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The usual equation to describe the conversion of a substrate in an enzymatic reaction. Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m1101301.png" /> be the concentration of some substrate which is converted by an enzyme <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m1101302.png" /> into a product <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m1101303.png" />. The reaction rate is proportional to <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m1101304.png" /> for small values of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m1101305.png" />, but there is a maximum rate, which is not surpassed even for large <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m1101306.png" />. These observations can be expressed by the equation
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<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m1101307.png" /></td> </tr></table>
+
{{TEX|auto}}
 +
{{TEX|done}}
  
In this case the first reaction parameter, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m1101308.png" />, describes the maximal conversion speed, and the second reaction parameter, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m1101309.png" />, is equal to the substrate concentration at which the reaction speed is exactly <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013010.png" />. For small values of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013011.png" />, the reaction rate is <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013012.png" />.
+
The usual equation to describe the conversion of a substrate in an enzymatic reaction. Let  $  S ( t ) $
 +
be the concentration of some substrate which is converted by an enzyme  $  E $
 +
into a product  $  P $.
 +
The reaction rate is proportional to  $  S ( t ) $
 +
for small values of $  S ( t ) $,  
 +
but there is a maximum rate, which is not surpassed even for large  $  S ( t ) $.  
 +
These observations can be expressed by the equation
  
The equation can be derived from the chemical equilibrium reactions between the substrate <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013013.png" /> and the enzyme <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013014.png" />, which combine to a compound <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013015.png" />. This compound is rearranged in an equilibrium reaction into a compound <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013016.png" />, which dissociates into the enzyme <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013017.png" /> and product <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013018.png" />. In a formula:
+
$$
 +
{
 +
\frac{d S ( t ) }{dt }
 +
} = - {
 +
\frac{k _ {a} \cdot S ( t ) }{K + S ( t ) }
 +
} .
 +
$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013019.png" /></td> </tr></table>
+
In this case the first reaction parameter,  $  k _ {a} $,
 +
describes the maximal conversion speed, and the second reaction parameter,  $  K $,
 +
is equal to the substrate concentration at which the reaction speed is exactly  $  { {k _ {a} } / 2 } $.  
 +
For small values of  $  S ( t ) $,
 +
the reaction rate is  $  k _ {a} \cdot K ^ {- 1 } \cdot S ( t ) $.
  
In this reaction, the step from <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013020.png" /> to <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013021.png" /> is usually far slower than the other reactions, and if the reaction from <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013022.png" /> to <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013023.png" /> is irreversible, or if the product <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013024.png" /> is removed by some transport mechanism, then the two reaction constants <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013025.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013026.png" /> can be discarded, and the rearrangement coefficient <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013027.png" /> determines a simplified reaction equation:
+
The equation can be derived from the chemical equilibrium reactions between the substrate  $  S $
 +
and the enzyme  $  E $,
 +
which combine to a compound  $  ES $.  
 +
This compound is rearranged in an equilibrium reaction into a compound  $  EP $,  
 +
which dissociates into the enzyme  $  E $
 +
and product $  P $.  
 +
In a formula:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013028.png" /></td> </tr></table>
+
$$
 +
S + E \rightleftarrows _  \beta  ^  \alpha  SE \rightleftarrows _  \mu  ^  \gamma  EP \rightleftarrows _  \sigma  ^  \lambda  E + P.
 +
$$
  
Putting <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013029.png" />, the following system of differential equations emerges from the reactions:
+
In this reaction, the step from $  SE $
 +
to  $  EP $
 +
is usually far slower than the other reactions, and if the reaction from  $  EP $
 +
to  $  E + P $
 +
is irreversible, or if the product  $  P $
 +
is removed by some transport mechanism, then the two reaction constants  $  \mu $
 +
and  $  \sigma $
 +
can be discarded, and the rearrangement coefficient  $  \gamma $
 +
determines a simplified reaction equation:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013030.png" /></td> </tr></table>
+
$$
 +
S + E \rightleftarrows _  \beta  ^  \alpha  SE { \mathop \rightarrow \limits ^  \gamma  } E + P.
 +
$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013031.png" /></td> </tr></table>
+
Putting  $  B = SE $,
 +
the following system of differential equations emerges from the reactions:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013032.png" /></td> </tr></table>
+
$$
 +
{
 +
\frac{dS ( t ) }{dt }
 +
} = \beta \cdot B - \alpha \cdot S ( t ) \cdot E,
 +
$$
  
These equations imply that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013033.png" /> is a constant, representing the maximal amount of either <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013034.png" /> or <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013035.png" />. Usually, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013036.png" /> is large with respect to fluctuations in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013037.png" /> and so the reaction enters a steady state, in which the concentrations of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013038.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013039.png" /> remain almost constant over a large time interval. In that case the proportion <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013040.png" /> equals <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013041.png" />, which yields
+
$$
 +
{
 +
\frac{dE }{dt }
 +
} = ( \beta + \gamma ) \cdot B - \alpha \cdot S ( t ) \cdot E,  
 +
$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013042.png" /></td> </tr></table>
+
$$
 +
{
 +
\frac{dB }{dt }
 +
} = \alpha \cdot S ( t ) \cdot E + ( \beta + \gamma ) \cdot B.
 +
$$
 +
 
 +
These equations imply that  $  E + B = E _ {0} $
 +
is a constant, representing the maximal amount of either  $  E $
 +
or  $  B $.  
 +
Usually,  $  S ( t ) $
 +
is large with respect to fluctuations in  $  E _ {0} $
 +
and so the reaction enters a steady state, in which the concentrations of  $  E $
 +
and  $  B $
 +
remain almost constant over a large time interval. In that case the proportion  $  E \simeq B $
 +
equals  $  \beta + \gamma \simeq \alpha \cdot S ( t ) $,
 +
which yields
 +
 
 +
$$
 +
B = {
 +
\frac{\alpha \cdot S ( t ) \cdot E _ {0} }{\gamma + \beta + \alpha \cdot S ( t ) }
 +
} ,
 +
$$
  
 
and
 
and
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013043.png" /></td> </tr></table>
+
$$
 +
{
 +
\frac{dS ( t ) }{dt }
 +
} = \beta \cdot B - \alpha \cdot S ( t ) \cdot E =
 +
$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013044.png" /></td> </tr></table>
+
$$
 +
=  
 +
- \gamma \cdot B = - {
 +
\frac{\alpha \cdot \gamma \cdot E _ {0} \cdot S ( t ) }{\gamma + \beta + \alpha \cdot S ( t ) }
 +
} .
 +
$$
  
The values of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013045.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013046.png" /> can be inferred from the differential equation. This gives:
+
The values of $  K $
 +
and $  k _ {a} $
 +
can be inferred from the differential equation. This gives:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013047.png" /></td> </tr></table>
+
$$
 +
K = {
 +
\frac{\gamma + \beta } \alpha
 +
} ,  k _ {a} = \gamma \cdot E _ {0} ,
 +
$$
  
 
and so
 
and so
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013048.png" /></td> </tr></table>
+
$$
 +
{
 +
\frac{dS ( t ) }{dt }
 +
} = - {
 +
\frac{k _ {a} \cdot S ( t ) }{K + S ( t ) }
 +
} .
 +
$$
  
The values can be fitted from observations of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013049.png" />, but it is impossible to also find the parameters <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013050.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013051.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013052.png" />, and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013053.png" /> from the curve of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/m/m110/m110130/m11013054.png" />.
+
The values can be fitted from observations of $  S ( t ) $,  
 +
but it is impossible to also find the parameters $  \alpha $,  
 +
$  \beta $,  
 +
$  \gamma $,  
 +
and $  E _ {0} $
 +
from the curve of $  S ( t ) $.
  
 
The equation was derived by L. Michaelis and M.L. Menten [[#References|[a1]]].
 
The equation was derived by L. Michaelis and M.L. Menten [[#References|[a1]]].

Latest revision as of 08:00, 6 June 2020


The usual equation to describe the conversion of a substrate in an enzymatic reaction. Let $ S ( t ) $ be the concentration of some substrate which is converted by an enzyme $ E $ into a product $ P $. The reaction rate is proportional to $ S ( t ) $ for small values of $ S ( t ) $, but there is a maximum rate, which is not surpassed even for large $ S ( t ) $. These observations can be expressed by the equation

$$ { \frac{d S ( t ) }{dt } } = - { \frac{k _ {a} \cdot S ( t ) }{K + S ( t ) } } . $$

In this case the first reaction parameter, $ k _ {a} $, describes the maximal conversion speed, and the second reaction parameter, $ K $, is equal to the substrate concentration at which the reaction speed is exactly $ { {k _ {a} } / 2 } $. For small values of $ S ( t ) $, the reaction rate is $ k _ {a} \cdot K ^ {- 1 } \cdot S ( t ) $.

The equation can be derived from the chemical equilibrium reactions between the substrate $ S $ and the enzyme $ E $, which combine to a compound $ ES $. This compound is rearranged in an equilibrium reaction into a compound $ EP $, which dissociates into the enzyme $ E $ and product $ P $. In a formula:

$$ S + E \rightleftarrows _ \beta ^ \alpha SE \rightleftarrows _ \mu ^ \gamma EP \rightleftarrows _ \sigma ^ \lambda E + P. $$

In this reaction, the step from $ SE $ to $ EP $ is usually far slower than the other reactions, and if the reaction from $ EP $ to $ E + P $ is irreversible, or if the product $ P $ is removed by some transport mechanism, then the two reaction constants $ \mu $ and $ \sigma $ can be discarded, and the rearrangement coefficient $ \gamma $ determines a simplified reaction equation:

$$ S + E \rightleftarrows _ \beta ^ \alpha SE { \mathop \rightarrow \limits ^ \gamma } E + P. $$

Putting $ B = SE $, the following system of differential equations emerges from the reactions:

$$ { \frac{dS ( t ) }{dt } } = \beta \cdot B - \alpha \cdot S ( t ) \cdot E, $$

$$ { \frac{dE }{dt } } = ( \beta + \gamma ) \cdot B - \alpha \cdot S ( t ) \cdot E, $$

$$ { \frac{dB }{dt } } = \alpha \cdot S ( t ) \cdot E + ( \beta + \gamma ) \cdot B. $$

These equations imply that $ E + B = E _ {0} $ is a constant, representing the maximal amount of either $ E $ or $ B $. Usually, $ S ( t ) $ is large with respect to fluctuations in $ E _ {0} $ and so the reaction enters a steady state, in which the concentrations of $ E $ and $ B $ remain almost constant over a large time interval. In that case the proportion $ E \simeq B $ equals $ \beta + \gamma \simeq \alpha \cdot S ( t ) $, which yields

$$ B = { \frac{\alpha \cdot S ( t ) \cdot E _ {0} }{\gamma + \beta + \alpha \cdot S ( t ) } } , $$

and

$$ { \frac{dS ( t ) }{dt } } = \beta \cdot B - \alpha \cdot S ( t ) \cdot E = $$

$$ = - \gamma \cdot B = - { \frac{\alpha \cdot \gamma \cdot E _ {0} \cdot S ( t ) }{\gamma + \beta + \alpha \cdot S ( t ) } } . $$

The values of $ K $ and $ k _ {a} $ can be inferred from the differential equation. This gives:

$$ K = { \frac{\gamma + \beta } \alpha } , k _ {a} = \gamma \cdot E _ {0} , $$

and so

$$ { \frac{dS ( t ) }{dt } } = - { \frac{k _ {a} \cdot S ( t ) }{K + S ( t ) } } . $$

The values can be fitted from observations of $ S ( t ) $, but it is impossible to also find the parameters $ \alpha $, $ \beta $, $ \gamma $, and $ E _ {0} $ from the curve of $ S ( t ) $.

The equation was derived by L. Michaelis and M.L. Menten [a1].

References

[a1] L. Michaelis, M.L. Menten, "Die Kinetik der Invertinwirkung" Biochem. Zeitschrift , 2 (1913) pp. 333–369
How to Cite This Entry:
Michaelis-Menten equation. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Michaelis-Menten_equation&oldid=19064
This article was adapted from an original article by E. Wattel (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article