# 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.

• What type of inhibition (competitive or otherwise) are you interested in? What do you mean by "roughly" in the context of the Hill equations? Hill-ish? Are the $K$s the same? – Karsten Theis May 28 '19 at 14:47
• @KarstenTheis Let's go with non-competitive inhibition for simplicity's sake, though if there exist modifications for competitive and uncompetitive inhibition I'd be interested in those as well. The Ks are not the same, and I definitely should have been clearer about that. I say "roughly" because I've absorbed the $K_i$ from the canonical form of the inhibition equation into $V_{max}$ in the numerator of the inhibition equation. – Bryce May 28 '19 at 18:15
• I also believe it would be more accurate to have some constants $c_i$ and $c_s$ to determine the proportional binding strengths of each compound. In the first two equations those could simply be absorbed into K, but in the third equation they might be important. – Bryce May 28 '19 at 18:22
• Could you edit the question to incorporate the information in the comments? And maybe use two different subscripts for the two different Ks? – Karsten Theis May 28 '19 at 21:18
• Should all be clearer now. Apologies on the original imprecision. I'm a physicist by training, and I suppose that preference for rough approximations slipped in. – Bryce May 28 '19 at 22:58

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.

• That makes perfect sense. Thanks! – Bryce May 30 '19 at 22:39