IN-PROGRESS¶

Enzyme Kinetics :¶

Our model: `E + S <-> ES` (with kinetic pars k1_forward and k1_reverse), and `ES -> E + P` (k2_forward)¶

In experiment `enzyme_1_a`, we were given k1_forward, k1_reverse and k2_forward... But what to do if we're just given kM and kcat¶

In Part 1, we'll "cheat" and use the actual value of k1_forward
In Part 2, we'll explore what happens if, lacking an actual value, we under-estimate k1_forward
In Part 3, we'll explore what happens if we over-estimate k1_forward

Background: please see experiment enzyme_1_a

THE REACTION:¶

the enzyme Adenosinedeaminase with the substrate 2,6-Diamino-9-β-D-deoxyribofuranosyl-9-H-purine,
and the initial concentration values choosen below, all satisfy the customary Michaelis-Menten assumptions that
[E] << [S] and that the reaction rate constants satisfy k1_reverse >> k2_forward

Source of kinetic parameters: page 16 of "Analysis of Enzyme Reaction Kinetics, Vol. 1", by F. Xavier Malcata, Wiley, 2023

TAGS : "uniform compartment", "chemistry", "numerical", "enzymes"¶

In [1]:

LAST_REVISED = "Nov. 4, 2024"
LIFE123_VERSION = "1.0.0.rc.0"      # Library version this experiment is based on

In [2]:

#import set_path                    # Using MyBinder?  Uncomment this before running the next cell!

In [3]:

#import sys
#sys.path.append("C:/some_path/my_env_or_install")   # CHANGE to the folder containing your venv or libraries installation!
# NOTE: If any of the imports below can't find a module, uncomment the lines above, or try:  import set_path   

import ipynbname
import pandas as pd

from life123 import check_version, ChemData, UniformCompartment, ReactionEnz, GraphicLog, PlotlyHelper

In [4]:

check_version(LIFE123_VERSION)

OK

In [ ]:

Assume we're only given values for kM and kcat¶

What values of k1_forward, k1_reverse and k2_forward are compatible with them?

In [5]:

# The following values are taken from experiment `enzyme_1`
kM = 8.27777777777777
kcat = 49

In [6]:

# k2_forward equals kcat ; not much to say here
k2_forward = kcat

By contrast kM = (k2_forward + k1_reverse) / k1_forward

We are given kM and k2_forward, which is the same value as the given kcat.

But how to solve for k1_forward and k1_reverse?? We have just 1 equation and 2 variables! The system of equations is "underdetermined" : what can we do?

We'll venture some guesses for k1_forward, and compute the corresponding k1_reverse that will satisfy the above equation.

k1_reverse = kM * k1_forward - kcat (kcat has same value as k2_forward)

In [7]:

# Example, using the k1_forward=18. from experiment `enzyme_1`
k1_forward = 18.
k1_reverse = kM * k1_forward - kcat 
k1_reverse

Out[7]:

99.99999999999986

Naturally, we're getting the same value we had for k1_reverse in that experiment. But what if k1_forward is quite different?

In [8]:

import numpy as np

In [9]:

def compute_k1_reverse(kM, kcat, k1_forward, verbose=True):
    k1_reverse = kM * k1_forward - kcat 
    if verbose:
        if np.allclose(k1_reverse, 0):
            print (f"k1_reverse: {k1_reverse} , K = INFINITE")
        else:
            K = k1_forward / k1_reverse
            print(f"k1_reverse: {k1_reverse} , K = {K}")
        
    return k1_reverse

In [10]:

compute_k1_reverse(kM, kcat, k1_forward=18.)

k1_reverse: 99.99999999999986 , K = 0.18000000000000024

Out[10]:

99.99999999999986

In [11]:

compute_k1_reverse(kM, kcat, k1_forward=0.1)

k1_reverse: -48.172222222222224 , K = -0.002075885134355899

Out[11]:

-48.172222222222224

The negative values aren't good! We must add a constraint that:¶

kM * k1_forward - kcat > 0
i.e. k1_forward > kcat/kM

In other words, k1_forward_min = kcat/kM

In [12]:

k1_forward_min = kcat / kM
k1_forward_min

Out[12]:

5.919463087248328

In [13]:

compute_k1_reverse(kM, kcat, k1_forward=k1_forward_min)

k1_reverse: 0.0 , K = INFINITE

Out[13]:

0.0

In [14]:

compute_k1_reverse(kM, kcat, k1_forward=6.)

k1_reverse: 0.6666666666666146 , K = 9.000000000000703

Out[14]:

0.6666666666666146

In [15]:

compute_k1_reverse(kM, kcat, k1_forward=10.)

k1_reverse: 33.7777777777777 , K = 0.29605263157894807

Out[15]:

33.7777777777777

In [16]:

compute_k1_reverse(kM, kcat, k1_forward=40.)

k1_reverse: 282.1111111111108 , K = 0.14178810555336763

Out[16]:

282.1111111111108

The fixed point k1_forward = k1_reverse occurs when k1_forward = kM * k1_forward - kcat , i.e.
kM * k1_forward - k1_forward = kcat
k1_forward * (kM - 1) = kcat
k1_forward = kcat / (kM - 1) Notice that the above makes sense only if kM > 1

In [17]:

# Our kM is indeed > 1, so all good:

k1_forward_fixed_pnt = kcat / (kM - 1)
k1_forward_fixed_pnt

Out[17]:

6.7328244274809235

In [18]:

compute_k1_reverse(kM, kcat, k1_forward=6.7328244274809235)

k1_reverse: 6.732824427480921 , K = 1.0000000000000004

Out[18]:

6.732824427480921

Indeed, a fixed point

In [19]:

k1_forward_choices = np.linspace(5.92, 40., 2000)    # Grid creation

In [20]:

k1_reverse_choices = compute_k1_reverse(kM, kcat, k1_forward=k1_forward_choices, verbose=False)

In [21]:

import plotly.express as px

In [22]:

px.line(x=k1_forward_choices, y=k1_reverse_choices, title="k1_forward vs. k1_reverse")

In [23]:

K_choices = k1_forward_choices / k1_reverse_choices

In [24]:

px.line(x=k1_forward_choices, y=np.log(K_choices), title="log(K) vs. k1_forward")

In [ ]:

In the scenario we're exploring, there's LITTLE ENZYME relative to initial substrate concentration¶

In [25]:

S0 = 20.
E0 = 1.

In [ ]:

1. Using the known, exact values for `k1_forward` and `k1_reverse`¶

Source: page 16 of "Analysis of Enzyme Reaction Kinetics, Vol. 1", by F. Xavier Malcata, Wiley, 2023

In [26]:

k1_forward = 18.
k1_reverse = 100.

In [27]:

chem_data = ChemData(names=["P", "ES"])

In [28]:

# Our Enzyme
chem_data.add_chemical(name="Adenosinedeaminase", label="E") 

# Our Substrate
chem_data.add_chemical(name="2,6-Diamino-9-β-D-deoxyribofuranosyl-9-H-purine", label="S");

In [29]:

chem_data.all_chemicals()

Out[29]:

	name	label
0	P	P
1	ES	ES
2	Adenosinedeaminase	E
3	2,6-Diamino-9-β-D-deoxyribofuranosyl-9-H-purine	S

Specify the Kinetic Parameters¶

In [30]:

# Elementary Reaction E + S <-> ES
chem_data.add_reaction(reactants=["E", "S"], products=["ES"],
                       forward_rate=k1_forward, reverse_rate=k1_reverse) 

# Elementary Reaction ES <-> E + P 
chem_data.add_reaction(reactants=["ES"], products=["E", "P"],
                       forward_rate=kcat, reverse_rate=0)

chem_data.describe_reactions()

Number of reactions: 2 (at temp. 25 C)
0: E + S <-> ES  (kF = 18 / kR = 100 / delta_G = 4,250.9 / K = 0.18) | 1st order in all reactants & products
1: ES <-> E + P  (kF = 49 / kR = 0) | 1st order in all reactants & products
Set of chemicals involved in the above reactions: {'E', 'S', 'P', 'ES'}

In [ ]:

In [31]:

# Here we use the "slower" preset for the variable steps, a conservative option prioritizing accuracy over speed
uc = UniformCompartment(chem_data=chem_data, preset="slower")
uc.set_conc(conc={"S": S0, "E": E0})      # Small ampount of enzyme `E`, relative to substrate `S`
uc.describe_state()

SYSTEM STATE at Time t = 0:
4 species:
  Species 0 (P). Conc: 0.0
  Species 1 (ES). Conc: 0.0
  Species 2 (E). Conc: 1.0
  Species 3 (S). Conc: 20.0
Set of chemicals involved in reactions: {'E', 'S', 'P', 'ES'}

In [32]:

uc.enable_diagnostics()   # To save diagnostic information about the simulation - in particular, the REACTION RATES

In [33]:

# Perform the reactions
uc.single_compartment_react(duration=1.2, initial_step=0.05)

Some steps were backtracked and re-done, to prevent negative concentrations or excessively large concentration changes
784 total step(s) taken
Number of step re-do's because of negative concentrations: 2
Number of step re-do's because of elective soft aborts: 1
Norm usage: {'norm_A': 706, 'norm_B': 710, 'norm_C': 706, 'norm_D': 706}

In [34]:

uc.plot_history(colors=['cyan', 'green', 'violet', 'red'], show_intervals=True, 
                title_prefix="Using exact values for k1_forward and k1_reverse")

plot_pandas() NOTICE: Excessive number of vertical lines (785) - only showing 1 every 6 lines

What is the initial rate of production of the final reaction product `P`?¶

One could take the numerical derivative (gradient) of the time values of [P] - but no need to! Reaction rates are computed in the course of the simulation, and stored alongside other diagnostic data, provided that diagnostics were enabled (as we did indeed enable)

In [35]:

history_with_rates_exact = uc.get_diagnostics().get_system_history_with_rxn_rates(rxn_index=1)   # We specify the reaction that involves `P`
history_with_rates_exact

Reaction:  ES <-> E + P

Out[35]:

	TIME	P	ES	E	S	caption	rate
0	0.000000	0.000000	0.000000	1.000000	20.000000		0.000000
1	0.000500	0.000000	0.180000	0.820000	19.820000		8.820000
2	0.000750	0.002205	0.246431	0.753569	19.751364		12.075109
3	0.000763	0.002356	0.249321	0.750679	19.748323		12.216716
4	0.000769	0.002432	0.250756	0.749244	19.746811		12.287060
...	...	...	...	...	...	...	...
779	1.166644	19.724207	0.029857	0.970143	0.245936		1.462999
780	1.173671	19.734488	0.028775	0.971225	0.236738		1.409959
781	1.180698	19.744395	0.027729	0.972271	0.227875		1.358736
782	1.187725	19.753943	0.026720	0.973280	0.219337		1.309274
783	1.194752	19.763144	0.025745	0.974255	0.211111		1.261520

784 rows × 7 columns

In [36]:

# Let's take a look at how the reaction rate varies with time
PlotlyHelper.plot_pandas(df=history_with_rates_exact, 
                         title="Reaction rate, dP/dt, over time",
                         x_var="TIME", fields="rate", 
                         x_label="time", y_label="dP/dt")

In [ ]:

2. Guessing a SMALLER value of `k1_forward` (UNDER-ESTIMATED)¶

In [37]:

chem_data = ChemData(names=["P", "ES"])

In [38]:

# Our Enzyme
chem_data.add_chemical(name="Adenosinedeaminase", label="E") 

# Our Substrate
chem_data.add_chemical(name="2,6-Diamino-9-β-D-deoxyribofuranosyl-9-H-purine", label="S");

In [39]:

k1_forward = 6.5    # Close to the smallest possible value

In [40]:

k1_reverse = compute_k1_reverse(kM, kcat, k1_forward=k1_forward)
k1_reverse

k1_reverse: 4.8055555555555 , K = 1.3526011560693798

Out[40]:

4.8055555555555

In [41]:

# Reaction E + S <-> ES , with 1st-order kinetics, 
# and a forward rate that is much faster than its revers one
chem_data.add_reaction(reactants=["E", "S"], products=["ES"],
                       forward_rate=k1_forward, reverse_rate=k1_reverse) 

# Reaction ES <-> E + P , with 1st-order kinetics
chem_data.add_reaction(reactants=["ES"], products=["E", "P"],
                       forward_rate=49.,reverse_rate=0)

chem_data.describe_reactions()

Number of reactions: 2 (at temp. 25 C)
0: E + S <-> ES  (kF = 6.5 / kR = 4.8056 / delta_G = -748.72 / K = 1.3526) | 1st order in all reactants & products
1: ES <-> E + P  (kF = 49 / kR = 0) | 1st order in all reactants & products
Set of chemicals involved in the above reactions: {'E', 'S', 'P', 'ES'}

In [42]:

# Here we use the "slower" preset for the variable steps, a conservative option prioritizing accuracy over speed
uc = UniformCompartment(chem_data=chem_data, preset="slower")
uc.set_conc(conc={"S": S0, "E": E0})      # Small ampount of enzyme `E`, relative to substrate `S`
uc.describe_state()

SYSTEM STATE at Time t = 0:
4 species:
  Species 0 (P). Conc: 0.0
  Species 1 (ES). Conc: 0.0
  Species 2 (E). Conc: 1.0
  Species 3 (S). Conc: 20.0
Set of chemicals involved in reactions: {'E', 'S', 'P', 'ES'}

In [43]:

uc.enable_diagnostics()   # To save diagnostic information about the simulation - in particular, the REACTION RATES

In [44]:

# Perform the reactions
uc.single_compartment_react(duration=1.2, initial_step=0.05)

Some steps were backtracked and re-done, to prevent negative concentrations or excessively large concentration changes
788 total step(s) taken
Number of step re-do's because of negative concentrations: 1
Number of step re-do's because of elective soft aborts: 2
Norm usage: {'norm_A': 704, 'norm_B': 709, 'norm_C': 704, 'norm_D': 704}

In [45]:

uc.plot_history(colors=['cyan', 'green', 'violet', 'red'], show_intervals=True, 
                title_prefix="Using smaller value for k1_forward")

plot_pandas() NOTICE: Excessive number of vertical lines (789) - only showing 1 every 6 lines

At first glance, not too different from before, overall, but let's take a closer look

In [46]:

history_with_rates_under = uc.get_diagnostics().get_system_history_with_rxn_rates(rxn_index=1)   # We specify the reaction that involves `P`
history_with_rates_under

Reaction:  ES <-> E + P

Out[46]:

	TIME	P	ES	E	S	caption	rate
0	0.000000	0.000000	0.000000	1.000000	20.000000		0.000000
1	0.000500	0.000000	0.065000	0.935000	19.935000		3.185000
2	0.000750	0.000796	0.094414	0.905586	19.904789		4.626306
3	0.000763	0.000854	0.095815	0.904185	19.903330		4.694958
4	0.000769	0.000883	0.096514	0.903486	19.902602		4.729203
...	...	...	...	...	...	...	...
783	1.171227	19.767701	0.026902	0.973098	0.205396		1.318201
784	1.178115	19.776782	0.025880	0.974120	0.197338		1.268138
785	1.185004	19.785518	0.024895	0.975105	0.189587		1.219867
786	1.191892	19.793921	0.023946	0.976054	0.182134		1.173332
787	1.198781	19.802003	0.023030	0.976970	0.174967		1.128479

788 rows × 7 columns

In [47]:

# Let's take a look at how the reaction rate varies with time
PlotlyHelper.plot_pandas(df=history_with_rates_under, 
                         title="Reaction rate, dP/dt, over time",
                         x_var="TIME", fields="rate", 
                         x_label="time", y_label="dP/dt")

In [ ]:

3. Guessing a LARGER value of `k1_forward` (OVER-ESTIMATED)¶

In [48]:

chem_data = ChemData(names=["P", "ES"])

In [49]:

# Our Enzyme
chem_data.add_chemical(name="Adenosinedeaminase", label="E") 

# Our Substrate
chem_data.add_chemical(name="2,6-Diamino-9-β-D-deoxyribofuranosyl-9-H-purine", label="S");

In [50]:

k1_forward = 40.    # Well above the known value of 18.

In [51]:

k1_reverse = compute_k1_reverse(kM, kcat, k1_forward=k1_forward)
k1_reverse

k1_reverse: 282.1111111111108 , K = 0.14178810555336763

Out[51]:

282.1111111111108

In [52]:

# Reaction E + S <-> ES , with 1st-order kinetics, 
# and a forward rate that is much faster than its revers one
chem_data.add_reaction(reactants=["E", "S"], products=["ES"],
                       forward_rate=k1_forward, reverse_rate=k1_reverse) 

# Reaction ES <-> E + P , with 1st-order kinetics
chem_data.add_reaction(reactants=["ES"], products=["E", "P"],
                       forward_rate=49.,reverse_rate=0)

chem_data.describe_reactions()

Number of reactions: 2 (at temp. 25 C)
0: E + S <-> ES  (kF = 40 / kR = 282.11 / delta_G = 4,842.4 / K = 0.14179) | 1st order in all reactants & products
1: ES <-> E + P  (kF = 49 / kR = 0) | 1st order in all reactants & products
Set of chemicals involved in the above reactions: {'E', 'S', 'P', 'ES'}

In [53]:

# Here we use the "slower" preset for the variable steps, a conservative option prioritizing accuracy over speed
uc = UniformCompartment(chem_data=chem_data, preset="slower")
uc.set_conc(conc={"S": S0, "E": E0})      # Small ampount of enzyme `E`, relative to substrate `S`
uc.describe_state()

SYSTEM STATE at Time t = 0:
4 species:
  Species 0 (P). Conc: 0.0
  Species 1 (ES). Conc: 0.0
  Species 2 (E). Conc: 1.0
  Species 3 (S). Conc: 20.0
Set of chemicals involved in reactions: {'E', 'S', 'P', 'ES'}

In [54]:

uc.enable_diagnostics()   # To save diagnostic information about the simulation - in particular, the REACTION RATES

In [55]:

# Perform the reactions
uc.single_compartment_react(duration=1.2, initial_step=0.05)

Some steps were backtracked and re-done, to prevent negative concentrations or excessively large concentration changes
864 total step(s) taken
Number of step re-do's because of negative concentrations: 2
Number of step re-do's because of elective soft aborts: 1
Norm usage: {'norm_A': 829, 'norm_B': 835, 'norm_C': 829, 'norm_D': 829}

In [56]:

uc.plot_history(colors=['cyan', 'green', 'violet', 'red'], show_intervals=True, 
                title_prefix="Using smaller value for k1_forward")

plot_pandas() NOTICE: Excessive number of vertical lines (865) - only showing 1 every 6 lines

At first glance, not too different from before, overall, but let's take a closer look

In [57]:

history_with_rates_over = uc.get_diagnostics().get_system_history_with_rxn_rates(rxn_index=1)   # We specify the reaction that involves `P`
history_with_rates_over

Reaction:  ES <-> E + P

Out[57]:

	TIME	P	ES	E	S	caption	rate
0	0.000000	0.000000	0.000000	1.000000	20.000000		0.000000
1	0.000500	0.000000	0.400000	0.600000	19.600000		19.600000
2	0.000750	0.004900	0.484489	0.515511	19.510611		23.739956
3	0.000763	0.005197	0.487513	0.512487	19.507291		23.888118
4	0.000769	0.005346	0.489003	0.510997	19.505651		23.961149
...	...	...	...	...	...	...	...
859	1.170211	19.713808	0.030426	0.969574	0.255766		1.490884
860	1.176184	19.722714	0.029500	0.970500	0.247787		1.445487
861	1.182158	19.731349	0.028612	0.971388	0.240039		1.401986
862	1.188131	19.739724	0.027735	0.972265	0.232542		1.358993
863	1.194105	19.747842	0.026901	0.973099	0.225257		1.318155

864 rows × 7 columns

In [58]:

# Let's take a look at how the reaction rate varies with time
PlotlyHelper.plot_pandas(df=history_with_rates_over, 
                         title="Reaction rate, dP/dt, over time",
                         x_var="TIME", fields="rate", 
                         x_label="time", y_label="dP/dt")

In [ ]:

Comparing the 3 scenarios¶

The concentrations of the intermediate `ES` over time¶

In [59]:

es_under = PlotlyHelper.plot_pandas(df=history_with_rates_under, 
                         title="underestimated k1_forward",
                         x_var="TIME", fields="ES", 
                         x_label="time", y_label="ES", colors = "yellow")

In [60]:

es_exact = PlotlyHelper.plot_pandas(df=history_with_rates_exact, 
                         title="exact k1_forward",
                         x_var="TIME", fields="ES", 
                         x_label="time", y_label="ES", colors = "green")

In [61]:

es_over = PlotlyHelper.plot_pandas(df=history_with_rates_over, 
                         title="overestimated k1_forward",
                         x_var="TIME", fields="ES", 
                         x_label="time", y_label="ES", colors = "purple")

In [62]:

 PlotlyHelper.combine_plots([es_under, es_exact, es_over], 
                            title="ES")

The concentrations of the product `P` over time¶

In [63]:

p_under = PlotlyHelper.plot_pandas(df=history_with_rates_under, 
                         title="underestimated k1_forward",
                         x_var="TIME", fields="P", 
                         x_label="time", y_label="P", colors = "yellow")

In [64]:

p_exact = PlotlyHelper.plot_pandas(df=history_with_rates_exact, 
                         title="exact k1_forward",
                         x_var="TIME", fields="P", 
                         x_label="time", y_label="P", colors = "cyan")

In [65]:

p_over = PlotlyHelper.plot_pandas(df=history_with_rates_over, 
                         title="overestimated k1_forward",
                         x_var="TIME", fields="P", 
                         x_label="time", y_label="P", colors = "purple")

In [66]:

 PlotlyHelper.combine_plots([p_under, p_exact, p_over], 
                            title="P")

In [ ]:

The Reaction Rate over time¶

In [67]:

r1_under = PlotlyHelper.plot_pandas(df=history_with_rates_under, 
                         title="underestimated k1_forward",
                         x_var="TIME", fields="rate", 
                         x_label="time", y_label="dP/dt", colors = "yellow")

In [68]:

r1_exact = PlotlyHelper.plot_pandas(df=history_with_rates_exact, 
                         title="exact k1_forward",
                         x_var="TIME", fields="rate", 
                         x_label="time", y_label="dP/dt", colors = "blue")

In [69]:

r1_over = PlotlyHelper.plot_pandas(df=history_with_rates_over, 
                         title="overestimated k1_forward",
                         x_var="TIME", fields="rate", 
                         x_label="time", y_label="dP/dt", colors = "purple")

In [70]:

 PlotlyHelper.combine_plots([r1_under, r1_exact, r1_over], 
                            title="Reaction Rate")

In [ ]:

IN-PROGRESS¶

Enzyme Kinetics :¶

Our model: E + S <-> ES (with kinetic pars k1_forward and k1_reverse), and ES -> E + P (k2_forward)¶

In experiment enzyme_1_a, we were given k1_forward, k1_reverse and k2_forward... But what to do if we're just given kM and kcat¶

THE REACTION:¶

TAGS : "uniform compartment", "chemistry", "numerical", "enzymes"¶

Assume we're only given values for kM and kcat¶

The negative values aren't good! We must add a constraint that:¶

In the scenario we're exploring, there's LITTLE ENZYME relative to initial substrate concentration¶

1. Using the known, exact values for k1_forward and k1_reverse¶

Specify the Kinetic Parameters¶

What is the initial rate of production of the final reaction product P?¶

2. Guessing a SMALLER value of k1_forward (UNDER-ESTIMATED)¶

3. Guessing a LARGER value of k1_forward (OVER-ESTIMATED)¶

Comparing the 3 scenarios¶

The concentrations of the intermediate ES over time¶

The concentrations of the product P over time¶

The Reaction Rate over time¶

Our model: `E + S <-> ES` (with kinetic pars k1_forward and k1_reverse), and `ES -> E + P` (k2_forward)¶

In experiment `enzyme_1_a`, we were given k1_forward, k1_reverse and k2_forward... But what to do if we're just given kM and kcat¶

1. Using the known, exact values for `k1_forward` and `k1_reverse`¶

What is the initial rate of production of the final reaction product `P`?¶

2. Guessing a SMALLER value of `k1_forward` (UNDER-ESTIMATED)¶

3. Guessing a LARGER value of `k1_forward` (OVER-ESTIMATED)¶

The concentrations of the intermediate `ES` over time¶

The concentrations of the product `P` over time¶