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.

  • 1
    $\begingroup$ 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? $\endgroup$ – Karsten Theis May 28 '19 at 14:47
  • $\begingroup$ @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. $\endgroup$ – Bryce May 28 '19 at 18:15
  • $\begingroup$ 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. $\endgroup$ – Bryce May 28 '19 at 18:22
  • $\begingroup$ Could you edit the question to incorporate the information in the comments? And maybe use two different subscripts for the two different Ks? $\endgroup$ – Karsten Theis May 28 '19 at 21:18
  • $\begingroup$ 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. $\endgroup$ – 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.

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

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.