Hill equation with inhibition and activation

The Hill equation for an activating enzymatic interaction with cooperative multiplicity $n$ is


$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = V\_\mathrm{max}\frac{[\ce{S}]^n}{K\_\ce{S} + [\ce{S}]^n}\tag{1}$$


and for an inhibitory interaction


$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = V\_\mathrm{max}\frac{K\_\ce{I}}{K\_\ce{I} + [\ce{I}]^m}\tag{2}$$


What would the equation be for competing activating and inhibiting compounds $\ce{S}$ and $\ce{I},$ respectively?


My sort of ansatz is that it would be of the form


$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = V\_\mathrm{max} \frac{c\_\ce{S}[\ce{S}]^n}{K\_\ce{SI} + c\_\ce{S}[\ce{S}]^n + c\_\ce{I}[\ce{I}]^m}\tag{3}$$


I spent about an hour trying to derive the actual answer, but I'm no expert in physical chemistry/biochemistry and increasingly have the sense I should turn to someone who is. 




**Mechanism of cooperative enzyme activity**


The easiest way to explain cooperativity is the example of hemoglobin, where four nearly identical subunits switch cooperatively between two states T and R of different binding affinity. When no ligand (L) is bound, hemoglobin is in the T state; once at least one ligand is bound, it is predominately in the R state.


$$\ce{T4 + 4 L <=> R\_4L + 3 L <=> R4L + 3 L <=> R4L2 + 2 L <=> R\_4L3 + L <=> R4L4}$$


The states with 1-3 ligands are less populated (because the R state has higher affinity for the ligand L thant the T state), and we will ignore them:


$$\ce{T4 + 4 L <=> R4L4}$$


From this, you can write the law of mass action and the binding isotherms.


If we apply this to an enzyme with a Hill coefficient of $n$, and call the active state of the enzyme E and the inactive state F, we would have the equilibrium:


$$\ce{F\_n + nS<=> E\_nS\_n}$$


For sufficiently high concentration of substrate S, all enzyme would be in complex with substrate, leading to a rate of $v\_\text{max}$.


**Mechanism of cooperative inhibition**


In the OP's question, the Hill coefficient for inhibition $m$ is different from that for enzyme activity ($n$). So we could have a regulatory protein with $m$ subunits attached to the enzyme. It binds cooperatively to inhibitor, which allosterically affects enzyme activity. The regulatory protein would also have two states, T and R, either free or bound to $m$ inhibitors:


$$\ce{ T\_m + m I <=> R\_mI\_m}$$


When the allosteric inhibitor is bound to ligand (i.e. in the R state), the enzyme lacks activity.


**Combination of the two**


Only a certain fraction of enzyme is bound to substrate, and only a certain fraction of enzyme-bound substrate is not inhibited by the regulatory protein. For this scenario, the enzyme activity is $v\_\text{max}$ times those two fractions.


$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = v\_\mathrm{max}\frac{[\ce{S}]^n}{K\_S + [\ce{S}]^n} \cdot \frac{K\_I}{K\_I + [\ce{I}]^m}\tag{2}$$


**What happens when there is competitive inhibition?**


It gets more complicated. Will the inhibitor also cause a switch in the enzymes conformation? In which direction? For the simpler (non-cooperative) Michaelis Menten model, a competitive inhibitor changes the apparent affinity of enzyme to ligand in a concentration-dependent manner (the enzyme partitions between free, ligand-bound and inhibitor-bound state). For the cooperative case, you would have to decide whether mixed states (such as $\ce{ES2I2}$) are populated; if they are, they would contribute to product formation, but less than $\ce{ES4}$ because not all active sites are turning over substrate.


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