k2.

		tau^-1(HALDANE) ~ k2
		ln tau1(HALDANE) ~ -4.6
		tau2^-1(HALDANE) ~ k1*s
		ln tau2(HALDANE) ~ -2.3 

		note: for HALDANE model, bimolecular reaction
		to form ES is rate-determining 
		at low [S] = 10-7 M << Km = 10-6 M;
		and buildup of ES complex is governed by k2, the
		rate of breakdown of ES to P;
		the situation is switched from that for the
		high [S] regime.


	
	select display for both MM and HALDANE results:
		ln concn vs ln time
		XY

	should be easy to pick out rate-determining step
	

	select display for both MM and HALDANE results:
		concn vs ln time
		rotate

	get used to 3-d display of results

ONE-STEP AND TWO-STEP EQUILIBRIUM BINDING

These mechanisms are those treated in "Solving...".


cd ~jrupley/kinetics/run.examples

xgobi one_step &	# the "&" puts the job in the background

xgobi two_step &

select display for both results:
	concn vs ln time
	XY

Note that the equilibrium levels for the one-step reaction are the same
as the levels for the transiently formed EU1 intermediate in the
two-step reaction.  This is in agreement with the rate constants used
in the simulations.

Note that in the second step of the two-step reaction, the enzyme drains
from the first-formed intermediate EU1 almost entirely into the more
stable form EU2.

tau1^-1 = k1*s + k-1 = 1.e8*1.e-3 + 1.e5 = 2.e5
ln tau1 = -2.3*5.3 = -12.2

tau2^-1 = tau1 * ((k1*k-2 + k1*k2)*s + k-1*k-2)
        = 1.e-5.3 * ((1.e8*1.e0 + 1.e8*1.e2)*1.e-3 + 1.e5*1.e0)
	= 1.e-5.3 * (1.e10*1.e-3 + 1.e5)
	= 1.e-5.3 * 1.e7
	= 1.e1.7
ln tau2 = -2.3*1.7
	= -3.91

To see the sequence of events of the two-step mechanism more clearly and
to define the tau values,

select display for both results:
	ln concn vs ln time
	XY

Note the position of the lines representing the log concentrations for 
ES and P, and the log time values of the breaks.

LYSOZYME

A considerably more complex mechanism, but perhaps typical of the complexity
of most enzyme reactions, in which there are expected to be multiple
isomerizations of enzyme-substrate species, as the system moves to the
transition state.


cd ~jrupley/kinetics/run.examples

xgobi lys_1 &	# the "&" puts the job in the background

select display for both results:
	concn vs ln time
	XY

mechanism:

EU2 <-> EU1 <-> E + S <-> ESB1 <-> ESB2 <-> ESG --> E + P

where EU1 and EU2 are abortive (nonproductive) complexes, and
ESB1, ESB2, and ESG are pre-equilibrium complexes along the reaction
path, with the RDS being breakdown of ESG to the glycosyl enzyme and
first product.

Note that the simulation shows that the enzyme flows from one ES or EU
form to the next in a sequence determined by the relative values of the
step rate constants and of the step equilibrium constants:

kf     1.e8*1.e-5  2.e2    1.4e1      .5       .04

	E -->  EU1  ->  EU2  ->  ESB2  ->  ESG  -> E + P
	       ESB1      |                  ^
			 |__________________|


To see the sequence of events and the tau values more clearly,

select display:
	ln concn vs ln time
	XY
	rotate
	identify

Note that after the pre-equilibria are established, both EU2 and ESG are
present at approximately the same concentration, ESB2 is at lower
concentration, and ESB1 and EU1 are more negligible.

LDH

Compare reactions:
	ldh_50_1	at 50 degC, from 10-3 M NAD+, lactate, LDH
	ldh_50_rev_1	at 50 degC, from 10-3 M NADH, pyruvate, LDH
	ldh_0_2		at 0 degC, from 10-3 M NAD+, lactate, LDH


cd ~jrupley/kinetics/run.examples

xgobi ldh_50_rev_1& xgobi ldh_50_1& xgobi ldh_0_2&

select display for all results:
	concn vs ln time
	XY
	identify some species


Note: 	same equilibrium position starting from different reactants at 50 degC;
	reaction equilibria are different at 0 degC.

Note at 50degC:
	equilibrium concentrations of EA = LDH.NAD+ and EY = LDH.NADH 
		are significant and comparable;
		this is striking in view of the equilibrium concentrations
		of NADH and NAD+ differing by several orders of magnitude;
		Km values evolve to match cell concentrations;
	equilibrium concentration of EAB is small but significantly >0:
		this is consistent with compulsory ordered rather than
		Theorell-Chance model;
		EAB = ternary complex of either LDH.NAD+.lactate or 
		LDH.NADH.pyruvate;
	equilibrium concentration of Y = NADH is very small: 
		[NADH] free is very low in cells, owing to strong binding by
		enzymes such as LDH, which itself is plentiful;
	in each reaction direction, there is a flow from the free enzyme, 
		into the first binary complex, into the ternary complex,
		into the second binary complex; 
		to see this, it may help to rotate the data in 3d and to
		display ln concn.
	because of the high and equal concentrations of substrate and enzyme, 
		there is essentially no free enzyme at equilibrium;
	consider what step is rate-determining;

CHYMOTRYPSIN

Compare reaction of amide and ester substrates.

cd ~jrupley/kinetics/run.examples

xgobi cht_amide& xgobi cht_ester&

select display for all results:
	concn vs ln time
	XY
	ln concn vs ln time
	rotate
	identify some species

Note:	concentrations of intermediates;
	P1 released before P2?