0
$\begingroup$

When deriving the Hill function (protein $X$ and $n$ molecules of the signal $S$), my book introduces the following rates:

$$ \text{collision rate} = k_{ON} \cdot X_n \cdot S^n\\ \text{dissociation rate} = k_{OFF} \cdot [nSX], $$

where $[nSX]$ is the complex. This translates into the following ODE:

$$ \frac{d}{dt} [nSX] = k_{ON} \cdot X_n \cdot S^n - k_{OFF} \cdot [nSX]. $$

This seems fine if I follow the $\textit{rules}$ of, for instance, this table.

On the other hand, during the class, my professor introduced the stoichiometric matrix. If we have the reaction $2A + B \leftrightarrow C$, then the corresponding system of ODEs would be:

$$ A' = -2r_1A^2B + 2r_2C\\ B' = -r_1A^2B + r_2C\\ C' = r_1A^2B - r_2C. $$

So, my question is: why in this case we put the $2$ in front of the reaction rates while for the Hill function we do not (put the $n$ in front of it)? Is it just because the $n$ molecules of signal $S$ are not consumed nor produced?

$\endgroup$
1
$\begingroup$

If the reaction is $\ce{aA + bB = pP + qQ} $ then we define the rates as

$$-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{p}\frac{d[P]}{dt}=+\frac{1}{q}\frac{d[Q]}{dt}$$

As to the Hill function it seems that the complex nSX is just one species.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.