# Biomod/2013/NanoUANL/Enzyme

(Difference between revisions)
 Revision as of 21:50, 11 October 2013 (view source)← Previous diff Revision as of 00:19, 12 October 2013 (view source)Next diff → Line 29: Line 29: '''1.2''' '''1.2''' - $[itex] - \tfrac{d[ES]}{dt}=k_1[E][S]-k_{-1}[ES]-k_2[ES] + \frac{d[ES]}{dt}=k_1[E][S]-k_{-1}[ES]-k_2[ES]$ [/itex] Line 37: Line 36: '''1.3''' '''1.3''' - $[itex] - [ES]= \tfrac{[E][S]}{K_M+[S]} + [ES]= \frac{[E][S]}{K_M+[S]}$ [/itex] Line 56: Line 54: '''1.5''' '''1.5''' - $\tfrac{d[E]}{dt}=-k_1[E][S]+k_{-1}[ES]$ + $\frac{d[E]}{dt}=-k_1[E][S]+k_{-1}[ES]$ + '''1.6''' '''1.6''' - $\tfrac{d[ES]}{dt}=k_1[E][S]-(k_{-1}+k_2)[ES]$ + $\frac{d[ES]}{dt}=k_1[E][S]-(k_{-1}+k_2)[ES]$ + '''1.7''' '''1.7''' - $\tfrac{d[E^0]}{dt}=\tfrac{d[P]}{dt}=k_2[ES]$ + $\frac{d[E^0]}{dt}=\tfrac{d[P]}{dt}=k_2[ES]$ where ''t'' is the elapsed time, the initial conditions are [ES]=0 and [E0]=0 at ''t''=0. To derive the rate equations that describe the corresponding single-molecule Michaelis-Menten kinetics, the concentrations in equations 5-7 are replaced by the probabilities P of finding the single enzyme molecule in the states E, ES, and E0 , leading to the equations: where ''t'' is the elapsed time, the initial conditions are [ES]=0 and [E0]=0 at ''t''=0. To derive the rate equations that describe the corresponding single-molecule Michaelis-Menten kinetics, the concentrations in equations 5-7 are replaced by the probabilities P of finding the single enzyme molecule in the states E, ES, and E0 , leading to the equations: '''1.8''' '''1.8''' - + $\frac{dP_E(t)}{dt}=-k_1^0P_E(t)+k_{-1}P_{ES}(t)$ - $\tfrac{dP_E(t)}{dt}=-k_1^0P_E(t)+k_{-1}P_{ES}(t)$ + '''1.9''' '''1.9''' - + $\frac{dP_{ES}(t)]}{dt}=k_1^0P_E(t)-(k_{-1}+k_2)P_{ES}(t)$ - $\tfrac{dP_{ES}(t)]}{dt}=k_1^0P_E(t)-(k_{-1}+k_2)P_{ES}(t)$ + '''1.10''' '''1.10''' - + $\frac{dP_E^0(t)}{dt}=k_2P_{ES}(t)$ - $\tfrac{dP_E^0(t)}{dt}=k_2P_{ES}(t)$ + These equations must satisfy the conditions ''PE''(0)=1, ''PES''(0)=0 and ''PE0''=0 at ''t''=0 (start of the reaction). Also, ''PE''(t) + ''PES''(t) + ''PE0''(t)=1. The rate constant ''k''10 can be taken as ''k''10=''k''1[S], assuming [S] is time-independent. Given that a single enzyme is unlikely to deplete all the substrate presence, [S] can be considered constant, virtually being unaffected. These equations must satisfy the conditions ''PE''(0)=1, ''PES''(0)=0 and ''PE0''=0 at ''t''=0 (start of the reaction). Also, ''PE''(t) + ''PES''(t) + ''PE0''(t)=1. The rate constant ''k''10 can be taken as ''k''10=''k''1[S], assuming [S] is time-independent. Given that a single enzyme is unlikely to deplete all the substrate presence, [S] can be considered constant, virtually being unaffected. Line 80: Line 77: Equations 8-10 become a system of linear first-order differential equations that can be solved exactly for ''PE''(t), ''PES''(t) and ''PE0''(t). Equations 8-10 become a system of linear first-order differential equations that can be solved exactly for ''PE''(t), ''PES''(t) and ''PE0''(t). - Knowing ''PE0''(t),  the probability that a turnover occurs between ''t'' and ''t +'' Δ''t'' is ''f(t)Δt, the same as Δ''PE0''(t). + Knowing ''PE0''(t),  the probability that a turnover occurs between ''t'' and ''t +'' Δ''t'' is ''f(t)Δt, the same as Δ''PE0''(t). Taking this into account, in the limit of infinitesimal Δ''t'': + + '''1.11''' + $f(t)=\frac{dP_E^0(t)}{dt}=k_2P_{ES}(t)$ + + Solving equations 8-10, and using equation 11: + + '''1.12''' + $+ f(t)=\frac{k_1k_2[S]}{2A}[exp(A+B)t-exp(B-A)t] +$ + + in which: + + '''1.13''' + $+ A=\sqrt((k_1[S]+k_{-1}+k_2)^2/4-k_1k_2[S]) +$ + + '''1.14''' + $+ B=\frac{-(k_1[S]+k_{-1}+k_2}{2} +$ + + when the substrate concentration dependence [S] has been shown throught the relation ''k10 = k1''[S]. + + With these equations, we have used different values for ''k-1'' and [S], using values found in literature. + + + + [Descripcion y link del codigo de simulink] + + [GRAFICAS Y ESO] + + + + At the first moment of ''f(t)'', $= \int_0^\infty dt tf(t)$, which would be the mean waiting time for the reaction, '''', and its reciprocal can be taken as the average reaction rate.  Starting eqtn 12: + + $\frac{1}{}=-\frac{(A^2-B^2)^2}{2Bk_1k_2[S]}$ + + ===Randomness parameter=== + The probability density ''f(t)'' completely characterizes single-enzyme kinetics, with the ''n''th moment being given by: + + '''1.15''' + $f(t)= \int_0^\infty dt f(t)t^n$ + + Although the first moment of ''f(t)'' can be described eqtn 11, higher moments of ''f(t)'' are usually calculated along with a "randomness parameter".  Implying no dynamic disorder, ''r'' is given by: + + '''1.16''' + $r=\frac{(k_1[S]+k_2+k_{-1})^2-2k_1k_2[S]}{(k1[S]+k_2+k_{-1})^2}$ + + As substrate concentration increases, ''r'' decreases, indicating the formation of an intermediate enzyme-substrate complex ES. At even higher concentrations, the catalytic step limits the reaction, as is the same when concentration is low and the substrate binding limits the rate.

## What is an enzyme?

In biological systems, chemical transformations are typically accelerated by enzymes, macromolecules capable of turning one or more compounds into others (substrates and products). The activity is determined greatly by their three-dimensional structure. Most enzymes are proteins, although several catalytic RNA molecules have been identified. They may also need to employ organic and inorganic cofactors for the reaction to occur. The process is based upon the diminishment of the activation energy needed for a reaction, greatly increasing its rate of reaction. The rate enhancement provided by these proteins can be as high as 10^19, while maintaining high substrate specificity.

Because of this, reaction rates are millions of times faster than un-catalyzed reactions. Enzymes are not consumed by the reactions that they take part in, and they do not alter the equilibrium. Enzyme activity can be affected by a wide variety of factors. Inhibitors and activators intervene directly in the reaction rate, environmental factors like temperature, pressure, pH and substrate concentration also play a part in these kinetics. For temperature and pH, usually exist a range of values for which the enzyme works better (optimal conditions). The enzyme activity lowers dramatically as you get farther away from this range of values. As for concentration, other kind of relationship is observed. With increasing concentration, enzyme activity increases, until we reach the most optimal performance. Further increase of concentration won’t have an impact on the enzyme activity.

Being able to determine these conditions allow us to manipulate the enzyme activity, thus achieving greater control over the reaction.

Enzyme information was gathered from research papers:

• Isoelectric Point: Isozymes range from 3.0 – 9.0 pH
• pH Dependence: range of 5.0 to 9.0, optimum (6.0 to 6.5)
• k1 = 1 x 107
• k-1 = between 20 and 50
• k2 = 200 at pH 6

### Enzyme kinetics

Michaelis-Menten kinetics is one of the oldest models for describing the catalytic activity of enzymes. The reaction cycle is divided into two basic steps: the reversible binding between the enzyme and substrate to form an intermediate complex, and the irreversible catalytic step to generate the product and release the enzyme; in which the first step is affected by the constants k1; and k-1, whereas the irreversible step only takes into account k2.

1.1

$E + S \leftrightarrow ES \rightarrow E^0 + P$

The rate of consumption can be expressed by the formation of the ES complex in the following equation:

1.2 $\frac{d[ES]}{dt}=k_1[E][S]-k_{-1}[ES]-k_2[ES]$

Using a steady-approximation and rearranging 1.2 we obtain:

1.3 $[ES]= \frac{[E][S]}{K_M+[S]}$

where KM is the Michaelis constant defined as

1.4

$\operatorname{K_M}= \tfrac{k_{-1} +k_2}{k_1}$

As it was mentioned in the introduction, single-enzyme studies have proven that the "traditional" enzyme kinetics do not apply, and a new approach is needed. Enzyme concentration is meaningless in a single-molecule level, so it is more appropriate to consider the probability PE(t) for the enzyme to find a catalytically active enzyme in a time t in the process. This is because the reaction is a stochastic event.

Therefore, the rate equations of each species are:

1.5 $\frac{d[E]}{dt}=-k_1[E][S]+k_{-1}[ES]$

1.6 $\frac{d[ES]}{dt}=k_1[E][S]-(k_{-1}+k_2)[ES]$

1.7 $\frac{d[E^0]}{dt}=\tfrac{d[P]}{dt}=k_2[ES]$

where t is the elapsed time, the initial conditions are [ES]=0 and [E0]=0 at t=0. To derive the rate equations that describe the corresponding single-molecule Michaelis-Menten kinetics, the concentrations in equations 5-7 are replaced by the probabilities P of finding the single enzyme molecule in the states E, ES, and E0 , leading to the equations:

1.8 $\frac{dP_E(t)}{dt}=-k_1^0P_E(t)+k_{-1}P_{ES}(t)$

1.9 $\frac{dP_{ES}(t)]}{dt}=k_1^0P_E(t)-(k_{-1}+k_2)P_{ES}(t)$

1.10 $\frac{dP_E^0(t)}{dt}=k_2P_{ES}(t)$

These equations must satisfy the conditions PE(0)=1, PES(0)=0 and PE0=0 at t=0 (start of the reaction). Also, PE(t) + PES(t) + PE0(t)=1. The rate constant k10 can be taken as k10=k1[S], assuming [S] is time-independent. Given that a single enzyme is unlikely to deplete all the substrate presence, [S] can be considered constant, virtually being unaffected.

Equations 8-10 become a system of linear first-order differential equations that can be solved exactly for PE(t), PES(t) and PE0(t).

Knowing PE0(t), the probability that a turnover occurs between t and t + Δt is f(t)Δt, the same as ΔPE0(t). Taking this into account, in the limit of infinitesimal Δt:

1.11 $f(t)=\frac{dP_E^0(t)}{dt}=k_2P_{ES}(t)$

Solving equations 8-10, and using equation 11:

1.12 $f(t)=\frac{k_1k_2[S]}{2A}[exp(A+B)t-exp(B-A)t]$

in which:

1.13 $A=\sqrt((k_1[S]+k_{-1}+k_2)^2/4-k_1k_2[S])$

1.14 $B=\frac{-(k_1[S]+k_{-1}+k_2}{2}$

when the substrate concentration dependence [S] has been shown throught the relation k10 = k1[S].

With these equations, we have used different values for k-1 and [S], using values found in literature.

[Descripcion y link del codigo de simulink]

[GRAFICAS Y ESO]

At the first moment of f(t), $ = \int_0^\infty dt tf(t)$, which would be the mean waiting time for the reaction, <t>, and its reciprocal can be taken as the average reaction rate. Starting eqtn 12:

$\frac{1}{}=-\frac{(A^2-B^2)^2}{2Bk_1k_2[S]}$

### Randomness parameter

The probability density f(t) completely characterizes single-enzyme kinetics, with the nth moment being given by:

1.15 $f(t)= \int_0^\infty dt f(t)t^n$

Although the first moment of f(t) can be described eqtn 11, higher moments of f(t) are usually calculated along with a "randomness parameter". Implying no dynamic disorder, r is given by:

1.16 $r=\frac{(k_1[S]+k_2+k_{-1})^2-2k_1k_2[S]}{(k1[S]+k_2+k_{-1})^2}$

As substrate concentration increases, r decreases, indicating the formation of an intermediate enzyme-substrate complex ES. At even higher concentrations, the catalytic step limits the reaction, as is the same when concentration is low and the substrate binding limits the rate.