eyeprolog

Competitive enzyme inhibition

How much does an inhibitor slow an enzyme down?

An enzyme converts a substrate at a rate that depends on how much substrate is present. A competitive inhibitor competes for the same binding site. EyeProlog applies the standard rate equations to one assay and decides whether the slowdown is significant.

Run it in the playground · Source

The question

The Michaelis–Menten equation gives an enzyme’s rate as Vmax × S / (Km + S), where S is the substrate concentration, Vmax the top speed, and Km the concentration at which the rate is half of that. A competitive inhibitor at concentration I with binding constant Ki effectively raises Km to Km × (1 + I / Ki). With this assay’s numbers, how large is the slowdown, and does it exceed 25%?

The measurements

assay(assay1, vmax_uM_s, 120.0).
assay(assay1, substrate_uM, 50.0).
assay(assay1, km_uM, 30.0).
assay(assay1, inhibitor_uM, 10.0).
assay(assay1, ki_uM, 5.0).
threshold(assay1, significant_inhibition_fraction, 0.25).

Concentrations are in micromolar (µM) and rates in µM per second. The threshold is part of the data, not hidden in a rule.

The equations as rules

competitive_multiplier(Assay, Multiplier) :-
  assay(Assay, inhibitor_uM, Inhibitor),
  assay(Assay, ki_uM, Ki),
  (Ratio is Inhibitor / Ki),
  (Multiplier is 1.0 + Ratio).

effective_km(Assay, Effectivekm) :-
  assay(Assay, km_uM, Km),
  competitive_multiplier(Assay, Multiplier),
  (Effectivekm is Km * Multiplier).

significant_inhibition(Assay) :-
  inhibition_fraction(Assay, Fraction),
  threshold(Assay, significant_inhibition_fraction, Limit),
  (Fraction > Limit).

Each named quantity is its own predicate, so each appears as its own step in the proof. The rates and the inhibition fraction follow the same pattern.

The answer

eyeprolog examples/competitive-enzyme-kinetics.pl

effectiveKm_uM(assay1, 90.0).
uninhibitedRate_uM_s(assay1, 75.0).
inhibitedRate_uM_s(assay1, 42.857142857142854).
inhibitionFraction(assay1, 0.4285714285714286).
status(assay1, significant_inhibition).
reason(assay1, "competitive inhibitor raises effective Km and lowers reaction rate").

The inhibitor triples the effective Km from 30 to 90 µM, which cuts the rate from 75 to about 42.9 µM/s: a 43% slowdown, well over the 25% threshold.

The reason

step(effective_km(assay1, 90.0),
     rule(8),
     ['Assay' = assay1, 'Effectivekm' = 90.0, 'Km' = 30.0, 'Multiplier' = 3.0],
     [assay(assay1, km_uM, 30.0), competitive_multiplier(assay1, 3.0), 90.0 is 30.0 * 3.0]).
step(competitive_multiplier(assay1, 3.0),
     rule(7),
     ['Assay' = assay1, 'Multiplier' = 3.0, 'Inhibitor' = 10.0, 'Ki' = 5.0, 'Ratio' = 2.0],
     [assay(assay1, inhibitor_uM, 10.0),
      assay(assay1, ki_uM, 5.0),
      2.0 is 10.0 / 5.0,
      3.0 is 1.0 + 2.0]).

The effective Km of 90 is 30 times a multiplier of 3, and the multiplier is 1 + 10 / 5. Every number in the conclusion traces back to a measured fact and one arithmetic step.

The check

condition('C1', resolution, ok, 18).
condition('C5', re_decision, ok, 11).
steps(29).
trusted(0).
verdict(checked).

All 18 rule steps resolve, and the checker redoes all 11 arithmetic steps. That confirms the calculation from the stated numbers. Whether the numbers were measured well, and whether the model fits this enzyme, is outside the proof.

Try it

Lower the inhibitor to 2.0 µM. The fraction drops to about 0.13, and status/2 and reason/2 disappear.
Double the substrate to 100 µM and predict whether the inhibition fraction rises or falls.
Add a second assay with its own facts. No rule needs to change.

Learn more in The Art of EyeProlog, Chapter 14.

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