This module, the enzyme kinetics simulator, is used to make simulations of
enzyme-catalysed reactions once the kinetic parameters of the enzyme (Km and ѵmax, but
also Ki and the Hill exponent) and the experimental conditions, such as initial concentration of substrate and concentration of inhibitor, are provided.
This application module
is provided with four panels:
Panel 1: You can see how the
concentration of substrate [S] (initially, [S]0) varies as it is consumed while the product concentration [P] increases.
Panel 2:
The initial rate (ѵ0) changes can also be observed as the reaction
progresses, thus visualising the effect of the decreasing initial substrate concentration on
the initial reaction rate.
Panel 3 shows the classical
Michaelis-Menten plot (ѵ0 vs. [S]0)
Panel 4 shows the corresponding Lineweaver-Burk plot.
In all panels, up to three different time courses are shown, for the sake of comparison.
Input data:
Test 3 conditions:
a
b
c
[S]0=
mM
ѵmax=
mM min⁻¹
Km=
mM
[I] =
mM
KIE=
mM
KIES=
mM
nHill =
It is not possible to study the effect of inhibitor simultaneouesly with Hill coefficients different from 1.
Panel 1
Panel 2
Panel 3
Panel 4
practice problems.
Queremos caracterizar una nueva quinasa que hemos aislado,
determinando su constante de Michaelis y su velocidad máxima. Por similitud con otras quinasas conocidas podemos asumir que la
Km estará en torno a 0.5 mM y la ѵmax alrededor de 10 mM/min. Para la
caracterización mediremos in vitro la velocidad inicial de conversión
de su sustrato. Para que esta medida sea fiable debemos asegurarnos de que
durante el tiempo de medida la velocidad de conversión sea realmente una
velocidad inicial. Dado que la medida requiere 3
minutos, ¿cuál sería la concentración inicial de sustrato mínima que nos
garantizaría condiciones de velocidad inicial?
Una enzima con una Km de 0.5 mM y una ѵmax de 10 mM/min se
encuentra en una mezcla de ensayo con una concentración 10 mM de su sustrato.
(a) ¿Cuánto tiempo necesitará para transformar el 80% del sustrato en
producto?
(b) ¿Cuál sería el valor de la Km de otra enzima que con la misma ѵmax requiriese un tiempo cuatro veces mayor para realizar la misma
transformación?
Dos isoenzimas A y B muestran valores de ѵmax de 5 y 10 mM/min, respectivamente. La primera tiene una Km de 0.5 mM. ¿Qué valor de Km debe tener la isoenzima B para proporcionar una velocidad inferior a la de la isoenzima A, en presencia de 10 mM de sustrato?
La ѵmax es directamente proporcional a la cantidad total de enzima. Si disponemos de dos enzimas con ѵmax de 10 mM/min, una con Km de 5 mM y otra con Km de 0.5 mM, ambas expuestas a
una concentración 10 mM de sustrato, ¿cuánto tenemos que incrementar la cantidad
de la primera enzima para conseguir mayor velocidad inicial que con la segunda?
¿Cuánto deben valer la Km y la ѵmax de una enzima para que la
representación de dobles inversas (1/ѵ frente a 1/[S]) corte al eje de abscisas en -4 y
al eje de ordenadas en el valor 0.4?
¿Cuánto tienen que cambiar la Km y la ѵmax de una enzima para
compensar el efecto de la adición al medio de una concentración 1 mM de
un inhibidor competitivo con una Ki de 0.1 mM?
¿Cuánto tienen que cambiar la Km y la ѵmax de una enzima para compensar el efecto de la adición al medio de una concentración 1 mM de un inhibidor acompetitivo con una constante de disociación de 0.1 mM?
A un medio que contiene una enzima con una Km de 0.5 mM y una ѵmax de 10 mM/min en condiciones de 10 mM del sustrato se le añade un
inhibidor acompetitivo (Ki = 0.5 mM) hasta alcanzar una concentración
de 1 mM. ¿Qué concentración de un inhibidor
competitivo, con la misma Ki, sería necesaria para que la velocidad inicial de la reacción
sea la misma que en el caso del inhibidor acompetitivo?
A una enzima con una Km de 0.5 mM y una ѵmax de 10 mM/min expuesta a
una concentración de su sustrato de 10 mM se le añade un inhibidor competitivo (Ki = 0.5 mM) hasta alcanzar la concentración de 1 mM. ¿Qué
concentración de sustrato sería necesaria para que la velocidad inicial sea la misma que
en ausencia de inhibidor?
A una enzima con una Km de 0.5 mM y una ѵmax de 10 mM/min expuesta a
una concentración de su sustrato de 10 mM se le añade un inhibidor
acompetitivo (Ki = 0.5 mM) hasta alcanzar la concentración de 1 mM. ¿Qué
concentración de sustrato sería necesaria para que la velocidad inicial sea la misma que
en ausencia de inhibidor?
Sabemos que dos enzimas trabajan a la misma velocidad in vivo sobre un
mismo sustrato y podemos suponer que la concentración de dicho sustrato
sea 2 mM. Sabemos que la primera enzima, con un
único centro activo, tiene Km = 0.5 mM y ѵmax = 8 mM/min, mientras la segunda,
que es multimérica,
tiene Km = 10 mM y ѵmax = 10 mM/min. ¿Cuántos
centros activos tiene la segunda enzima?
¿En qué condiciones una enzima no alostérica con ѵmax = 1 mM/min
y Km = 0.5 mM catalizará más rápidamente que otra enzima con
cuatro centros activos con cooperatividad con ѵmax = 3 mM/min y Km = 0.5 mM?
¿Qué cantidad de inhibidor competitivo con una constante de disociación
de 0.5 habría que añadir a una enzima alostérica con dos centros activos,
con una ѵmax de 3 mM/min y una Km de 0.5 mM, para que catalice a la
misma velocidad una concentración de sustrato de 1 mM que una enzima
no alostérica con los mismos parámetros cinéticos?
We wish to characterise an unknown kinase we have isolated,
by determining its Michaelis constant and its maximum velocity. Based on homology with known kinases, we may assume that Km will be around 0.5 mM and ѵmax around 10 mM/min. For the characterisation we will measure in vitro the initial velocity of substrate conversion. For this measure to be reliable, we must assure that during the measurement the velocity of conversion is truly an initial velocity. Since the time required for measurement will be 3
minutes, which is the minimal initial concentration of substrate that will guarantee initial velocity conditions?
An enzyme with a Km of 0.5 mM and a ѵmax of 10 mM/min is present in a medium with 10 mM concentration of its substrate:
(a) How long will it need to transform 80% of the substrate in product?
(b) Which would be the Km value of another enzyme which, with the same ѵmax, requires a 4-times longer time to perform the same transformation?
Two isoenzymes A and B display ѵmax values of 5 and 10 mM/min, respectively. The first one has a Km of 0.5 mM. What value of Km should isoenzyme B have in order to provide a velocity lower than isoenzyme A, in the presence of 10 mM substrate?
The ѵmax is directly proportional to the total amount of enzyme. If we have two enzymes with ѵmax of 10 mM/min, both exposed to a 10 mM substrate concentration, one with Km of 5 mM and the other with Km of 0.5 mM, in which proportion must we increase the amount of the first enzyme to achieve a higher velocity than the second enzyme?
How much should the values of Km and ѵmax be for an enzyme so that the double reciprocal plot (1/ѵ versus 1/[S]) cuts the abscissas at -4 and the ordinates at 0.4?
How much should Km and ѵmax change for an enzyme to compensate for the effect of adding 1 mM of a competitive inhibitor whose Ki is 0.1 mM?
How much should Km and ѵmax change for an enzyme to compensate for the effect of adding 1 mM of an uncompetitive inhibitor whose dissociation constant is 0.1 mM?
To a solution containing an enzyme with Km of 0.5 mM and ѵmax of 10 mM/min as well as 10 mM substrate we add a noncompetitive inhibitor (Ki = 0.5 mM) up to a concentration of 1 mM. Which concentration of a competitive inhibitor, with the same Ki, would be necessary for achieving the same initial velocity of the reaction as in the case of uncompetitive inhibitor?
To a solution containing an enzyme with Km of 0.5 mM and ѵmax of 10 mM/min as well as 10 mM substrate we add a competitive inhibitor (Ki = 0.5 mM) up to a concentration of 1 mM. Which concentration of the substrate would be needed for the initial velocity to be the same as in the absence of the inhibitor?
To a solution containing an enzyme with Km of 0.5 mM and ѵmax of 10 mM/min as well as 10 mM substrate we add an uncompetitive inhibitor (Ki = 0.5 mM) up to a concentration of 1 mM. Which concentration of the substrate would be needed for the initial velocity to be the same as in the absence of the inhibitor?
We know that two enzymes perform at the same velocity in vivo on the same substrate, and we may assume that the concentration of this substrate is 2 mM.
We know that the first enzyme, with a single active site, displays Km = 0.5 mM and ѵmax = 8 mM/min, while the second,
which is multimeric, displays Km = 10 mM and ѵmax = 10 mM/min. How many active sites has the second enzyme?
Under which conditions a non-allosteric enzyme with ѵmax = 1 mM/min
and Km = 0.5 mM will catalyse more rapidly than a second enzyme with ѵmax = 3 mM/min, Km = 0.5 mM, four active sites and cooperativity?
What amount of a competitive inhibitor with a dissociation constant of 0.5 should we need to add to an allosteric enzyme with two active sites, a
ѵmax of 3 mM/min and a Km of 0.5 mM so that it catalyses a substrate concentration of 1 mM at the same velocity as a non-allosteric enzyme with the same values of kinetic parameters?
Symbols and abbreviations:
Inhibition
and mixed
[S] = concentration of substrate
[S]₀ = initial concentration of substrate
ѵ = velocity of the reaction (rate of conversion of S into P, in terms of concentration per time unit; ѵ = d[P]/dt)
ѵ₀ = initial velocity (i.e. when reaction has just started and the amount of P is minimal)
ѵmax = maximum (initial) velocity possible for a given concentration of enzyme
Km = Michaelis constant of the enzyme
[I] = concentration of inhibitor
KIE = equilibrium constant for dissociation of the inhibitor from the enzyme, IE ⇄ I + E
KIES = equilibrium constant for dissociation of the inhibitor from the enzyme:substrate complex, IES ⇄ I + ES
competitive inhibition: just KIE
uncompetitive inhibition: just KIES
ideal non competitive inhibition: KIE = KIES
mixed inhibition: both KIE and KIES but different
nHill = Hill index or Hill coefficient (n≠1 indicates cooperativity)
Equations to plot the graphs are based on the work by Prof. Néstor Torres (Universidad de La Laguna, Spain)
A simple simulator to teach enzyme kinetics dynamics. Application in a problem-solving exercise. N. Torres and G. Santos (2017) doi:10.1080/23752696.2017.1307693
In this version the solving of differential equations is done with the Numeric Javascript library (www.numericjs.com) which uses the Dormand-Prince RK numeric integration method.